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how to fix mysqli_query() expects parameter 1 [duplicate]

mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or resultWarning: mysqli_query() expects parameter 1 to be mysqli boolean givenHow can I prevent SQL injection in PHP?I call session_start() the script hangs and nothing happensmysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or resultHow do I check if a string contains a specific word?How to fix “Headers already sent” error in PHPDisplaying First Name issues in PHPPassword System Not Doing AnythingMySQL php connect failureHow to convert some old MySQL code to MySQLi?mysqli expects parameter 1

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

-3

1

This question already has an answer here:

• 
  Warning: mysqli_query() expects parameter 1 to be mysqli boolean given
  
  3 answers
  
  


• 
  mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result
  
  32 answers
  
  

after changing the host I face the problem.
connected

Warning: mysqli_query() expects parameter 1 to be mysqli, boolean
given in /home/myterst/public_html/video/index.php on line 11

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result,
null given in /home/myterst/public_html/video/index.php on line 12

<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
 include('include/config.php');
 session_start();

if (isset($_POST['login'])) 

 $email = $_POST['email'];
 $password = $_POST['password'];

 $qry = mysqli_query($conn,"select * from status_users where email='$email' and password='$password'");
 $result = mysqli_fetch_array($qry);

 if($result) 
 $_SESSION['email'] = $result['email'];

 echo "<script>window.location='dashboard.php';</script>";

 else 

 $danger = "Invalid Username or Password";
 

?>

php mysqli

share|improve this question

edited Mar 23 at 21:13

mickmackusa

24k103861

asked Mar 23 at 5:55

user3645256user3645256

12

marked as duplicate by Alon Eitan, Progman, mickmackusa php
Users with the  php badge can single-handedly close php questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Mar 23 at 21:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Where are you initializing $conn?
  
  – Raul Sauco
  Mar 23 at 6:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The connection is not working so that var is false, add error checking on the connection code, is most likely wrong creds.
  
  – Lawrence Cherone
  Mar 23 at 6:03
  

  
  
  
  

add a comment | 

-3

1

This question already has an answer here:

• 
  Warning: mysqli_query() expects parameter 1 to be mysqli boolean given
  
  3 answers
  
  


• 
  mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result
  
  32 answers
  
  

after changing the host I face the problem.
connected

Warning: mysqli_query() expects parameter 1 to be mysqli, boolean
given in /home/myterst/public_html/video/index.php on line 11

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result,
null given in /home/myterst/public_html/video/index.php on line 12

<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
 include('include/config.php');
 session_start();

if (isset($_POST['login'])) 

 $email = $_POST['email'];
 $password = $_POST['password'];

 $qry = mysqli_query($conn,"select * from status_users where email='$email' and password='$password'");
 $result = mysqli_fetch_array($qry);

 if($result) 
 $_SESSION['email'] = $result['email'];

 echo "<script>window.location='dashboard.php';</script>";

 else 

 $danger = "Invalid Username or Password";
 

?>

php mysqli

share|improve this question

edited Mar 23 at 21:13

mickmackusa

24k103861

asked Mar 23 at 5:55

user3645256user3645256

12

marked as duplicate by Alon Eitan, Progman, mickmackusa php
Users with the  php badge can single-handedly close php questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Mar 23 at 21:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Where are you initializing $conn?
  
  – Raul Sauco
  Mar 23 at 6:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The connection is not working so that var is false, add error checking on the connection code, is most likely wrong creds.
  
  – Lawrence Cherone
  Mar 23 at 6:03
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  Warning: mysqli_query() expects parameter 1 to be mysqli boolean given
  
  3 answers
  
  


• 
  mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result
  
  32 answers
  
  

after changing the host I face the problem.
connected

Warning: mysqli_query() expects parameter 1 to be mysqli, boolean
given in /home/myterst/public_html/video/index.php on line 11

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result,
null given in /home/myterst/public_html/video/index.php on line 12

<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
 include('include/config.php');
 session_start();

if (isset($_POST['login'])) 

 $email = $_POST['email'];
 $password = $_POST['password'];

 $qry = mysqli_query($conn,"select * from status_users where email='$email' and password='$password'");
 $result = mysqli_fetch_array($qry);

 if($result) 
 $_SESSION['email'] = $result['email'];

 echo "<script>window.location='dashboard.php';</script>";

 else 

 $danger = "Invalid Username or Password";
 

?>

php mysqli

share|improve this question

edited Mar 23 at 21:13

mickmackusa

24k103861

asked Mar 23 at 5:55

user3645256user3645256

12

<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
 include('include/config.php');
 session_start();

if (isset($_POST['login'])) 

 $email = $_POST['email'];
 $password = $_POST['password'];

 $qry = mysqli_query($conn,"select * from status_users where email='$email' and password='$password'");
 $result = mysqli_fetch_array($qry);

 if($result) 
 $_SESSION['email'] = $result['email'];

 echo "<script>window.location='dashboard.php';</script>";

 else 

 $danger = "Invalid Username or Password";
 

?>

share|improve this question

edited Mar 23 at 21:13

mickmackusa

24k103861

asked Mar 23 at 5:55

user3645256user3645256

12

share|improve this question

edited Mar 23 at 21:13

mickmackusa

24k103861

asked Mar 23 at 5:55

user3645256user3645256

12

edited Mar 23 at 21:13

mickmackusa

24k103861

edited Mar 23 at 21:13

mickmackusa

24k103861

asked Mar 23 at 5:55

user3645256user3645256

12

asked Mar 23 at 5:55

user3645256user3645256

12

1 Answer
1

active

oldest

votes

-1

Try this.. I hope it will works sure

 $qry = "select * from status_users where email='$email' and password='$password'";

 $result = mysqli_query($conn,$qry);

 $row = mysqli_fetch_array($result,MYSQLI_NUM);

share|improve this answer

answered Mar 23 at 8:01

SUDAR MANISUDAR MANI

494

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Downvoting because of sql injection, you should not learn the OP bad practices
  
  – Alon Eitan
  Mar 23 at 8:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Try-this answers are low value on Stackoverflow because they do very little to educate/empower thousands of future researchers.
  
  – mickmackusa
  Mar 23 at 21:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for help. still 1 error Notice: Undefined variable: result in /home/proaucco/public_html/video/index.php on line 13
  
  – user3645256
  Mar 24 at 15:17
  

  
  
  
  

add a comment | 

-1

Try this.. I hope it will works sure

 $qry = "select * from status_users where email='$email' and password='$password'";

 $result = mysqli_query($conn,$qry);

 $row = mysqli_fetch_array($result,MYSQLI_NUM);

share|improve this answer

answered Mar 23 at 8:01

SUDAR MANISUDAR MANI

494

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Downvoting because of sql injection, you should not learn the OP bad practices
  
  – Alon Eitan
  Mar 23 at 8:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Try-this answers are low value on Stackoverflow because they do very little to educate/empower thousands of future researchers.
  
  – mickmackusa
  Mar 23 at 21:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for help. still 1 error Notice: Undefined variable: result in /home/proaucco/public_html/video/index.php on line 13
  
  – user3645256
  Mar 24 at 15:17
  

  
  
  
  

add a comment | 

-1

Try this.. I hope it will works sure

 $qry = "select * from status_users where email='$email' and password='$password'";

 $result = mysqli_query($conn,$qry);

 $row = mysqli_fetch_array($result,MYSQLI_NUM);

share|improve this answer

answered Mar 23 at 8:01

SUDAR MANISUDAR MANI

494

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Downvoting because of sql injection, you should not learn the OP bad practices
  
  – Alon Eitan
  Mar 23 at 8:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Try-this answers are low value on Stackoverflow because they do very little to educate/empower thousands of future researchers.
  
  – mickmackusa
  Mar 23 at 21:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for help. still 1 error Notice: Undefined variable: result in /home/proaucco/public_html/video/index.php on line 13
  
  – user3645256
  Mar 24 at 15:17
  

  
  
  
  

add a comment | 

Try this.. I hope it will works sure

 $qry = "select * from status_users where email='$email' and password='$password'";

 $result = mysqli_query($conn,$qry);

 $row = mysqli_fetch_array($result,MYSQLI_NUM);

share|improve this answer

answered Mar 23 at 8:01

SUDAR MANISUDAR MANI

494

 $qry = "select * from status_users where email='$email' and password='$password'";

 $result = mysqli_query($conn,$qry);

 $row = mysqli_fetch_array($result,MYSQLI_NUM);

share|improve this answer

answered Mar 23 at 8:01

SUDAR MANISUDAR MANI

494

answered Mar 23 at 8:01

SUDAR MANISUDAR MANI

494

answered Mar 23 at 8:01

SUDAR MANISUDAR MANI

494