ajaxForm無法正常工作並且不返回錯誤

Question

當我想使用jQuery ajaxForm時遇到一個非常奇怪的問題。

我想設置表格並上傳進度百分比較高的文件。

我的ajaxForm函數根本不觸發。

這是我的代碼，但由於沒有錯誤，所以我無法找出問題出在哪里。

 function recieve(res,obj) { var frm = res.substr(1,res.length-8); $('#'+frm).find('.error').each(function(){ $(this).remove(); }); if(obj['alert']!=undefined) $(res).html(obj['alert']); if(obj['field']!=undefined) { for(var i in obj['field']) { /*$("#"+i).next('.error').remove();*/ if(obj['field'][i]!='') $("#"+i).after('<div class="error avesome OC OBC">'+obj['field'][i]+'</div>'); } } if(obj['msgbox']!=undefined) alert(obj['msgbox']); if(obj['location']!=undefined) document.location = obj['location']; } function sendAjax(form,response,loader,progress) { var frm = $(form); frm.ajaxForm({ dataType: 'json', data: frm.serialize()+'&_ajax=1', beforeSend: function(){ $("input[type='submit']").attr('disabled','disabled'); $(progress).width('0%').parent('.progress').removeClass('hidd'); }, uploadProgress: function(event, position, total, percentComplete){ var pVel = percentComplete + '%'; $(progress).width(pVel); }, complete: function(data){ recieve(response,unserialize(data)); $(progress).parent('.progress').addClass('hidd'); $("input[type='submit']").attr('disabled',null); } }); return false; } 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> <form id="setting_form" action="http://127.0.0.1/marketing/users/setting/" method="post" onsubmit="return sendAjax('#setting_form','#setting_form_result','#setting_form_loader','#setting_form_progress');"> <input type="hidden" name="_submit" value="1" /> <input type="hidden" name="data[setting][id]" id="setting_id" value="11" /> <input type="hidden" name="data[setting][data-token]" id="setting_data-token" value="8022735" /> <input type="hidden" name="data[setting][token]" id="setting_token" value="90e18fe55fbc38708456606f4b2b3f96" /> <input type="submit" name="data[setting][submit]" id="setting_submit" value="send" /> <div id="setting_form_progress" class="bar fade"></div> <div id="setting_form_result"></div> </form> 

Answer 1

我猜你想念這個庫http://malsup.com/jquery/form/在jquery之后插入代碼

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script src="http://malsup.github.com/jquery.form.js"></script>