Question

如果点击div（非提交按钮）而不刷新页面，如何发布和返回图像？

  <html>
    <head>
      <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    </head>
    <body>

  <input multiple class="fileInput" type="file" id="files" name="files[]"/>
  <br>
  <div id="submit" > click me to pass the photo to upload.php</div>
  <div name="photoreturn"></div>
</body>
</html>



<style>
#submit{
    background-color: #ffcc99;
    display:inline-block;
    padding:5px;
    border-radius: 3px;
    cursor:pointer;
}
</style>

Answer 1

使用AJAX。每个人都这样做。

$("body").on("click", "#submit", function(e) {
  e.preventDefault();
  post_image();
});

function post_image() {
  var img = find("input#files").val();

   $.ajax({
     url: './your_dir_here/upload.php',
     type: 'POST',
     data: img,
     dataType: 'json',
     success: function(data) {
        $("#photoreturn").fadeIn("fast");
        alert("img uploaded");
     },
   }); // End of ajax call 
}

类似的东西。

注意：我还没有尝试过运行代码，您可能会收到错误，但这就是想法。祝你好运！

Answer 2

您可以通过在div上添加onclick事件来调用ajax函数

tableMainContent.removeAllViews();
new Handler().postDelayed(new Runnable() {
    @Override
    public void run() {
        //Add view for table layout here
    }
}, 2000);

<div id="submit" onclick="uploadImage()" > click me to pass the photo to upload.php</div>

ajax function

单击div（不提交按钮）时如何在不刷新页面的情况下发布照片？

2 个答案: