Question

我正在运行以下命令来上传和删除任何文件：

$(document).on('click','.close',function(){
  $(this).parents('span').remove();
})

document.getElementById("usp-files").onchange = function () {
    document.getElementById("uploadFile").value = this.value;
};

document.getElementById('usp-files').onchange = uploadOnChange;

function uploadOnChange() {
  var filename = this.value;
  var lastIndex = filename.lastIndexOf("\\");
  if (lastIndex >= 0) {
      filename = filename.substring(lastIndex + 1);
  }
  var files = $('#usp-files')[0].files;
  for (var i = 0; i < files.length; i++) {
   $("#upload_prev").append('<span>'+files[i].name+'<p class="close">X</p></span>');
  }
  document.getElementById('filename').value = filename;
}

HTML

  <input type="hidden" id="uploadFile" placeholder="Add files from My Computer" class="steptextboxq3"/>
  <div class="fileUpload btn btn-primary">
    <span>Browse</span>
    <input name="usp-files[]" id="usp-files" type="file" maxlength="999999" data-required="false" placeholder="Upload your images here" class="upload sp-input usp-input-files my_images select-file multiple" multiple="multiple" name="browsefile">
  </div>
  <input id="filename" type="hidden" />
  <div id="upload_prev"></div>

从外观上看，我们可以看到它正在删除它，但是当我将其发送到服务器时，它会保存我上传的所有文件，即使我已经在前端删除了它们。

jsFiddle here

不从多文件中删除值

0 个答案: