CODEWITHSUNDEEP
javascriptphpjqueryjsonajax
user2684521
user2684521

Reputation: 380

Unable to alert a json object after a ajax call

I have a php file with an array that has a few values. I also have a html page with a single input field. I have a listener that listens for changes in the input field, when a change happens a ajax call is made to the php file with an array. The php file uses json_encode() to convert the array to json.

Once this is done the data is returned to my html page and I am trying to alert the data.

However I am not able to, I get no errors, simply nothing happens and I am not sure why.

Here is my code.

<input type="text" name="searchDB" id="searchDB" placeholder="Type Address Here" autofocus=""/>
 <script>
            $(document).ready(function(){
                $("#searchDB").change(function(){
                    $.ajax({
                        type:"GET",
                        url: "request_9_dnm_db.php",
                        data: "hi="+$("#searchDB").val(),
                        dataType: "json",
                        success:function(msg){
                            var obj = jQuery.parseJSON(msg);
                            alert(obj.name);
                        }//Success

                    });//Ajax Call
                });//SearchDB Change
            });//document.ready
        </script>

And Here is my php

<?php

    $boom = array("user"=>array( array(
     "name"=>"Bob",
     "last"=>"Smith"), array(
         "name"=>"Jon",
         "last"=>"Snow"
     )
    )  
 );

 $coded = json_encode($boom);
return $coded;

Upvotes: 0

Views: 285

Answers (2)

Pranav C Balan
Pranav C Balan

Reputation: 115222

You are already added dataType: "json" so you don't need to parse it again using var obj = jQuery.parseJSON(msg);

$.ajax({
    type:"GET",
    url: "request_9_dnm_db.php",
    data: "hi="+$("#searchDB").val(),
    dataType: "json",
    success:function(obj){
        alert(obj.name);
    }//Success
});//Ajax Call

Also in your php you need to echo the json encoded string instead of returning

echo $coded;

Upvotes: 0

Praveen Kumar Purushothaman
Praveen Kumar Purushothaman

Reputation: 167182

You are using return in PHP in a wrong place. Give this instead:

echo $coded;

The above code should be used instead of return $coded;.

Upvotes: 3

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