2015-10-08 17:12
浏览 422

'SyntaxError:意外的输入结束'。 解析JSON时出错?

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){
            count = JSON.parse(req.responseText);
    }"GET","../scripts/php/returnSocialCount.php?type=" + type + "&fileSrc=" + fileSrc,false);
    if(count !== null){
        return count;
        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?


  • 写回答
  • 好问题 提建议
  • 追加酬金
  • 关注问题
  • 邀请回答

2条回答 默认 最新

  • drnf09037160 2015-10-08 17:20

    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:

    解决 无用
    打赏 举报

相关推荐 更多相似问题