Cannot detect json_encode result from mysql in ajax success

Question

I am trying to get the JSON response in success AJAX from MySql.
This is the Error i am getting in my console -> Uncaught SyntaxError: Unexpected token <

My AJAX Code.

    function getMessage(siteid)
    {   
    document.getElementById("add").disabled = true;
    $.ajax({
        type: "POST",               
        url: "checkSite.php",   
        data: { siteid : siteid }, //6 digit number
        success: function(data) {
            alert(data);
            var obj = jQuery.parseJSON(data);
            if(objData.status == "not available"){
                alert("Not Available");
                document.getElementById("add").disabled = false;
            }
            if(objData.status == "available"){
                alert("Available");
                document.getElementById("add").disabled = false;
            }
        },
        error: function(result){
            alert("error");
        }
    }); 
}

My checkSite.php

$siteid = $_POST['siteid']; 
if($siteid !=''){
   $sql = "SELECT * FROM  tablename WHERE siteid='$siteid'";
   $result2 = mysql_query($sql);
   if(mysql_num_rows($result2) > 0){        
       echo json_encode(array('status' => 'not available'));
   }
   else{            
    echo json_encode(array('status' => 'available'));
   }        

}

Its showing me parse Error. jQuery.parseJSON(data) is not at all working.

Rikesh · Accepted Answer · 2014-11-14 10:24:21Z

2

There is a syntax error on this line

data: { siteid : siteid }

add a comma at the end

data: { siteid : siteid },
                         ^

edited Nov 14, 2014 at 10:24

Rikesh

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answered Nov 14, 2014 at 10:23

Anish

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7 Comments

Matarishvan Over a year ago

Thanks. but i have missed it here.. But my error is in success.

Matarishvan Over a year ago

No Anish.. i have issues in Success AJAX.

Matarishvan Over a year ago

This is my colsole.log(data) <!DOCTYPE html> <html> <head> </head> {"status":"available"}

Jai Over a year ago

you are not returning a valid json, it has a doctype also with the json, so invalid json. so what you can do just either remove the doctype or move your php code to a php file not on a php webpage. I prefer the latter.

Matarishvan Over a year ago

Jai, Please suggest me how to do the same. Thanks

|

snapster · Accepted Answer · 2014-11-14 10:43:51Z

1

try something like this and console.log and check the result

$.ajax({            

    url         :   'checkSite.php',
    dataType    :   'json',
    cache       :   false,
    contentType :   false,
    processData :   false,
    data        :   { siteid : siteid },                           
    type        :   'post',
    success     :   function(data){ 
                    console.log (data);
   }
});

answered Nov 14, 2014 at 10:43

snapster

313 bronze badges

Comments

vanarajcs · Accepted Answer · 2014-11-14 10:43:28Z

Try this

$.ajax({
                type: "POST",               
                url: "checkSite.php",   
                data: { siteid : siteid }, //6 digit number
                dataType : 'json'
                success: function(data) {
                    if(data.status == "not available"){
                        alert("Not Available");
                        document.getElementById("add").disabled = false;
                    }
                    if(data.status == "available"){
                        alert("Available");
                        document.getElementById("add").disabled = false;
                    }
                },
                error: function(result){
                    alert("error");
                }
            }); 
        })

Saketh · Accepted Answer · 2014-11-15 07:19:15Z

You did not have else block for "if($siteid !='')" case, so you will get empty from this service call which can't be parse by JSON in success callback. for your case of exception this if condition"if($siteid !='')" seemed to be failing.

So, you checksite.php should be like this shown below.

$siteid = $_POST['siteid']; 
if($siteid !=''){
  if(siteid =='1'){        
       echo json_encode(array('status' => 'not available'));
   }
   else{            
    echo json_encode(array('status' => 'available'));
   }        

}
else
{
     echo json_encode(array('status' => 'not a valid site id'));
}

EDIT:

This can be a content type issue also, use dataType as json, and you don't require to parse json explicitly, you can skip the line "jQuery.parseJSON(data);"

for your case its taking html I believe, its might add html/header tags so JSON.parse could not be parsed. see below, your modified code.

$.ajax({
    type: "POST",               
    url: "checkSite.php",   
    data: { siteid : siteid }, //6 digit number
   dataType : 'json',
   accepts: "application/json; charset=utf-8",
   success: function(data) {
        console.log(data);
        var objData= data;
        if(objData.status == "not available"){
            alert("Not Available");
            document.getElementById("add").disabled = false;
        }
        if(objData.status == "available"){
            alert("Available");
            document.getElementById("add").disabled = false;
        }
    },
    error: function(result){
        alert("error");
    }
});

Hope this will solve your problem!!

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