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dependant dropdown not populating
jQuery $.ajax(), $.post sending “OPTIONS” as REQUEST_METHOD in Firefoxhow to populate dropdown with dynamic date based on 2 other dropdown boxesjQuery Autocomplete ajax does not create dropdown of selections from phpUse AJAX or pre-load: dynamic changes of items in select elementphp script echoing part of the php instead of what intendedHow to populate a textbox with value of country dropdown each time user makes a selection?Populating dropdown list from Text file, value is being set as incrementing number instead of contentHow do I populate one select from a choice in another select using jquery and a json arrayGet text from dynamic dropdown and use it in another fileajax jquery json phpmailer with smtp
0
I am trying to populate a drop down box depending which county has been selected the second drop down should populate with the provinces of the chosen county.
I don't understand why the second dropdown is not populating. I am getting a JSON response in the console so the PHP is correct. I am sure it is something silly but I just cant see it. Thanks in advance.
index.php page
<?php
include "config.php";
?>
<!doctype html>
<html>
<head>
<title>dropdown</title>
<link href="style.css" rel="stylesheet" type="text/css">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="jquery-1.12.0.min.js" type="text/javascript"></script>
<script type="text/javascript">
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var len = response.length;
$( "#province" ).empty();
for ( var i = 0; i < len; i++ )
var id = response[ i ][ 'provinceid' ];
var name = response[ i ][ 'provincename' ];
$( "#province" ).append( "<option value='" + id + "'>" + name + "</option>" );
);
);
);
</script>
</head>
<body>
<div>Country</div>
<select id="country">
<option value="0">- Select -</option>
<?php
// Fetch Country
$stmt = $conn->prepare('SELECT * FROM countries');
$stmt->execute();
while($countries = $stmt->fetch())
$countryid = $countries['id'];
$countryname_en = $countries['countryname_en'];
// Option
echo "<option value='".$countryid."' >".$countryname_en."</option>";
?>
</select>
<div class="clear"></div>
<div>Province</div>
<select id="province">
<option value="0">- Select -</option>
</select>
</body>
</html>
PHP
<?php
include "config.php";
var_dump($_POST);
$countryid = $_POST['countryid'];
$countryid = "CA";
$stmt = $conn->prepare('SELECT * FROM provincestates WHERE countryid = :countryid');
$stmt->execute(array(
':countryid' => $countryid
));
/*
echo "<pre>";
echo "prov is:" . $province_array = array();
echo "</pre>";
*/
while ($province = $stmt->fetch(PDO::FETCH_ASSOC))
$provinceid = $province['provincestatecode'];
$provincename = $province['provincestatename_en'];
$province_array[] = array(
"provinceid" => $provinceid,
"provincename" => $provincename
);
echo json_encode($province_array);
php json ajax dropdown
edited Mar 21 at 16:05
RamRaider
18.4k31935
asked Mar 21 at 15:54
TwiggitTwiggit
428
add a comment |
0
I am trying to populate a drop down box depending which county has been selected the second drop down should populate with the provinces of the chosen county.
I don't understand why the second dropdown is not populating. I am getting a JSON response in the console so the PHP is correct. I am sure it is something silly but I just cant see it. Thanks in advance.
index.php page
<?php
include "config.php";
?>
<!doctype html>
<html>
<head>
<title>dropdown</title>
<link href="style.css" rel="stylesheet" type="text/css">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="jquery-1.12.0.min.js" type="text/javascript"></script>
<script type="text/javascript">
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var len = response.length;
$( "#province" ).empty();
for ( var i = 0; i < len; i++ )
var id = response[ i ][ 'provinceid' ];
var name = response[ i ][ 'provincename' ];
$( "#province" ).append( "<option value='" + id + "'>" + name + "</option>" );
);
);
);
</script>
</head>
<body>
<div>Country</div>
<select id="country">
<option value="0">- Select -</option>
<?php
// Fetch Country
$stmt = $conn->prepare('SELECT * FROM countries');
$stmt->execute();
while($countries = $stmt->fetch())
$countryid = $countries['id'];
$countryname_en = $countries['countryname_en'];
// Option
echo "<option value='".$countryid."' >".$countryname_en."</option>";
?>
</select>
<div class="clear"></div>
<div>Province</div>
<select id="province">
<option value="0">- Select -</option>
</select>
</body>
</html>
PHP
<?php
include "config.php";
var_dump($_POST);
$countryid = $_POST['countryid'];
$countryid = "CA";
$stmt = $conn->prepare('SELECT * FROM provincestates WHERE countryid = :countryid');
$stmt->execute(array(
':countryid' => $countryid
));
/*
echo "<pre>";
echo "prov is:" . $province_array = array();
echo "</pre>";
*/
while ($province = $stmt->fetch(PDO::FETCH_ASSOC))
$provinceid = $province['provincestatecode'];
$provincename = $province['provincestatename_en'];
$province_array[] = array(
"provinceid" => $provinceid,
"provincename" => $provincename
);
echo json_encode($province_array);
php json ajax dropdown
edited Mar 21 at 16:05
RamRaider
18.4k31935
asked Mar 21 at 15:54
TwiggitTwiggit
428
What doesconsole.log(response)return inside your success function?
– EternalHour
Mar 21 at 17:56
Once you can get a response with the proper data structure, here is a working example. jsfiddle.net/hadxLser
– EternalHour
Mar 21 at 18:01
add a comment |
0
0
0
I am trying to populate a drop down box depending which county has been selected the second drop down should populate with the provinces of the chosen county.
I don't understand why the second dropdown is not populating. I am getting a JSON response in the console so the PHP is correct. I am sure it is something silly but I just cant see it. Thanks in advance.
index.php page
<?php
include "config.php";
?>
<!doctype html>
<html>
<head>
<title>dropdown</title>
<link href="style.css" rel="stylesheet" type="text/css">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="jquery-1.12.0.min.js" type="text/javascript"></script>
<script type="text/javascript">
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var len = response.length;
$( "#province" ).empty();
for ( var i = 0; i < len; i++ )
var id = response[ i ][ 'provinceid' ];
var name = response[ i ][ 'provincename' ];
$( "#province" ).append( "<option value='" + id + "'>" + name + "</option>" );
);
);
);
</script>
</head>
<body>
<div>Country</div>
<select id="country">
<option value="0">- Select -</option>
<?php
// Fetch Country
$stmt = $conn->prepare('SELECT * FROM countries');
$stmt->execute();
while($countries = $stmt->fetch())
$countryid = $countries['id'];
$countryname_en = $countries['countryname_en'];
// Option
echo "<option value='".$countryid."' >".$countryname_en."</option>";
?>
</select>
<div class="clear"></div>
<div>Province</div>
<select id="province">
<option value="0">- Select -</option>
</select>
</body>
</html>
PHP
<?php
include "config.php";
var_dump($_POST);
$countryid = $_POST['countryid'];
$countryid = "CA";
$stmt = $conn->prepare('SELECT * FROM provincestates WHERE countryid = :countryid');
$stmt->execute(array(
':countryid' => $countryid
));
/*
echo "<pre>";
echo "prov is:" . $province_array = array();
echo "</pre>";
*/
while ($province = $stmt->fetch(PDO::FETCH_ASSOC))
$provinceid = $province['provincestatecode'];
$provincename = $province['provincestatename_en'];
$province_array[] = array(
"provinceid" => $provinceid,
"provincename" => $provincename
);
echo json_encode($province_array);
php json ajax dropdown
edited Mar 21 at 16:05
RamRaider
18.4k31935
asked Mar 21 at 15:54
TwiggitTwiggit
428
I am trying to populate a drop down box depending which county has been selected the second drop down should populate with the provinces of the chosen county.
I don't understand why the second dropdown is not populating. I am getting a JSON response in the console so the PHP is correct. I am sure it is something silly but I just cant see it. Thanks in advance.
index.php page
<?php
include "config.php";
?>
<!doctype html>
<html>
<head>
<title>dropdown</title>
<link href="style.css" rel="stylesheet" type="text/css">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="jquery-1.12.0.min.js" type="text/javascript"></script>
<script type="text/javascript">
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var len = response.length;
$( "#province" ).empty();
for ( var i = 0; i < len; i++ )
var id = response[ i ][ 'provinceid' ];
var name = response[ i ][ 'provincename' ];
$( "#province" ).append( "<option value='" + id + "'>" + name + "</option>" );
);
);
);
</script>
</head>
<body>
<div>Country</div>
<select id="country">
<option value="0">- Select -</option>
<?php
// Fetch Country
$stmt = $conn->prepare('SELECT * FROM countries');
$stmt->execute();
while($countries = $stmt->fetch())
$countryid = $countries['id'];
$countryname_en = $countries['countryname_en'];
// Option
echo "<option value='".$countryid."' >".$countryname_en."</option>";
?>
</select>
<div class="clear"></div>
<div>Province</div>
<select id="province">
<option value="0">- Select -</option>
</select>
</body>
</html>
PHP
<?php
include "config.php";
var_dump($_POST);
$countryid = $_POST['countryid'];
$countryid = "CA";
$stmt = $conn->prepare('SELECT * FROM provincestates WHERE countryid = :countryid');
$stmt->execute(array(
':countryid' => $countryid
));
/*
echo "<pre>";
echo "prov is:" . $province_array = array();
echo "</pre>";
*/
while ($province = $stmt->fetch(PDO::FETCH_ASSOC))
$provinceid = $province['provincestatecode'];
$provincename = $province['provincestatename_en'];
$province_array[] = array(
"provinceid" => $provinceid,
"provincename" => $provincename
);
echo json_encode($province_array);
php json ajax dropdown
php json ajax dropdown
edited Mar 21 at 16:05
RamRaider
18.4k31935
asked Mar 21 at 15:54
TwiggitTwiggit
428
edited Mar 21 at 16:05
RamRaider
18.4k31935
asked Mar 21 at 15:54
TwiggitTwiggit
428
edited Mar 21 at 16:05
RamRaider
18.4k31935
edited Mar 21 at 16:05
RamRaider
18.4k31935
edited Mar 21 at 16:05
RamRaider
18.4k31935
18.4k31935
asked Mar 21 at 15:54
TwiggitTwiggit
428
asked Mar 21 at 15:54
TwiggitTwiggit
428
asked Mar 21 at 15:54
TwiggitTwiggit
428
428
What doesconsole.log(response)return inside your success function?
– EternalHour
Mar 21 at 17:56
Once you can get a response with the proper data structure, here is a working example. jsfiddle.net/hadxLser
– EternalHour
Mar 21 at 18:01
add a comment |
What doesconsole.log(response)return inside your success function?
– EternalHour
Mar 21 at 17:56
Once you can get a response with the proper data structure, here is a working example. jsfiddle.net/hadxLser
– EternalHour
Mar 21 at 18:01
What does
console.log(response) return inside your success function?– EternalHour
Mar 21 at 17:56
What does
console.log(response) return inside your success function?– EternalHour
Mar 21 at 17:56
Once you can get a response with the proper data structure, here is a working example. jsfiddle.net/hadxLser
– EternalHour
Mar 21 at 18:01
Once you can get a response with the proper data structure, here is a working example. jsfiddle.net/hadxLser
– EternalHour
Mar 21 at 18:01
add a comment |
1 Answer
1
active
oldest
votes
0
I'm not overly familiar wit jQuery but it seems you need to parse the response using JSON.parse ( or native jQuery method ? ) and then you ought to be able to access the data OK. The following might point you in the right direction - or it might not as it is not tested...
<script>
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var json=JSON.parse( response ); /* PARSE the data */
$( "#province" ).empty();
for( var i in json )/* access using object notation */
var obj=json[ i ];
var id=obj.provinceid;
var name=obj.provincename;
$( "#province" ).append( new Option(id,name) );
);
);
);
</script>
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
Hi RamRaider, thanks for the quick reply but the dropdown still does not populate and it looks like the exact same payload I was getting before but I learned something new so thank you.
– Twiggit
Mar 21 at 16:43
He doesn't needJSON.parse,dataType: 'json'parses it automatically.
– EternalHour
Mar 21 at 17:57
@EternalHour - thank you for the clarification. As I mentioned I don't know jQuery but perhaps should have known ( had suspicions ) that jQuery was smarter than your average bear and that would have been covered. I learned something there though so thanks.
– RamRaider
Mar 21 at 19:10
Like you I'm just here to help.
– EternalHour
Mar 21 at 21:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
I'm not overly familiar wit jQuery but it seems you need to parse the response using JSON.parse ( or native jQuery method ? ) and then you ought to be able to access the data OK. The following might point you in the right direction - or it might not as it is not tested...
<script>
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var json=JSON.parse( response ); /* PARSE the data */
$( "#province" ).empty();
for( var i in json )/* access using object notation */
var obj=json[ i ];
var id=obj.provinceid;
var name=obj.provincename;
$( "#province" ).append( new Option(id,name) );
);
);
);
</script>
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
Hi RamRaider, thanks for the quick reply but the dropdown still does not populate and it looks like the exact same payload I was getting before but I learned something new so thank you.
– Twiggit
Mar 21 at 16:43
He doesn't needJSON.parse,dataType: 'json'parses it automatically.
– EternalHour
Mar 21 at 17:57
@EternalHour - thank you for the clarification. As I mentioned I don't know jQuery but perhaps should have known ( had suspicions ) that jQuery was smarter than your average bear and that would have been covered. I learned something there though so thanks.
– RamRaider
Mar 21 at 19:10
Like you I'm just here to help.
– EternalHour
Mar 21 at 21:28
add a comment |
0
I'm not overly familiar wit jQuery but it seems you need to parse the response using JSON.parse ( or native jQuery method ? ) and then you ought to be able to access the data OK. The following might point you in the right direction - or it might not as it is not tested...
<script>
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var json=JSON.parse( response ); /* PARSE the data */
$( "#province" ).empty();
for( var i in json )/* access using object notation */
var obj=json[ i ];
var id=obj.provinceid;
var name=obj.provincename;
$( "#province" ).append( new Option(id,name) );
);
);
);
</script>
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
Hi RamRaider, thanks for the quick reply but the dropdown still does not populate and it looks like the exact same payload I was getting before but I learned something new so thank you.
– Twiggit
Mar 21 at 16:43
He doesn't needJSON.parse,dataType: 'json'parses it automatically.
– EternalHour
Mar 21 at 17:57
@EternalHour - thank you for the clarification. As I mentioned I don't know jQuery but perhaps should have known ( had suspicions ) that jQuery was smarter than your average bear and that would have been covered. I learned something there though so thanks.
– RamRaider
Mar 21 at 19:10
Like you I'm just here to help.
– EternalHour
Mar 21 at 21:28
add a comment |
0
0
0
I'm not overly familiar wit jQuery but it seems you need to parse the response using JSON.parse ( or native jQuery method ? ) and then you ought to be able to access the data OK. The following might point you in the right direction - or it might not as it is not tested...
<script>
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var json=JSON.parse( response ); /* PARSE the data */
$( "#province" ).empty();
for( var i in json )/* access using object notation */
var obj=json[ i ];
var id=obj.provinceid;
var name=obj.provincename;
$( "#province" ).append( new Option(id,name) );
);
);
);
</script>
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
I'm not overly familiar wit jQuery but it seems you need to parse the response using JSON.parse ( or native jQuery method ? ) and then you ought to be able to access the data OK. The following might point you in the right direction - or it might not as it is not tested...
<script>
$( document ).ready( function ()
$( "#country" ).change( function ()
var countryid = $( this ).val();
$.ajax(
url: 'getUsers.php',
type: 'POST',
data:
countryid: countryid
,
dataType: 'json',
success: function ( response )
var json=JSON.parse( response ); /* PARSE the data */
$( "#province" ).empty();
for( var i in json )/* access using object notation */
var obj=json[ i ];
var id=obj.provinceid;
var name=obj.provincename;
$( "#province" ).append( new Option(id,name) );
);
);
);
</script>
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
answered Mar 21 at 16:15
RamRaiderRamRaider
18.4k31935
18.4k31935
Hi RamRaider, thanks for the quick reply but the dropdown still does not populate and it looks like the exact same payload I was getting before but I learned something new so thank you.
– Twiggit
Mar 21 at 16:43
He doesn't needJSON.parse,dataType: 'json'parses it automatically.
– EternalHour
Mar 21 at 17:57
@EternalHour - thank you for the clarification. As I mentioned I don't know jQuery but perhaps should have known ( had suspicions ) that jQuery was smarter than your average bear and that would have been covered. I learned something there though so thanks.
– RamRaider
Mar 21 at 19:10
Like you I'm just here to help.
– EternalHour
Mar 21 at 21:28
add a comment |
Hi RamRaider, thanks for the quick reply but the dropdown still does not populate and it looks like the exact same payload I was getting before but I learned something new so thank you.
– Twiggit
Mar 21 at 16:43
He doesn't needJSON.parse,dataType: 'json'parses it automatically.
– EternalHour
Mar 21 at 17:57
@EternalHour - thank you for the clarification. As I mentioned I don't know jQuery but perhaps should have known ( had suspicions ) that jQuery was smarter than your average bear and that would have been covered. I learned something there though so thanks.
– RamRaider
Mar 21 at 19:10
Like you I'm just here to help.
– EternalHour
Mar 21 at 21:28
Hi RamRaider, thanks for the quick reply but the dropdown still does not populate and it looks like the exact same payload I was getting before but I learned something new so thank you.
– Twiggit
Mar 21 at 16:43
Hi RamRaider, thanks for the quick reply but the dropdown still does not populate and it looks like the exact same payload I was getting before but I learned something new so thank you.
– Twiggit
Mar 21 at 16:43
He doesn't need
JSON.parse, dataType: 'json' parses it automatically.– EternalHour
Mar 21 at 17:57
He doesn't need
JSON.parse, dataType: 'json' parses it automatically.– EternalHour
Mar 21 at 17:57
@EternalHour - thank you for the clarification. As I mentioned I don't know jQuery but perhaps should have known ( had suspicions ) that jQuery was smarter than your average bear and that would have been covered. I learned something there though so thanks.
– RamRaider
Mar 21 at 19:10
@EternalHour - thank you for the clarification. As I mentioned I don't know jQuery but perhaps should have known ( had suspicions ) that jQuery was smarter than your average bear and that would have been covered. I learned something there though so thanks.
– RamRaider
Mar 21 at 19:10
Like you I'm just here to help.
– EternalHour
Mar 21 at 21:28
Like you I'm just here to help.
– EternalHour
Mar 21 at 21:28
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What does
console.log(response)return inside your success function?– EternalHour
Mar 21 at 17:56
Once you can get a response with the proper data structure, here is a working example. jsfiddle.net/hadxLser
– EternalHour
Mar 21 at 18:01