dongzhao5970 2019-05-07 09:43
浏览 215

图像上传后JQuery自动提交表单 - 不提交按钮

I would like to auto-submit form after image upload. The form is being processed on the same page. so i dont have if(isset($_POST["submit"])) {

<div class="container">
  <h1>jQuery Image Upload 
    <small>with preview</small>
  </h1>
  <div class="avatar-upload">
    <form id="formImageUpload" action="#" method="post" enctype="multipart/form-data">
      <div class="avatar-edit">
        <input type='file' id="imageUpload" name="imageUpload" accept=".png, .jpg, .jpeg" onchange="this.form.submit()"  />
        <label for="imageUpload"></label>
      </div>
    </form>

    <div class="avatar-preview">
      <div id="imagePreview" style="background-image: url(http://i.pravatar.cc/500?img=7);" onchange="this.form.submit()" >
      </div>
    </div>
  </div>
</div>

jquery file is:

function readURL(input) {
  if (input.files && input.files[0]) {
    var reader = new FileReader();
    reader.onload = function(e) {
        $('#imagePreview').css('background-image', 'url('+e.target.result +')');
        $('#imagePreview').hide();
        $('#imagePreview').fadeIn(650);
        $("#formImageUpload" ).submit();
    }
    reader.readAsDataURL(input.files[0]);
  }
}


$("#imageUpload").change(function() {   
   readURL(this);
});
  • 写回答

1条回答 默认 最新

  • douyuan4357 2019-05-07 12:42
    关注

    So, your answer is like this:

    <div class="container">
      <h1>jQuery Image Upload <small>with preview</small></h1>
      <div class="avatar-upload">
        <form id="formImageUpload" action="#" method="post" enctype="multipart/form-data">
          <div class="avatar-edit">
            <input type='file' id="imageUpload" name="imageUpload" accept=".png, .jpg, .jpeg" onchange="readURL(this)"  />
            <label for="imageUpload"></label>
          </div>
        </form>
    
        <div class="avatar-preview">
          <div id="imagePreview" style="background-image: url(http://i.pravatar.cc/500?img=7);" ></div>
        </div>
      </div>
    </div>
    

    JS:

    function readURL(input) {
      if (input.files && input.files[0]) {
        var reader = new FileReader();
        reader.onload = function(e) {
          $('#imagePreview').css('background-image', 'url('+e.target.result +')');
          $('#imagePreview').hide();
          $('#imagePreview').fadeIn(650);
          $("#formImageUpload" ).submit();
        }
        reader.readAsDataURL(input.files[0]);
      }
    }
    
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