javascript - 如何通过ajax load()函数发送图像文件并通过php插入mysql db显示在div中

Question

这是我将图像文件发送到 upload.php 脚本文件的形式，图像文件应显示在 'onsuccessmsg' div 中

<form action="upload.php" method="POST" id="uploadform">
   <input type="file" file="file" id="file_to_send"/>
   <input type="submit" value="Upload"/><br/><br/>
   Message :
   <div id="onsuccessmsg" style="border:5px solid #CCC;padding:15px;"></div>
  </form>

这是 JavaScript 代码:

<script>

$(document).ready(function(){

$("#uploadform").on('submit',function(){


        $("#onsuccessmsg").load("upload.php",{},function(res){

        });

});
});
</script>

这是 upload.php 文件:

<?php

if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST"){

 $name = $_FILES['file']['name'];
 $size = $_FILES['file']['size'];
 $tmp  = $_FILES['file']['tmp_name'];

 if(move_uploaded_file($tmp, "./uploads/".$imgn))
    {
     echo "File Name : ".$_FILES['file']['name'];
     echo "<br/>File Temporary Location : ".$_FILES['file']['tmp_name'];
     echo "<br/>File Size : ".$_FILES['file']['size'];
     echo "<br/>File Type : ".$_FILES['file']['type'];
     echo "<br/>Image : <img style='margin-left:10px;' src='uploads/".$imgn."'>";
    }
}

?>

Onsubmit 表单如何将 name 和 id 值发送到 upload.php 文件并在 onsuccessmsg div 中回显图像文件，我很困惑，请帮助我。

Answer 1

通过这段代码可以得到图片名称

$('#submit').click(function()
{
    var str=$("#image").val();
    var arr=str.split("\\");
    var image=arr[2];
    alert(image);
});