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Reused while loop only working for one selection box


Emulate a do-while loop in Python?How to POST empty <select .. multiple> HTML elements if empty?Syntax for a single-line Bash infinite while loopHow to style a <select> dropdown with only CSS?Set selected option of select boxHow do I make a placeholder for a 'select' box?$_POST is empty when submitting a value that was posted to the formHow to pass select box value to PHP variable on same page?Disable <select> option when selected in another <select> boxIssue filling selection box with PHP from database






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








0















I need to generate a list of teams from the cups selection box at the top of my page, this code allows me to run a while loop to generate the correct results for the first selection box, however, when i replicate the code in the second it shows no results?



 <form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>



<input type="submit" name="submit" value="Generate" class="submit">
<?php

if (isset($_POST['submit'])) //checking if submit button was clicked

include_once 'action/dbcon.php';

$cname = $_POST['cupname'];

if (empty($cname))
header("Location: tables.php?field=empty"); //return them if fields are empty
exit();
else
$sql = "SELECT * FROM teams WHERE cup_name='$cname'";
$show_teams = mysqli_query($conn, $sql);
$numberCheck = mysqli_num_rows($show_teams);

if ($numberCheck < 8)
header("Location: tables.php?tables=1"); //Take to cup page if there arent enough teams in the cup



?>









share|improve this question



















  • 1





    Warning: Your code is vulnerable to SQL Injection attacks. You should use parameterised queries and prepared statements to help prevent attackers from compromising your database by using malicious input values. bobby-tables.com gives an explanation of the risks, as well as some examples of how to write your queries safely using PHP / mysqli. Never insert unsanitised data directly into your SQL. The way your code is written now, someone could easily steal, incorrectly change, or even delete your data.

    – ADyson
    Mar 27 at 23:56

















0















I need to generate a list of teams from the cups selection box at the top of my page, this code allows me to run a while loop to generate the correct results for the first selection box, however, when i replicate the code in the second it shows no results?



 <form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>



<input type="submit" name="submit" value="Generate" class="submit">
<?php

if (isset($_POST['submit'])) //checking if submit button was clicked

include_once 'action/dbcon.php';

$cname = $_POST['cupname'];

if (empty($cname))
header("Location: tables.php?field=empty"); //return them if fields are empty
exit();
else
$sql = "SELECT * FROM teams WHERE cup_name='$cname'";
$show_teams = mysqli_query($conn, $sql);
$numberCheck = mysqli_num_rows($show_teams);

if ($numberCheck < 8)
header("Location: tables.php?tables=1"); //Take to cup page if there arent enough teams in the cup



?>









share|improve this question



















  • 1





    Warning: Your code is vulnerable to SQL Injection attacks. You should use parameterised queries and prepared statements to help prevent attackers from compromising your database by using malicious input values. bobby-tables.com gives an explanation of the risks, as well as some examples of how to write your queries safely using PHP / mysqli. Never insert unsanitised data directly into your SQL. The way your code is written now, someone could easily steal, incorrectly change, or even delete your data.

    – ADyson
    Mar 27 at 23:56













0












0








0








I need to generate a list of teams from the cups selection box at the top of my page, this code allows me to run a while loop to generate the correct results for the first selection box, however, when i replicate the code in the second it shows no results?



 <form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>



<input type="submit" name="submit" value="Generate" class="submit">
<?php

if (isset($_POST['submit'])) //checking if submit button was clicked

include_once 'action/dbcon.php';

$cname = $_POST['cupname'];

if (empty($cname))
header("Location: tables.php?field=empty"); //return them if fields are empty
exit();
else
$sql = "SELECT * FROM teams WHERE cup_name='$cname'";
$show_teams = mysqli_query($conn, $sql);
$numberCheck = mysqli_num_rows($show_teams);

if ($numberCheck < 8)
header("Location: tables.php?tables=1"); //Take to cup page if there arent enough teams in the cup



?>









share|improve this question














I need to generate a list of teams from the cups selection box at the top of my page, this code allows me to run a while loop to generate the correct results for the first selection box, however, when i replicate the code in the second it shows no results?



 <form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
while ($team = mysqli_fetch_assoc($show_teams)) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>



<input type="submit" name="submit" value="Generate" class="submit">
<?php

if (isset($_POST['submit'])) //checking if submit button was clicked

include_once 'action/dbcon.php';

$cname = $_POST['cupname'];

if (empty($cname))
header("Location: tables.php?field=empty"); //return them if fields are empty
exit();
else
$sql = "SELECT * FROM teams WHERE cup_name='$cname'";
$show_teams = mysqli_query($conn, $sql);
$numberCheck = mysqli_num_rows($show_teams);

if ($numberCheck < 8)
header("Location: tables.php?tables=1"); //Take to cup page if there arent enough teams in the cup



?>






php html while-loop






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 27 at 23:53









Brad ArcherBrad Archer

164 bronze badges




164 bronze badges










  • 1





    Warning: Your code is vulnerable to SQL Injection attacks. You should use parameterised queries and prepared statements to help prevent attackers from compromising your database by using malicious input values. bobby-tables.com gives an explanation of the risks, as well as some examples of how to write your queries safely using PHP / mysqli. Never insert unsanitised data directly into your SQL. The way your code is written now, someone could easily steal, incorrectly change, or even delete your data.

    – ADyson
    Mar 27 at 23:56












  • 1





    Warning: Your code is vulnerable to SQL Injection attacks. You should use parameterised queries and prepared statements to help prevent attackers from compromising your database by using malicious input values. bobby-tables.com gives an explanation of the risks, as well as some examples of how to write your queries safely using PHP / mysqli. Never insert unsanitised data directly into your SQL. The way your code is written now, someone could easily steal, incorrectly change, or even delete your data.

    – ADyson
    Mar 27 at 23:56







1




1





Warning: Your code is vulnerable to SQL Injection attacks. You should use parameterised queries and prepared statements to help prevent attackers from compromising your database by using malicious input values. bobby-tables.com gives an explanation of the risks, as well as some examples of how to write your queries safely using PHP / mysqli. Never insert unsanitised data directly into your SQL. The way your code is written now, someone could easily steal, incorrectly change, or even delete your data.

– ADyson
Mar 27 at 23:56





Warning: Your code is vulnerable to SQL Injection attacks. You should use parameterised queries and prepared statements to help prevent attackers from compromising your database by using malicious input values. bobby-tables.com gives an explanation of the risks, as well as some examples of how to write your queries safely using PHP / mysqli. Never insert unsanitised data directly into your SQL. The way your code is written now, someone could easily steal, incorrectly change, or even delete your data.

– ADyson
Mar 27 at 23:56












1 Answer
1






active

oldest

votes


















0















The problem is that fetch_assoc() returns the next row in the resultset, unless there are no more left (in which case it returns NULL - see docs).



Your first while loop runs until there are no more results. Therefore when you start the second while loop, there are still...you guessed it: no more results. You already used them up. Therefore the first call to fetch_assoc() in the second loop returns NULL immediately, and so the loop condition is never met, and the loop never executes.



There are two different ways to solve this:



1) Reset the result pointer to the start of the resultset. Somewhere between the two loops, write



mysqli_data_seek($show_teams, 0);


See docs for more info.



2) Read all the data into a PHP array, which you can then loop through as many times as you like using foreach:



 <?php
$teams = array();
while ($team = mysqli_fetch_assoc($show_teams))
$teams[] = $team;

?>

<form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?><!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>





share|improve this answer

























  • Both methods are throwing Notice: Undefined variable: show_teams in C:xampphtdocsfriendlyfixturestables.php on line 96 and Warning: mysqli_data_seek() expects parameter 1 to be mysqli_result, null given in C:xampphtdocsfriendlyfixturestables.php on line 101

    – Brad Archer
    Mar 28 at 0:17












  • According to the code in your question, $show_teams is your query result object. So that's why I used that name. Maybe you changed it in the meantime or something?

    – ADyson
    Mar 28 at 0:18












  • none of the variable names were changed hence why im extremely confused, dont suppose you have any clue?

    – Brad Archer
    Mar 28 at 0:26











  • I can't see how you run the query and assign the result, so it's hard to be sure. Can you add that to the code in the question?

    – ADyson
    Mar 28 at 0:29











  • Not to worry, i placed the data seek elsewhere, now the variable is being recognised. However, i do also appreciate the array solution, thanks for the help!

    – Brad Archer
    Mar 28 at 0:35










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0















The problem is that fetch_assoc() returns the next row in the resultset, unless there are no more left (in which case it returns NULL - see docs).



Your first while loop runs until there are no more results. Therefore when you start the second while loop, there are still...you guessed it: no more results. You already used them up. Therefore the first call to fetch_assoc() in the second loop returns NULL immediately, and so the loop condition is never met, and the loop never executes.



There are two different ways to solve this:



1) Reset the result pointer to the start of the resultset. Somewhere between the two loops, write



mysqli_data_seek($show_teams, 0);


See docs for more info.



2) Read all the data into a PHP array, which you can then loop through as many times as you like using foreach:



 <?php
$teams = array();
while ($team = mysqli_fetch_assoc($show_teams))
$teams[] = $team;

?>

<form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?><!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>





share|improve this answer

























  • Both methods are throwing Notice: Undefined variable: show_teams in C:xampphtdocsfriendlyfixturestables.php on line 96 and Warning: mysqli_data_seek() expects parameter 1 to be mysqli_result, null given in C:xampphtdocsfriendlyfixturestables.php on line 101

    – Brad Archer
    Mar 28 at 0:17












  • According to the code in your question, $show_teams is your query result object. So that's why I used that name. Maybe you changed it in the meantime or something?

    – ADyson
    Mar 28 at 0:18












  • none of the variable names were changed hence why im extremely confused, dont suppose you have any clue?

    – Brad Archer
    Mar 28 at 0:26











  • I can't see how you run the query and assign the result, so it's hard to be sure. Can you add that to the code in the question?

    – ADyson
    Mar 28 at 0:29











  • Not to worry, i placed the data seek elsewhere, now the variable is being recognised. However, i do also appreciate the array solution, thanks for the help!

    – Brad Archer
    Mar 28 at 0:35















0















The problem is that fetch_assoc() returns the next row in the resultset, unless there are no more left (in which case it returns NULL - see docs).



Your first while loop runs until there are no more results. Therefore when you start the second while loop, there are still...you guessed it: no more results. You already used them up. Therefore the first call to fetch_assoc() in the second loop returns NULL immediately, and so the loop condition is never met, and the loop never executes.



There are two different ways to solve this:



1) Reset the result pointer to the start of the resultset. Somewhere between the two loops, write



mysqli_data_seek($show_teams, 0);


See docs for more info.



2) Read all the data into a PHP array, which you can then loop through as many times as you like using foreach:



 <?php
$teams = array();
while ($team = mysqli_fetch_assoc($show_teams))
$teams[] = $team;

?>

<form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?><!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>





share|improve this answer

























  • Both methods are throwing Notice: Undefined variable: show_teams in C:xampphtdocsfriendlyfixturestables.php on line 96 and Warning: mysqli_data_seek() expects parameter 1 to be mysqli_result, null given in C:xampphtdocsfriendlyfixturestables.php on line 101

    – Brad Archer
    Mar 28 at 0:17












  • According to the code in your question, $show_teams is your query result object. So that's why I used that name. Maybe you changed it in the meantime or something?

    – ADyson
    Mar 28 at 0:18












  • none of the variable names were changed hence why im extremely confused, dont suppose you have any clue?

    – Brad Archer
    Mar 28 at 0:26











  • I can't see how you run the query and assign the result, so it's hard to be sure. Can you add that to the code in the question?

    – ADyson
    Mar 28 at 0:29











  • Not to worry, i placed the data seek elsewhere, now the variable is being recognised. However, i do also appreciate the array solution, thanks for the help!

    – Brad Archer
    Mar 28 at 0:35













0














0










0









The problem is that fetch_assoc() returns the next row in the resultset, unless there are no more left (in which case it returns NULL - see docs).



Your first while loop runs until there are no more results. Therefore when you start the second while loop, there are still...you guessed it: no more results. You already used them up. Therefore the first call to fetch_assoc() in the second loop returns NULL immediately, and so the loop condition is never met, and the loop never executes.



There are two different ways to solve this:



1) Reset the result pointer to the start of the resultset. Somewhere between the two loops, write



mysqli_data_seek($show_teams, 0);


See docs for more info.



2) Read all the data into a PHP array, which you can then loop through as many times as you like using foreach:



 <?php
$teams = array();
while ($team = mysqli_fetch_assoc($show_teams))
$teams[] = $team;

?>

<form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?><!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>





share|improve this answer













The problem is that fetch_assoc() returns the next row in the resultset, unless there are no more left (in which case it returns NULL - see docs).



Your first while loop runs until there are no more results. Therefore when you start the second while loop, there are still...you guessed it: no more results. You already used them up. Therefore the first call to fetch_assoc() in the second loop returns NULL immediately, and so the loop condition is never met, and the loop never executes.



There are two different ways to solve this:



1) Reset the result pointer to the start of the resultset. Somewhere between the two loops, write



mysqli_data_seek($show_teams, 0);


See docs for more info.



2) Read all the data into a PHP array, which you can then loop through as many times as you like using foreach:



 <?php
$teams = array();
while ($team = mysqli_fetch_assoc($show_teams))
$teams[] = $team;

?>

<form class="seed-form">
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?><!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>
<select name="team" required>
<option value='Holder' disabled selected>Select Team</option> <!--Placeholder for Select-->
<?php
foreach ($teams as $team) ?>
<option value="<?php echo $team["team_id"]; ?>"><?php echo $team["team_name"]; ?></option>
<?php ?> <!--FILLS SELECT BOX WITH TEAMS FROM THAT CUP-->
</select>






share|improve this answer












share|improve this answer



share|improve this answer










answered Mar 28 at 0:03









ADysonADyson

29.3k12 gold badges28 silver badges46 bronze badges




29.3k12 gold badges28 silver badges46 bronze badges















  • Both methods are throwing Notice: Undefined variable: show_teams in C:xampphtdocsfriendlyfixturestables.php on line 96 and Warning: mysqli_data_seek() expects parameter 1 to be mysqli_result, null given in C:xampphtdocsfriendlyfixturestables.php on line 101

    – Brad Archer
    Mar 28 at 0:17












  • According to the code in your question, $show_teams is your query result object. So that's why I used that name. Maybe you changed it in the meantime or something?

    – ADyson
    Mar 28 at 0:18












  • none of the variable names were changed hence why im extremely confused, dont suppose you have any clue?

    – Brad Archer
    Mar 28 at 0:26











  • I can't see how you run the query and assign the result, so it's hard to be sure. Can you add that to the code in the question?

    – ADyson
    Mar 28 at 0:29











  • Not to worry, i placed the data seek elsewhere, now the variable is being recognised. However, i do also appreciate the array solution, thanks for the help!

    – Brad Archer
    Mar 28 at 0:35

















  • Both methods are throwing Notice: Undefined variable: show_teams in C:xampphtdocsfriendlyfixturestables.php on line 96 and Warning: mysqli_data_seek() expects parameter 1 to be mysqli_result, null given in C:xampphtdocsfriendlyfixturestables.php on line 101

    – Brad Archer
    Mar 28 at 0:17












  • According to the code in your question, $show_teams is your query result object. So that's why I used that name. Maybe you changed it in the meantime or something?

    – ADyson
    Mar 28 at 0:18












  • none of the variable names were changed hence why im extremely confused, dont suppose you have any clue?

    – Brad Archer
    Mar 28 at 0:26











  • I can't see how you run the query and assign the result, so it's hard to be sure. Can you add that to the code in the question?

    – ADyson
    Mar 28 at 0:29











  • Not to worry, i placed the data seek elsewhere, now the variable is being recognised. However, i do also appreciate the array solution, thanks for the help!

    – Brad Archer
    Mar 28 at 0:35
















Both methods are throwing Notice: Undefined variable: show_teams in C:xampphtdocsfriendlyfixturestables.php on line 96 and Warning: mysqli_data_seek() expects parameter 1 to be mysqli_result, null given in C:xampphtdocsfriendlyfixturestables.php on line 101

– Brad Archer
Mar 28 at 0:17






Both methods are throwing Notice: Undefined variable: show_teams in C:xampphtdocsfriendlyfixturestables.php on line 96 and Warning: mysqli_data_seek() expects parameter 1 to be mysqli_result, null given in C:xampphtdocsfriendlyfixturestables.php on line 101

– Brad Archer
Mar 28 at 0:17














According to the code in your question, $show_teams is your query result object. So that's why I used that name. Maybe you changed it in the meantime or something?

– ADyson
Mar 28 at 0:18






According to the code in your question, $show_teams is your query result object. So that's why I used that name. Maybe you changed it in the meantime or something?

– ADyson
Mar 28 at 0:18














none of the variable names were changed hence why im extremely confused, dont suppose you have any clue?

– Brad Archer
Mar 28 at 0:26





none of the variable names were changed hence why im extremely confused, dont suppose you have any clue?

– Brad Archer
Mar 28 at 0:26













I can't see how you run the query and assign the result, so it's hard to be sure. Can you add that to the code in the question?

– ADyson
Mar 28 at 0:29





I can't see how you run the query and assign the result, so it's hard to be sure. Can you add that to the code in the question?

– ADyson
Mar 28 at 0:29













Not to worry, i placed the data seek elsewhere, now the variable is being recognised. However, i do also appreciate the array solution, thanks for the help!

– Brad Archer
Mar 28 at 0:35





Not to worry, i placed the data seek elsewhere, now the variable is being recognised. However, i do also appreciate the array solution, thanks for the help!

– Brad Archer
Mar 28 at 0:35








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