return back data from php to html

Question

I'm newbie of php and html and I've a problem. I have this form to upload a file :

    <h2>Upload file</h2>                
    <form name="uploadFile" action = "upload.php" method = "POST" enctype = "multipart/form-data">
    <input type = "file" name = "document"/>
    <input type = "submit"/>

In my external upload.php file if the upload is successful I echo a string with the name of the file and a message. If I show the value with the echo function the value is right, but I don't want a redirect to upload.php page, I want to show it back in HTML.

How can I print the return value of my upload.php file in HTML form page?

If you do not want to redirect the user to another page you can handle the POSTdata in the current script. To check if the form is submitted use an if like this: if(isset($_POST['document'])) — datagutten
– datagutten, Commented May 21, 2017 at 9:06
Use ajax and return the response from upload.php to show it on html page — Nimish
– Nimish, Commented May 21, 2017 at 9:07
Whenever you ask anything in the comments. Use @name_of_person .Otherwise he/she will not get notification of comments. Now your problem is solved or you want me to write code ? — Nimish
– Nimish, Commented May 21, 2017 at 10:32

Nirav Madariya · Accepted Answer · 2017-05-21 11:01:21Z

2

Please change your index.php file as shown below :

<?php
      if(isset($_FILES['document'])) {
        echo $_FILES["document"]["name"];
        echo "<p id='results'></p>";
?>
<script>
        $(document).ready(function() {
            $.ajax({
                url: "data.json",
                dataType: "text",
                success: function(data) {
                    $('#results').html(data);
               }
           });
        });
</script>

<?php } ?>
<h2>Upload file</h2>                
<form name="uploadFile" method = "POST" enctype = "multipart/form-data">
    <input type = "file" name = "document"/>
    <input type = "submit"/>
</form>

This prints the filename of the file you uploaded.
When you work with file upload, PHP saves name, file size, and all other attribute to $_FILES variable which is a Global Variable.

As you said you want to read other file as soon as file uploaded, this might help.

edited May 21, 2017 at 11:01

answered May 21, 2017 at 10:18

Nirav Madariya

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6 Comments

pnappa Over a year ago

yes, It works thanks. I need also to return back another information from the upload.php external script and not only filename. Any idea?

pnappa Over a year ago

a string there is content in the document.

Nirav Madariya Over a year ago

If this answer helps, mark it as an answer, and also upvote it, so that others with the same problem can benefit in future.

Nirav Madariya Over a year ago

you mean a string that the uploaded file contains?

Nirav Madariya Over a year ago

Let us continue this discussion in chat.

|

Bara' ayyash · Accepted Answer · 2017-05-21 09:13:59Z

0

You can redirect to the php page with GET variable like this: upload.php?status=success And then you can check for the status variable

answered May 21, 2017 at 9:13

Bara' ayyash

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8 Comments

RST Over a year ago

Not downvoting but this doesn't help.

pnappa Over a year ago

can you please example?

Bara' ayyash Over a year ago

after finishing uploading you redirect to upload.php right? make a redirect to upload.php?status=success instead. and then you can check the status like this $status= isset($_GET['status']) ? $_GET['status'] : "noStatus";

Bara' ayyash Over a year ago

@RST , i have used it before in collage projects :) and it works when i need to get status after doing a request

RST Over a year ago

As far as I understand his scenario, his upload.php is where is determined whether or not the upload was successful. You are already sending 'success'. If you can add some code to explain what you are doing, it might help.

|

manian · Accepted Answer · 2017-05-21 10:02:22Z

0

Add your php code for uploading at the top of the file where your form is and name it as .PHP file. Please try the below code,

<?php
  if(isset($_POST['document'])) {
    //do your post actions here
    //validate your input
    //upload the file
    //print success or error message
  }
?>
<h2>Upload file</h2>                
<form name="uploadFile" method = "POST" enctype = "multipart/form-data">
  <input type = "file" name = "document"/>
  <input type = "submit"/>

edited May 21, 2017 at 10:02

answered May 21, 2017 at 9:25

manian

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6 Comments

Nirav Madariya Over a year ago

when your action attribute have upload.php, You'll never be back here with $_POST['document'] not null, It will always redirect to upload.php on submit.

manian Over a year ago

@NiravMadariya, assume that the file itself upload.php or I will have the action empty. I will correct my answer. Sorry, I missed it. Thanks for the catch.

pnappa Over a year ago

how in my index.php (where the form is) I show the return data of upload.php form action? In the php if code you posted, how can I print the return data from the upload.php?

manian Over a year ago

@pnappa, I just corrected my answer based on NiravMadariya's comment. Removed the action attribute to post it to the same page.

pnappa Over a year ago

I removed the action attribute, and I add php. The upload works, the page is still the same (without redirect) but the data is not shown. What I have to echo, the $document variable in the php code?

|

PEHLAJ · Accepted Answer · 2017-05-21 13:26:39Z

As Nirav's posted an answer. Use it with some modification.

<?php
if(isset($_FILES['document'])) {
    echo $_FILES["document"]["name"];
    // when uploaded the file assign status=1;
    $file_name=$_FILES["document"]["name"];
    $status="1";
}
?>
<script type="text/javascript">
var status = <?php echo $status; ?>;
var filename = <?php echo $file_name; ?>;
$(document).ready(function(){
    if(status){
        $.ajax({
        url: "upload.php",
        type: "POST",
        data: {
            status:status,
            filename: filename
        }
        success: function(result){
                    $('#element').html(result);
                }
        });
    }
    else{
        $('#element').html("failed"); 
    }

});
</script>

HTML

<h2>Upload file</h2>                
<form name="uploadFile" action="upload.php" method = "POST" enctype = "multipart/form-data">
  <input type = "file" name = "document"/>
  <input type = "submit"/>
</form>
<div id="element"><!--Here Show message from upload.php--></div>

Upload.php

<?php
if(isset($_POST['status']) && !empty($_POST['status'])){

    echo $_POST['filename']." uploaded successfully";
}
?>

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