HTML阻止重定向上传php

I need to prevent the page redirected to the upload php when click upload button.

How can I do this in below code.

<form id="myForm"  action="http://example/DB_1/AccessWeb/file_upload.php" method="post" enctype="multipart/form-data">
    Select image to upload:
    <input type="file" name="fileToUpload" id="fileToUpload1">
</form>

<button onclick="myFunction()">  Upload
</button>

<script>

function myFunction(){

  document.getElementById("myForm").submit();


}
</script>

A very basic, quickly written example of how to send a file - using ajax to the same page so that the user doesn't get redirected. This is plain vanilla javascript rather than jQuery.

The callback function can do more than print the response - it could, for instance, be used to update the DOM with new content based upon the success/failure of the upload.

<?php
    $field='fileToUpload';

    if( $_SERVER['REQUEST_METHOD']=='POST' && !empty( $_FILES ) ){
        $obj=(object)$_FILES[ $field ];

        $name=$obj->name;
        $tmp=$obj->tmp_name;
        $size=$obj->size;
        $error=$obj->error;
        $type=$obj->type;

        if( $error==UPLOAD_ERR_OK ){
            /* 
                This is where you would process the uploaded file
                with various tests to ensure the file is OK before
                saving to disk.

                What you send back to the user is up to you - it could
                be json,text,html etc etc but here the ajax callback 
                function simply receives the name of the file chosen.
            */
            echo $name;
        } else {
            echo "bad foo!";
        }
        exit();
    }
?>
<!doctype html>
<html>
    <head>
        <title>File Upload - using ajax</title>
        <script>
            document.addEventListener('DOMContentLoaded',function(e){
                var bttn=document.getElementById('bttn');
                bttn.onclick=function(e){
                    /* Assign a new FormData object using the buttons parent ( the form ) as the argument */
                    var data=new FormData( e.target.parentNode );
                    var xhr=new XMLHttpRequest();
                    xhr.onload=function(e){
                        document.getElementById('status').innerHTML=this.response;
                    }
                    xhr.onerror=function(e){
                        alert(e);
                    }
                    xhr.open('POST',location.href,true);
                    xhr.send(data);
                };
            },false);
        </script>
    </head>
    <body>
        <form method='post' enctype='multipart/form-data'>
            Select image to upload:
            <input type='file' name='fileToUpload'>
            <input type='button' id='bttn' value='Upload' />
        </form><div id='status'></div>
    </body>
</html>

Using JQuery AJAX methods will allow you to send and receive information to a specified url without the need to refresh your page.

You will need to include the JQuery library in your HTML page aswell. You can either download it and put it in your project folder or include an online library here, like so:

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

So your form will now look like this:

<form id="myForm" method="post" >
        Select image to upload:
        <input type="file" name="fileToUpload" id="fileToUpload1">
        <input type="submit">
</form>

Then you can use this code to simply upload your image to your file upload page (tested and working for myself):

 <script>
$(document).ready(function ()
    {
        $("#myForm").submit(function (e)
            {
                //Stops submit button from refreshing page.
                e.preventDefault();

                var form_data = new FormData(this);

                $.ajax({
                    url: 'http://example/DB_1/AccessWeb/file_upload.php', //location of where you want to send image
                    dataType: 'json', // what to expect back from the PHP script, if anything
                    cache: false,
                    contentType: false,
                    processData: false,
                    data: form_data,
                    type: 'post',
                    success: function (response)
                        {
                            alert('success');
                        },
                    error: function ()
                        {
                            alert('failure');
                        }
                });
            });
    });
 </script>

use AJAX With jQuery

$("#myForm").submit(function() 
{

    var formData = new FormData(this);

    $.post($(this).attr("action"), formData, function(response) {
        //Handle the response here
    });

    return false;
});