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嗨，我是 PHP 新手：我曾尝试使用随机生成的 nos 上传图像文件。然后我想在视图 php 中返回上传的文件名并在 jquery 函数中设置。这里是代码上传文件。

    // controller file this file only when i click upload calling
// controller file
       function uploadfileview(){

      $targeturl =  getcwd()."/uploads/";

      if($_SERVER['REQUEST_METHOD'] == "POST"){
      $checkexisting =  getcwd()."/uploads/".$_FILES['userfile']['name'][0];
      $random_number = rand(0,1000000);
//checking the file is existing or not            
if(file_exists($checkexisting)) {
        if(move_uploaded_file($_FILES['userfile']['tmp_name'][0], $targeturl.$random_number.'-'.$_FILES['userfile']['name'][0])){
          echo json_encode(array(
            'files' => $random_number.'-'.$_FILES['userfile']['name'][0],
            'post' => $_POST,
            'fileurl' => getcwd()."/uploads/".$random_number.'-'.$_FILES['userfile']['name'][0]
            ));
            $bb = $random_number.'-'.$_FILES['userfile']['name'][0];
          }
      }else{
        if(move_uploaded_file($_FILES['userfile']['tmp_name'][0], $targeturl.$random_number.'-'.$_FILES['userfile']['name'][0])){
          echo json_encode(array(
            'files' => $random_number.'-'.$_FILES['userfile']['name'][0],
            'post' => $_POST,
            'fileurl' => getcwd()."/uploads/".$random_number.'-'.$_FILES['userfile']['name'][0]
            ));
          }
        $data['testing'] = $random_number.'-'.$_FILES['userfile']['name'][0];
       return  $this->$data('reimbursement');
      }
     // exit;
     }
     }

这是ajax上传form.function。

// for uploaded the file name
jQuery(function(){
   var button = $('#uploader-button'), interval;
    new AjaxUpload( button, {
          action: baseUrl + "expensereimbursement/uploadfileview",
          name: 'userfile[]',
          multiple: true,
          onSubmit : function(file , ext){
      // Allow only images. You should add security check on the server-side.
              if (ext && /^(jpg|png|jpeg|pdf)$/.test(ext)){

              } else {
                  // extension is not allowed
                  $('#bq-reimbursement-form .error-ajax').text('Error: Invalid File Format').css('color','red').show().fadeOut(5000);
                  // cancel upload
                  return false;
              }
              // change button text, when user selects file
              button.text('Uploading');

              // If you want to allow uploading only 1 file at time,
              // you can disable upload button
              this.disable();

              // Uploding -> Uploading. -> Uploading...
              interval = window.setInterval(function(){
                  var text = button.text();
                  if (text.length < 13){
                      button.text(text + '.');
                  } else {
                      button.text('Uploading');
                  }
              }, 200);
          },
          onComplete: function(file, response){
              button.text('Upload');

              window.clearInterval(interval);

              // enable upload button
              this.enable();
              var obj = $.parseJSON(response);
              if(obj.error){
            $('#bq-reimbursement-form .error-ajax').text('Error: File Already Exist!').css('color','red').show().fadeOut(5000);
              }else {
            var url = "https://docs.google.com/viewer?url=" + obj.fileurl;
            var html = '<li><input type="text" name="imageName[]" value="'+file +'" class="display-type" ><a class="filenames" target="_blank"  href="'+url+'">'+file +'</a><span class="close-button display-type">x</span></li>';
            $('#upload-file-bq .files').append(html);

            }
          }
    });
    });



var html = '<li><input type="text" name="imageName[]" value="'+file +'" class="display-type" ><a class="filenames" target="_blank"  href="'+url+'">'+file +'</a><span class="close-button display-type">x</span></li>'; 

//   file = my uploaded file name to be return from controller.

这里文件只有我想返回上传的文件名，任何人都可以帮助我。这是否可能在 jquery 中获取值。

4

1 回答 1

0

CodeIgniter 带有一个用于上传文件的类。你应该使用它。通过查看Upload Class的 CodeIgniter 文档。您可以在上传文件后执行此操作。 

$aUploadData = $this->upload->data()

$aUploadData然后将包含有关您上传的文件的数据数组。要返回您上传的文件的名称，您可以执行以下操作

$aUploadData = $this->upload->data()
return $aUploadData['file_name']
于 2013-08-27T10:33:24.817 回答