CODEWITHSUNDEEP
phpjavascriptajax
shazia
shazia

Reputation: 179

unable to print array returned by AJAX with javascript

function getWriters(cat, lev, id)   
{
    if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp = new XMLHttpRequest();
    }
    else { // code for IE6, IE5
        xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200)

        {

            document.getElementById("writer").innerHTML = xmlhttp.responseText;
            var writer = eval(xmlhttp.responseText);
            document.write(writer)
        }


    }

    xmlhttp.open("GET", "order.php?op=2&id=0&cat=" + cat + "&lev=" + lev, true);
    xmlhttp.send();


}

xmlhttp.responseText returns

Array ( [0] => Array ( [name] => Unassigned [user_id] => 2 [writing_level] => [writing_category] => ) [1] => Array ( [name] => arsalan [user_id] => 3 [writing_level] => [writing_category] => ) [2] => Array ( [name] => Shazia [user_id] => 4 [writing_level] => [writing_category] => ) [3] => Array ( [name] => janea [user_id] => 5 [writing_level] => [writing_category] => ) [4] => Array ( [name] => s [user_id] => 6 [writing_level] => [writing_category] => ) [5] => Array ( [name] => iuiui [user_id] => 8 [writing_level] => [writing_category] => ) [6] => Array ( [name] => demo [user_id] => 9 [writing_level] => [writing_category] => ) [7] => Array ( [name] => wewe [user_id] => 10 [writing_level] => [writing_category] => ) [8] => Array ( [name] => Muhammad Zoyeb [user_id] => 11 [writing_level] => [writing_category] => ) [9] => Array ( [name] => Atif Rauf Alvi [user_id] => 12 [writing_level] => [writing_category] => ) [10] => Array ( [name] => demo-1 [user_id] => 13 [writing_level] => [writing_category] => ) [11] => Array ( [name] => ffff** [user_id] => 14 [writing_level] => High School,Masters [writing_category] => Literature and Language,Social Sciences ) )

i am getting syntax error at document.write(writer); if i remove eval() there is no syntax error.

Can anyone explain how to fix this or some other way to convert the string returned to array.

Thanks

Upvotes: 1

Views: 1373

Answers (3)

Andy E
Andy E

Reputation: 344605

The string being returned from the server isn't valid JSON or JavaScript syntax, so you can't pass it to eval(). It looks like you're using PHP's print_r function on the array at the server — you need to use json_encode (PHP >5.2):

echo json_encode($myArray);

Then you can parse it locally with JSON.parse (recommended) or eval.

Upvotes: 3

napolux
napolux

Reputation: 16084

You're returning the PHP "representation" of an array and by evaluating this string you fall into the syntax error.

Try this:

  • Return the array from the PHP as a comma-separated string
  • Explode the string in JS and put it into an array (if you need a JS array, otherwise print it and stop)

To split a string use:

var response = xmlhttp.responseText;
var responseArray = response.split(",");

Upvotes: 0

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385204

It looks like your JSON response is the stringifed version of a PHP array. Why not return JSON from your PHP, then Javascript picks up the array natively.

Upvotes: 0

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