php
  • jquery
  • ajax
  • 2014-02-07 42 views 0 likes 
    0

    你好我的代碼如下我從網上得到我需要移動使用jquery ajax所有字段,而不是文件輸入這裏是PHP頁面圖像的代碼上傳完美的,但其他普通文本輸入不傳遞到服務器端我怎樣才能通過所有表單域使用pekeup圖像上傳

    function UploadFile(){ 
        var error = true; 
        if (options.theme=="bootstrap"){ 
        var htmlprogress = '<div class="file"><div class="filename"></div><div class="progress progress-striped"><div class="bar pekeup-progress-bar" style="width: 0%;"><span class="badge badge-info"></span></div></div></div>'; 
        } 
        if (options.theme=="custom"){ 
        var htmlprogress = '<div class="file"><div class="filename"></div><div class="progress-pekeupload"><div class="bar-pekeupload pekeup-progress-bar" style="width: 0%;"><span></span></div></div></div>'; 
        } 
        obj.next('a').next('div').prepend(htmlprogress); 
        var formData = new FormData(); 
        formData.append(options.field, obj[0].files[0]); 
        formData.append('data', options.data); 
        $.ajax({ 
         url: options.url, 
         type: 'POST', 
         data: formData, 
           // dataType: 'json', 
         success: function(data){ 
          alert(data) 
          var percent = 100; 
          obj.next('a').next('div').find('.pekeup-progress-bar:first').width(percent+'%'); 
          obj.next('a').next('div').find('.pekeup-progress-bar:first').text(percent+"%"); 
          if (data==1){ 
           if (options.multi==false){ 
           obj.attr('disabled','disabled'); 
           } 
           options.onFileSuccess(file,data); 
          } 
          else{ 
           options.onFileError(file,data); 
           obj.next('a').next('div').find('.file:first').remove(); 
           if((options.theme == "bootstrap")&&(options.showErrorAlerts==true)){ 
           obj.next('a').next('div').prepend('<div class="alert alert-error"><button type="button" class="close" data-dismiss="alert">&times;</button> '+data+'</div>'); 
           bootstrapclosenotification(); 
           } 
           if((options.theme == "custom")&&(options.showErrorAlerts==true)){ 
    
           obj.next('a').next('div').prepend('<div class="alert-pekeupload"><button type="button" class="close" data-dismiss="alert">&times;</button> '+data+'</div>'); 
           customclosenotification(); 
           } 
           error = false; 
          } 
         }, 
         xhr: function() { // custom xhr 
           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; 
          }, 
         cache: false, 
           contentType: false, 
           processData: false 
         }); 
        return error; 
    } 
    
    +0

    嘗試更換'VAR FORMDATA =新FORMDATA();'和'VAR FORMDATA =新FORMDATA(的document.getElementById( 'your_form_id' ));' – Ravi

    回答

    0

    更換

    `var formData = new FormData();` 
    

    與以下

    `var formData = new FormData(document.getElementById('put_your_form_id_here'));` 
    

    你可以在這裏閱讀更多(瀏覽發送使用FORMDATA對象的主題文件) https://developer.mozilla.org/en-US/docs/Web/Guide/Using_FormData_Objects

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