Load another page in PHP using a loader while using a simple HTML form

Question

I'm quite new to PHP. So I'm here to briefly explain my question.

Recently, I started building my site for an online tool, so it has an input box like this. (I've not designed it fully yet.) (Page URL: index.php)

I made it to fetch some datas from a social media site! As you can see, there is written "Enter Page URL". Users must have to put the URL of that page.

I should put the HTML content also! Here's it.

<form method="get" action="download.php">
<input type="text" name="url" placeholder="Enter Page URL" required />
<input type="submit" name="submit" value="Download" />
</form>

The URL of the action page is download.php. Due to heavy PHP codes inside the download.php, the page responds very slow! Sometimes even it even takes upto 5 seconds to load. Only I know what is there, sharing everything isn't a good idea at all! Users might be thinking, it's their network issue or a site problem or something like that! Being irritated, they're leaving there.

Instead of loading the another page, I wanna put the download.php page inside the index.php page with a loader.

Content inside the download.php page looks almost like this.

What if I put the download page in the index page with a loader like this? (Loader Image URL: http://webplayer.d8u.in/loading.gif)

For example, if I return back from a playing video in YouTube mobile website, a loader appears & then suddenly we start seeing the video list! Including another pages in PHP can be done by include or file_get_contents(), but I dunno how to execute this with a loader (not by refreshing the current page). I guess there're so many answers of this in the internet, but I couldn't find any of them! Asking here was a better idea for me.

So, how can I do this using jQuery & PHP?

did you try sending by Ajax?

einsh10
– einsh10

2018-01-23 20:39:51 +00:00
Commented Jan 23, 2018 at 20:39 — einsh10
– einsh10, Commented Jan 23, 2018 at 20:39
No, I didn't try it yet!

Subhajit Gorai
– Subhajit Gorai

2018-01-23 20:40:30 +00:00
Commented Jan 23, 2018 at 20:40 — Subhajit Gorai
– Subhajit Gorai, Commented Jan 23, 2018 at 20:40

Rajib Chy · Accepted Answer · 2018-11-08 21:32:25Z

2

This is not the exact code, but I think you will get the main idea on how to compete your code with ajax

function sendmyform(){
    var val = $('#myfield').val();
    $('#loadingbar').show();
    $.ajax({
        method: "POST",
        url: "download.php",
        data: { name: val }
    })
    .done(function( msg ) {
        //msg here is the returned data from dowload.php
        $('#loadingbar').hide();
        $('#myfield').val('');
    });
};

HTML:

<div id="loadingbar" style="display:none">Loading...</div>
<input id="myfield" type="text" name="url" placeholder="Enter Page URL" required />
<input type="button" onclick="sendmyform()" value="Download" />

something like this :) more documentation here http://api.jquery.com/jquery.ajax/

edited Nov 8, 2018 at 21:32

Rajib Chy

88012 silver badges23 bronze badges

answered Jan 23, 2018 at 20:53

einsh10

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Comments

Kuba Zabielski · Accepted Answer · 2018-01-23 20:47:35Z

1

If you want change content of the page without refreshing page you have to use javascript exacly the AJAX. Read about https://www.w3schools.com/xml/ajax_intro.asp

answered Jan 23, 2018 at 20:47

Kuba Zabielski

1562 silver badges10 bronze badges

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