'SyntaxError: Unexpected end of input'. error on parsing JSON?

Question

I've got this AJAX function:

function fetchSocialCount(type,fileSrc){
    var count = null;
    var req = new XMLHttpRequest();
    req.onload = function(){
        if(req.status === 200 && req.readyState === 4){
            console.log(req.responseText);
            count = JSON.parse(req.responseText);
        }
    }
    req.open("GET","../scripts/php/returnSocialCount.php?type=" + type + "&fileSrc=" + fileSrc,false);
    req.send();
    if(count !== null){
        return count;
    }
    else{
        console.log("The AJAX request has failed.");
    }
}

But when I run it, I get a 'SyntaxError: Unexpected end of input'. error. The Chrome Dev Tools also covers the count = JSON.parse(req.responseText); part in blue, so I'm guessing it's something wrong with the JSON I'm recieving. From the PHP script, I'm sending a pretty complex JSON object, so I tried to send a really basic one and it worked without problems.

This is the part of the PHP script that sends the response:

echo '{"likes":'.$json->videoListArray[$i]->likes . ',"dislikes":' . $json->videoListArray[$i]->dislikes . '}';

Is there something wrong with the syntax of the echo? What's the problem?

Thanks.

You can use the browser developer tools to see what the actual response text looks like. — Pointy
– Pointy, Commented Oct 8, 2015 at 17:14
@Pointy Thank you. Do you know where might that option be in Chrome? — bool3max
– bool3max, Commented Oct 8, 2015 at 17:15
press F12 then choose the console tab. then try doing an ajax request again it will appear there. — melvnberd
– melvnberd, Commented Oct 8, 2015 at 17:17
I wouldn't advise creating the JSON like that. Create an associative array in normal PHP syntax and then use echo json_encode($array); to create the JSON and return it to the ajax call. You run the risk of introducing a syntax error by creating this manually... — War10ck
– War10ck, Commented Oct 8, 2015 at 17:17
try : echo '{"likes":"'.$json->videoListArray[$i]->likes . '","dislikes":"' . $json->videoListArray[$i]->dislikes . '"}'; — Mimouni
– Mimouni, Commented Oct 8, 2015 at 17:18

Slicktrick · Accepted Answer · 2015-10-08 17:25:00Z

3

You need to use json_encode (PHP json_encode manual page) and echo it like so:

echo json_encode(array('likes'=>$json->videoListArray[$i]->likes, 'dislikes'=>$json->videoListArray[$i]->dislikes));

My guess is you have a comma, semi-colon, or some other character in your response that is causing the issue. json_encode will make sure everything is formatted nicely.

EDIT - War10ck beat me to the punch.

answered Oct 8, 2015 at 17:25

Slicktrick

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War10ck · Accepted Answer · 2015-10-08 17:27:38Z

1

It appears that there is probably a syntax error in your JSON that you're returning. It's not usually a good idea to manually create JSON for this very reason. Try creating an associative array and then use json_encode() to return the desired data to the ajax call:

// Create the array
$data = array();

// Add the data
$data['likes'] = $json->videoListArray[$i]->likes;
$data['dislikes'] = $json->videoListArray[$i]->dislikes;

// Generate JSON from the array and return the desired results
echo json_encode($data);

Reference Documentation:

edited Oct 8, 2015 at 17:27

answered Oct 8, 2015 at 17:20

War10ck

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2 Comments

bool3max Over a year ago

Thank you. I didn't know associative arrays encode to JSON objects.

War10ck Over a year ago

@aCodingN00b You're welcome. Any array structure in php usually encodes to JSON pretty well, even nested arrays or plan 0 based index arrays. If you ever return JSON to your server, you can also use json_decode($return_data); to generate a php object from the returned data that you can use. Alternately, using json_decode($returned_data, TRUE); will modify the returned data back into a php associative array, so you have several options. I'll link the PHP doc references in my answer for these two functions.

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