CODEWITHSUNDEEP
phpjqueryajaxjson
Piotr
Piotr

Reputation: 161

jQuery $.ajax request of dataType json will not retrieve data from PHP script

I've been looking all over for the solution but I cannot find anything that works. I am trying to get a bunch of data from the database and then via AJAX autocomplete input fields in a form. To do this I've decided to use json, because why not, right? Alternatively I've been thinking to just send back a delimited string and then tokenise it, which in hind-sight would've been much easier and spared me the headache... Since I've decided to use json though, I guess I should stick with it and find out what went wrong! What happens is that when the get_member_function() is executed, an error pops up in an alert dialogue and reads "[object Object]". I've tried this also using the GET request, and by setting the contentType to ”application/json; charset=utf-8″. Alas, no dice. Can anyone please suggest what I am doing wrong? Take care, Piotr.

My javascript/jQuery function is as follows:

function get_member_info()
   {

   var url = "contents/php_scripts/admin_scripts.php"; 
   var id = $( "select[ name = member ] option:selected" ).val();

   $.ajax(
   {

      type: "POST",
      dataType: "json",
      url: url,
      data: { get_member: id },
      success: function( response ) 
      { 

          $( "input[ name = type ]:eq( " + response.type + " )" ).attr( "checked", "checked" );
          $( "input[ name = name ]" ).val( response.name );
          $( "input[ name = fname ]" ).val( response.fname );
          $( "input[ name = lname ]" ).val( response.lname );
          $( "input[ name = email ]" ).val( response.email );
          $( "input[ name = phone ]" ).val( response.phone );
          $( "input[ name = website ]" ).val( response.website );
          $( "#admin_member_img" ).attr( "src", "images/member_images/" + response.image );

      },
      error: function( error )
      {

         alert( error );

      }

   } );

}

and the relevant code in "contents/php_scripts/admin_scripts.php" is as follows:

   if( isset( $_POST[ "get_member" ] ) )
   {

      $member_id = $_POST[ "get_member" ];
      $query = "select * from members where id = '$member_id'";

      $result = mysql_query( $query );

      $row = mysql_fetch_array( $result );

      $type = $row[ "type" ];
      $name = $row[ "name" ];
      $fname = $row[ "fname" ];
      $lname = $row[ "lname" ];
      $email = $row[ "email" ];
      $phone = $row[ "phone" ];
      $website = $row[ "website" ];
      $image = $row[ "image" ];

      $json_arr = array( "type" => $type, "name" => $name, "fname" => $fname, "lname" => $lname, "email" => $email, "phone" => $phone, "website" => $website, "image" => $image );

      echo json_encode( $json_arr );

   }

Upvotes: 16

Views: 246555

Answers (8)

Tim
Tim

Reputation: 1700

Well, it might help someone. I was stupid enough to put var_dump('testing'); in the function I was requesting JSON from to be sure the request was actually received. This obviously also echo's as part for the expected json response, and with dataType set to json defined, the request fails.

Upvotes: 0

nishant
nishant

Reputation: 11

  session_start();
include('connection.php');

/* function msg($subjectname,$coursename,$sem)
    {
    return '{"subjectname":'.$subjectname.'"coursename":'.$coursename.'"sem":'.$sem.'}';
    }*/ 
$title_id=$_POST['title_id'];
$result=mysql_query("SELECT * FROM `video` WHERE id='$title_id'") or die(mysql_error());
$qr=mysql_fetch_array($result);
$subject=$qr['subject'];
$course=$qr['course'];
$resultes=mysql_query("SELECT * FROM course JOIN subject ON course.id='$course' AND subject.id='$subject'");
$qqr=mysql_fetch_array($resultes);
$subjectname=$qqr['subjectname'];
$coursename=$qqr['coursename'];
$sem=$qqr['sem'];
$json = array("subjectname" => $subjectname, "coursename" => $coursename, "sem" => $sem,);
header("Content-Type: application/json", true);
echo json_encode( $json_arr );


 $.ajax({type:"POST",    
                  dataType: "json",    
                   url:'select-title.php',
                   data:$('#studey-form').serialize(),
                   contentType: "application/json; charset=utf-8",
                   beforeSend: function(x) {
        if(x && x.overrideMimeType) {
            x.overrideMimeType("application/j-son;charset=UTF-8");
        }
    },
                   success:function(response)
                  {
                    var response=$.parseJSON(response)
                    alert(response.subjectname);
                    $('#course').html("<option>"+response.coursename+"</option>"); 
                    $('#subject').html("<option>"+response.subjectname+"</option>");

                  },
                  error: function( error,x,y)
                  {

                  alert( x,y );

              }
                   });

Upvotes: 1

Muhammad Saqlain Arif
Muhammad Saqlain Arif

Reputation: 544

Try this...  
  <script type="text/javascript">

    $(document).ready(function(){

    $("#find").click(function(){
        var username  = $("#username").val();
            $.ajax({
            type: 'POST',
            dataType: 'json',
            url: 'includes/find.php',
            data: 'username='+username,
            success: function( data ) {
            //in data you result will be available...
            response = jQuery.parseJSON(data);
//further code..
            },

    error: function(xhr, status, error) {
        alert(status);
        },
    dataType: 'text'

    });
        });

    });



    </script>

    <form name="Find User" id="userform" class="invoform" method="post" />

    <div id ="userdiv">
      <p>Name (Lastname, firstname):</p>
      </label>
      <input type="text" name="username" id="username" class="inputfield" />
      <input type="button" name="find" id="find" class="passwordsubmit" value="find" />
    </div>
    </form>
    <div id="userinfo"><b>info will be listed here.</b></div>

Upvotes: 1

Ada Bellash Mon Tr&#233;sor
Ada Bellash Mon Tr&#233;sor

Reputation: 97

If you are using a newer version (over 1.3.x) you should learn more about the function parseJSON! I experienced the same problem. Use an old version or change your code

success=function(data){
  //something like this
  jQuery.parseJSON(data)
}

Upvotes: 0

Peter
Peter

Reputation: 2306

Try using jQuery.parseJSON when you get the data back.

type: "POST",
dataType: "json",
url: url,
data: { get_member: id },
success: function(data) { 
    response = jQuery.parseJSON(data);
    $("input[ name = type ]:eq(" + response.type + " )")
        .attr("checked", "checked");
    $("input[ name = name ]").val( response.name);
    $("input[ name = fname ]").val( response.fname);
    $("input[ name = lname ]").val( response.lname);
    $("input[ name = email ]").val( response.email);
    $("input[ name = phone ]").val( response.phone);
    $("input[ name = website ]").val( response.website);
    $("#admin_member_img")
        .attr("src", "images/member_images/" + response.image);
},
error: function(error) {
    alert(error);
}

Upvotes: 10

Alnitak
Alnitak

Reputation: 339955

The $.ajax error function takes three arguments, not one:

error: function(xhr, status, thrown)

You need to dump the 2nd and 3rd parameters to find your cause, not the first one.

Upvotes: 5

McHerbie
McHerbie

Reputation: 2995

I think I know this one...

Try sending your JSON as JSON by using PHP's header() function:

/**
 * Send as JSON
 */
header("Content-Type: application/json", true);

Though you are passing valid JSON, jQuery's $.ajax doesn't think so because it's missing the header.

jQuery used to be fine without the header, but it was changed a few versions back.

ALSO

Be sure that your script is returning valid JSON. Use Firebug or Google Chrome's Developer Tools to check the request's response in the console.

UPDATE

You will also want to update your code to sanitize the $_POST to avoid sql injection attacks. As well as provide some error catching.

if (isset($_POST['get_member'])) {

    $member_id = mysql_real_escape_string ($_POST["get_member"]);

    $query = "SELECT * FROM `members` WHERE `id` = '" . $member_id . "';";

    if ($result = mysql_query( $query )) {

       $row = mysql_fetch_array($result);

       $type = $row['type'];
       $name = $row['name'];
       $fname = $row['fname'];
       $lname = $row['lname'];
       $email = $row['email'];
       $phone = $row['phone'];
       $website = $row['website'];
       $image = $row['image'];

       /* JSON Row */
       $json = array( "type" => $type, "name" => $name, "fname" => $fname, "lname" => $lname, "email" => $email, "phone" => $phone, "website" => $website, "image" => $image );

    } else {

        /* Your Query Failed, use mysql_error to report why */
        $json = array('error' => 'MySQL Query Error');

    }

     /* Send as JSON */
     header("Content-Type: application/json", true);

    /* Return JSON */
    echo json_encode($json);

    /* Stop Execution */
    exit;

}

Upvotes: 24

Gelmir
Gelmir

Reputation: 1857

In addition to McHerbie's note, try json_encode( $json_arr, JSON_FORCE_OBJECT ); if you are on PHP 5.3...

Upvotes: 2

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