dongzhabo2796 2014-08-07 03:40
浏览 108

语法错误:输入parseJSON的意外结束

I'm on about hour 5 of this and figure it's time to ask for help. I'm trying to use AJAX+php to upload an image and some text data on a form to a database. The total system is:

an input page with a form, social.php a php processing page, postMsg.php and a javascript function postMsg() that posts the form to the php processing page and is supposed to return the results to a div on social.php

The problem is that the $.parseJSON(data) command in the javascript results in an "unexpected end of input" error:

Failed:  
SyntaxError {stack: (...), message: "Unexpected end of input"}
(index):156
Uncaught SyntaxError: Unexpected end of input jquery.js:4235
b.extend.parseJSON jquery.js:4235
(anonymous function) (index):158
c jquery.js:4611
p.fireWith jquery.js:4687
k jquery.js:10335
r

I thought there was an issue with my javascript, but I code-checked it and it's fine:

     function postMsg() {
        console.log("submit event");
        var fd = new FormData(document.getElementById("commentSubmit"));
        fd.append("label", "WEBUPLOAD");
        document.getElementById('progressBar').style.display = 'block';

        $.ajax({
          url: "postMsg.php",
          type: "POST",
          xhr: function() {  // Custom XMLHttpRequest
            var myXhr = $.ajaxSettings.xhr();
            if(myXhr.upload){ // Check if upload property exists
                myXhr.upload.addEventListener('progress',progressHandlingFunction, false); // For handling the progress of the upload
            }
            return myXhr;
          },
          data: fd,
          enctype: 'multipart/form-data',
          processData: false,  // tell jQuery not to process the data
          contentType: false   // tell jQuery not to set contentType
        }).done(function( data ) {
            console.log("PHP Output:");
            console.log( data );

            try {responseData = $.parseJSON(data)}
            catch (e) {console.log('Failed: ', e, data);}

            var items = $.parseJSON(data);
            document.getElementById('progressBar').style.display = 'none';
        });
        return false;
    }

Then I thought there was an issue with my php, but replaced it all with a simple command and it still resulted in the same error:

$json_array = array('selfie'=>'hello');

Then I thought there might be an issue with my input form, so I rewrote that, but it's still returning the same error:

echo '<div data-role="fieldcontain" style="margin-top: -30px;margin-bottom: -30px;border-bottom: 0px;">
                <form method="post" name="commentSubmit" id="commentSubmit">
                  <div style="width=100%; font-size:.9em;" data-role="fieldcontain">
                  <label class="ui-input-text" for="msg_txt">Chip in:</label>';

//          $selfie = get_selfie($uid);
        echo '<div style="margin-top: 10px; margin-bottom: 10px; display: block; font-size:.9em">';

echo '<input name="file" type="file">';
        echo '<textarea style="width:100% text-align:left; font-weight:normal;" class="ui-btn ui-btn-inline ui-btn-icon-notext ui-btn-up-c" data-iconshadow="true" data-shadow="true" data-corners="false" cols="23" rows="1" name="msg_txt" id="msg_txt"></textarea>';
        echo '<a style="border-radius:8px" class="ui-btn ui-btn-inline ui-btn-icon-notext ui-corner-right ui-controlgroup-last ui-btn-up-c" title="My button" data-wrapperels="span" data-iconshadow="true" onclick="postMsg();" data-shadow="true" data-corners="true" data-role="button" data-icon="search" data-iconpos="notext" data-theme="c" data-inline="true"><span class="ui-btn-inner ui-corner-right ui-controlgroup-last"><span class="ui-btn-text">My button</span><span class="ui-icon ui-icon-search ui-icon-shadow">&nbsp;</span></span></a>';
        echo '<div id="photoUploaded" style="display: none;
position: relative;
text-align: center;
background-color: white;
opacity: .5;
color: black;
float: right; 
vertical-align: middle;
font-family: sans-serif;
border-radius: 10px;
padding: 10px;">photo loaded</div>';

                  echo '<input name="refresh" value="1" id="refresh" type="hidden">
                    <input name="uname" value="'.get_name($uid).'" id="uname" type="hidden">
                    <input name="uid" value="'.$uid.'" id="uname" type="hidden">

                </form>

Any ideas?

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1条回答 默认 最新

  • doumu9019 2015-03-14 14:39
    关注

    This plagued me for sometime as most of the answers to this same question get caught running down rabbit holes. Plus, the answer is hidden DEEP within the jquery documentation for the ajax object. Without further ado:

    jQuery.ajax:dataType

    "json": Evaluates the response as JSON and returns a JavaScript object. The JSON data is parsed in a strict manner; any malformed JSON is rejected and a parse error is thrown. As of jQuery 1.9, an empty response is also rejected; the server should return a response of null or {} instead. (See json.org for more information on proper JSON formatting.)

    Ergo, you must always at least return an empty JSON object or null from any request in order for jquery to accept it as valid.

    评论

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