javascript
2012-04-07 78 views 0 likes 
0

它不断给我一个未终止的字符串litereral错误但下面我看不到问题。如何修复如下:我总是收到一个未终止的字符串文字错误

var $fileImage = $("<form action='upload.php' method='post' enctype='multipart/form-data' target='upload_target' onsubmit='startUpload();' > 
    <p class='f1_upload_process'>Loading...<br/><img src='Images/loader.gif' /><br/></p><p class='f1_upload_form'><br/><label> 
    File: <input name='fileImage' type='file' class='fileImage' /></label><label><input type='submit' name='submitBtn' class='sbtn' value='Upload' /></label> 
    </p> <iframe class='upload_target' name='upload_target' src='#' style='width:0;height:0;border:0px solid #fff;'></iframe></form>"); 

$image.append($fileImage); 

回答

4

JavaScript不是PHP。换行符在字符串中是不允许的。

在每行的末尾添加\,或者使用" + "连接字符串。

// Using \, scroll to the right to see how to use it 
var $fileImage = $("<form action='upload.php' method='post' enctype='multipart/form-data' target='upload_target' onsubmit='startUpload();' >\ 
    <p class='f1_upload_process'>Loading...<br/><img src='Images/loader.gif' /><br/></p><p class='f1_upload_form'><br/><label>\ 
    File: <input name='fileImage' type='file' class='fileImage' /></label><label><input type='submit' name='submitBtn' class='sbtn' value='Upload' /></label>\ 
    </p> <iframe class='upload_target' name='upload_target' src='#' style='width:0;height:0;border:0px solid #fff;'></iframe></form>"); 

// Using string concatenation 
var $fileImage = $("<form action='upload.php' method='post' enctype='multipart/form-data' target='upload_target' onsubmit='startUpload();' >" + 
    "<p class='f1_upload_process'>Loading...<br/><img src='Images/loader.gif' /><br/></p><p class='f1_upload_form'><br/><label>" + 
    "File: <input name='fileImage' type='file' class='fileImage' /></label><label><input type='submit' name='submitBtn' class='sbtn' value='Upload' /></label>" + 
    "</p> <iframe class='upload_target' name='upload_target' src='#' style='width:0;height:0;border:0px solid #fff;'></iframe></form>"); 

由于您使用jQuery的,我建议通过使用jQuery阅读和维护的代码:

var $form = $('<form>', { 
    action: 'upload.php', 
    method: 'post', 
    enctype: 'multipart/form-data', 
    target: 'upload_target' 
}).submit(startUpload); 
// Et cetera. 
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