Combine php scripts with html bootstrap button

Question

I have some bootstrap button on page.html:

<button type="button" class="btn">Basic</button>

And some PHP script script.php.

Is there are any way to combine them, so then user clicks button, script executes and passes the result back to page.html?

I've tried to google this question - but there are advises to do it with Ajax. But I want to keep the code as simple as possible.

If you want to keep it basic just link to the script? <a href="script.php" class="btn">Basic</a> — ɴᴀᴛᴇ ᴅᴇᴠ
– ɴᴀᴛᴇ ᴅᴇᴠ, Commented Dec 14, 2017 at 12:28
It's basic html, as I understand. Any way to link it with bootstrap button? — Alex
– Alex, Commented Dec 14, 2017 at 12:35
What you want to do is exactly what AJAX is here for. Any HTML solution need to reload the page since PHP is a server side scripting and so you need the server to parse the PHP to render it. With AJAX you do it without reloading the entire page. — user2399035
– user2399035, Commented Dec 14, 2017 at 12:58
here is some simple ajax example w3schools.com/jquery/tryit.asp?filename=tryjquery_ajax_ajax — Kristian
– Kristian, Commented Dec 14, 2017 at 13:20

Nawaz Ghori · Accepted Answer · 2017-12-14 13:00:56Z

1

Just write button inside a <form> and you specify the desired target URL in the action attribute of <form>.

In case you have form you can use this

<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" rel="stylesheet" />
<form action="/script.php">
  <button type="submit" class="btn">Basic</button>
</form>

or use this for the basic one

<a href="/script" class="btn">Basic</a>

edited Dec 14, 2017 at 13:00

Nawaz Ghori

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answered Dec 14, 2017 at 12:48

Carlvic Lim

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Comments

Lajos Arpad · Accepted Answer · 2017-12-14 13:47:50Z

1

You need to do this with AJAX. Example of AJAX function:

function sendRequest(type, url, callback, async, params) {
    if (async !== false) async = true;
    var xhttp = new XMLHttpRequest();
    xhttp.onreadystatechange = callback;
    xhttp.open(type, url, async);
    xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xhttp.send(params);
}

and this is how you can call it:

sendRequest('POST', 'yoururl', function(response) {
    if (this.readyState === 4) {
        //Handle the response
    }
}, true);

answered Dec 14, 2017 at 13:47

Lajos Arpad

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