CODEWITHSUNDEEP
phpajax
user1657720
user1657720

Reputation: 33

retrieve mysql result to jquery

My problem is that I can't retrieve the result of the mysql result via ajax, please help

ajax code:

 $.ajax({
    type: "POST",
    url: "do_find_courses.php",
    //data:{question_id:question_id,answer:answer},
    data:{user_id:user_id}, dataType:'json',
    success:function(msg) {
        alert ('asdasd')
        // $("#quiz_form,#demo1").addClass("hide");
        // $('#result').show();
        $('p').html(msg);
    }
 });

PHP code:

 $final=array();
 $sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
 $row_courses = mysql_fetch_array($sql_courses);            
 $result=$row_courses['course_name'];
 //array_push($final,$result);
 //print_r($result);
 echo json_encode($result);

Upvotes: 1

Views: 48

Answers (2)

Sina R.
Sina R.

Reputation: 1788

Its better to send it with a key value like this:

And its better to use console.log(variable); to check variable's content

ajax code:

 $.ajax({
    type: "POST",
    url: "do_find_courses.php",
    //data:{question_id:question_id,answer:answer},
    data:{user_id:user_id}, dataType:'json',
    success:function(msg) {
        alert ('asdasd');
                    console.log(msg);//You should check output of this in browser console
        // $("#quiz_form,#demo1").addClass("hide");
        // $('#result').show();
        $('p').html(msg.cname);
    }
 });

PHP code:

 $final=array();
 $sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
 $row_courses = mysql_fetch_array($sql_courses);            
 $result=$row_courses['course_name']; // this will have first_coures_name (an string)
 $final['cname']=$result;
 //print_r($result);
 echo json_encode($final); //the output should be this {'cname':'first_course_name'}

Upvotes: 0

Razor Jack
Razor Jack

Reputation: 834

Change PHP Code as below

$final = array();
$sql_courses = mysql_query("SELECT course_id, course_name FROM course") or die(mysql_error());
$row_courses = mysql_fetch_array($sql_courses);

echo json_encode($row_courses);

change php code as below:

 $.ajax({
     type: "POST",
     url: "do_find_courses.php",
     //data:{question_id:question_id,answer:answer},
     data: {
         user_id: user_id
     },
     dataType: 'json',
     success: function (msg) {
         $('p').html(msg.course_name);
     }
 });

Upvotes: 1

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