SyntaxError: JSON.parse: unexpected end of data

Question

I try add data to database using ajax. I prepare data:

$(".panel").on('click','.buttons_save',function() {
    groupIdD = this.id;
    groupNameD=$("#groupName").val();
    groupDescriptionD=$("#groupDescription").val();
    paramGroupD=$("#paramGroup").val();
    paramOrderD=$("#paramOrder").val();

    $.ajax({
        dataType: "json",
        type: "POST",
        url: ajaxSetGroupParams,
        data: { 
            groupId: groupIdD,
            paramGroup: groupNameD,
            groupName: groupDescriptionD,
            groupDescription: paramGroupD,
            paramOrder: paramOrderD
        }
    })

    .done(function() {
        alert( "success" );
    })
    .fail(function(XMLHttpRequest, textStatus, errorThrown) {
        console.log(XMLHttpRequest);
        console.log(textStatus);
        console.log(errorThrown);
    })


});

And I add this data to database (everything is ok, data are add to database). $polaczenie_z_baza=$_SESSION['db']['connectObjs']['system'];

$id_group=$_REQUEST['groupId'];
$paramGroup=$_REQUEST['paramGroup'];
$groupName=$_REQUEST['groupName'];
$groupDescription=$_REQUEST['groupDescription'];
if (isset($_REQUEST['paramOrder'])) {
    $paramOrder=$_REQUEST['paramOrder'];
}
else {
    $paramOrder=0;
}

$qUpdateData    = '
UPDATE
`grupy_parametrow_konfiguracyjnych`
SET
`id_grupa_nadrzedna` ='.$paramGroup.',
`nazwa_grupy` = '.$groupName.',
`opis_grupy` = '.$groupDescription.',
`kolejnosc` ='.$paramOrder.'
WHERE
`id_grupa_parametrow` = '.$id_group; 

$result = stdIUDQuery( $qUpdateData, $polaczenie_z_baza);

But always I have fail information:

parsererror
SyntaxError: JSON.parse: unexpected end of data

return window.JSON.parse( data );

I hope this is only a minimal example, if not you should take care for SQL Injection techniques. — David J
– David J, Commented Apr 2, 2014 at 8:35
This is modified example. I have data validation and prepared statments. — marcin375
– marcin375, Commented Apr 2, 2014 at 8:39

user3477804 · Accepted Answer · 2014-04-02 08:33:32Z

3

This will be happening because your server isn't returning JSON, but you're telling jQuery that it is. Since you're not using the contents of the response from the server in your done handler, you can simply remove the dataType: "json", from your $.ajax call.

answered Apr 2, 2014 at 8:33

user3477804

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1 Comment

marcin375 Over a year ago

Thanks a lot! Now everything is ok.

Brian Sanchez · Accepted Answer · 2020-08-26 19:30:22Z

0

3 ideas also to debug the error.

A firefox

save the response in notepad as idea.json
open in firefox
it will show you if is good json or not.

B console.log(variable.constructor)

check if is string or json with console.log(variable_name.constructor);

C check the correct line

check all JSON.parse lines, my problem was for 20 mins checking the wrong line.

answered Aug 26, 2020 at 19:30

Brian Sanchez

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