send array with JSON

Question

i'm using this code for post text of option tag :

$("#major_names").change(function () {
    $.post('find_lesson.php',  { dars:$("#major_names option:selected").text() },
        function(data){ 
            if (data.success)
                        $("div").append(data.doroos);
            else
                alert('mm');
        },'json'); 
return false; 
});

now in find_lesson file i use this code for fetch any record from database

<?php

$lessonName=$_POST['lesson '];
$query= mysql_query("SELECT * FROM at_*** WHERE title = '{$lessonName}'");
$result= mysql_fetch_array($query);

$sql= mysql_query("SELECT * FROM  *** JOIN at_lessons ON  . . .");
while($result=mysql_fetch_assoc($sql))
{
$data ['doroos']= $result['title'];
}
$data['success']=true;


echo json_encode($data);

?>

mysql command is correct but after send array i have getting 1 record. please help me

Prasanth · Accepted Answer · 2012-08-08 08:29:32Z

5

Change

$data ['doroos']= $result['title'];

to

$data ['doroos'][] = $result['title'];

You are basically overwriting the doroos. Instead, use [] to add every $result['title'] to doroos.

answered Aug 8, 2012 at 8:29

Prasanth

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2 Comments

DolDurma Over a year ago

hello. after change it i getting this error in firebug : NS_ERROR_XPC_BAD_CONVERT_JS: Could not convert JavaScript argument arg 0 [nsIDOMDocumentFragment.appendChild] [Break On This Error] fragment.appendChild( script );

Prasanth Over a year ago

It's a javascript error. do alert(data.doroos) and tell what you get.

retromuz · Accepted Answer · 2012-08-08 08:34:34Z

0

You should do something like below instead. Otherwise "doroos" will only have the last title record from the fetched resultset.

<?php

$lessonName=$_POST['lesson '];
$query= mysql_query("SELECT * FROM at_*** WHERE title = '{$lessonName}'");
$result= mysql_fetch_array($query);

$sql= mysql_query("SELECT * FROM  *** JOIN at_lessons ON  . . .");

$doroos = array();
$x = 0;
while($result=mysql_fetch_assoc($sql))
{    
     $doroos[x++] = $result['title'];
}
$data ['doroos']= $doroos;
$data['success']=true;


echo json_encode($data);

?>

answered Aug 8, 2012 at 8:34

retromuz

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2 Comments

DolDurma Over a year ago

i have get this error now: NS_ERROR_XPC_BAD_CONVERT_JS: Could not convert JavaScript argument arg 0 [nsIDOMDocumentFragment.appendChild] [Break On This Error] fragment.appendChild( script ); jquery-latest.js (line 6497) POST aou.ir/ivu/assessment/find_dars.php 200 OK 35ms jquery-latest.js (line 8240) HeadersPostResponseHTMLJSONCookies {"doroos":[],"success":true}

DolDurma Over a year ago

after change code to this: $data['doroos'] = array(); $x = 0; while($result=mysql_fetch_assoc($sql)) { $data['doroos'][$x++] = $result['title']; } $data['success']=true; echo json_encode($data); ?> i can get this result in firebug: {"doroos":["\u0627\u0646\u0633\u0627\u0646 \u0634\u0646\u0627\u0633\u06cc"],"success":true} but i can't show that with this line: if (data.success) $("div").append(data.doroos);

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