Advertisement
Advertisement


How to call a PHP function on the click of a button


Question

I have created a page called functioncalling.php that contains two buttons, Submit and Insert. As a beginner in PHP, I want to test which function is executed when a button gets clicked. I want the output to come on the same page. So I created two functions, one for each button. The source code for functioncalling.php is as follows:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
    <body>
        <form action="functioncalling.php">
            <input type="text" name="txt" />
            <input type="submit" name="insert" value="insert" onclick="insert()" />
            <input type="submit" name="select" value="select" onclick="select()" />
        </form>

        <?php
            function select(){
               echo "The select function is called.";
            }
            function insert(){
               echo "The insert function is called.";
            }
        ?>

The problem here is that I don't get any output after any of the buttons are clicked.

Where exactly am I going wrong?

2020/07/24
1
152
7/24/2020 9:32:53 PM

Accepted Answer

Yes, you need Ajax here. Please refer to the code below for more details.

 

Change your markup like this

<input type="submit" class="button" name="insert" value="insert" />
<input type="submit" class="button" name="select" value="select" />

 

jQuery:

$(document).ready(function(){
    $('.button').click(function(){
        var clickBtnValue = $(this).val();
        var ajaxurl = 'ajax.php',
        data =  {'action': clickBtnValue};
        $.post(ajaxurl, data, function (response) {
            // Response div goes here.
            alert("action performed successfully");
        });
    });
});

In ajax.php

<?php
    if (isset($_POST['action'])) {
        switch ($_POST['action']) {
            case 'insert':
                insert();
                break;
            case 'select':
                select();
                break;
        }
    }

    function select() {
        echo "The select function is called.";
        exit;
    }

    function insert() {
        echo "The insert function is called.";
        exit;
    }
?>
2019/07/11
94
7/11/2019 6:46:59 PM

Button clicks are client side whereas PHP is executed server side, but you can achieve this by using Ajax:

$('.button').click(function() {
  $.ajax({
    type: "POST",
    url: "some.php",
    data: { name: "John" }
  }).done(function( msg ) {
    alert( "Data Saved: " + msg );
  });
});

In your PHP file:

<?php
    function abc($name){
        // Your code here
    }
?>
2019/07/11

You should make the button call the same page and in a PHP section check if the button was pressed:

HTML:

<form action="theSamePage.php" method="post">
    <input type="submit" name="someAction" value="GO" />
</form>

PHP:

<?php
    if($_SERVER['REQUEST_METHOD'] == "POST" and isset($_POST['someAction']))
    {
        func();
    }
    function func()
    {
        // do stuff     
    }
?>
2016/12/06

You cannot call PHP functions like clicking on a button from HTML. Because HTML is on the client side while PHP runs server side.

Either you need to use some Ajax or do it like as in the code snippet below.

<?php
    if ($_GET) {
        if (isset($_GET['insert'])) {
            insert();
        } elseif (isset($_GET['select'])) {
            select();
        }
    }

    function select()
    {
       echo "The select function is called.";
    }

    function insert()
    {
       echo "The insert function is called.";
    }
?>

You have to post your form data and then check for appropriate button that is clicked.

2020/04/13

To show $message in your input:

<?php
    if(isset($_POST['insert'])){
        $message= "The insert function is called.";
    }
    if(isset($_POST['select'])){
        $message="The select function is called.";
    }
?>


<form  method="post">
    <input type="text" name="txt" value="<?php if(isset($message)){ echo $message;}?>" >
    <input type="submit" name="insert" value="insert">
    <input type="submit" name="select" value="select" >
</form>

To use functioncalling.php as an external file you have to include it somehow in your HTML document.

2019/07/11

Try this:

if($_POST['select'] and $_SERVER['REQUEST_METHOD'] == "POST"){
    select();
}
if($_POST['insert'] and $_SERVER['REQUEST_METHOD'] == "POST"){
    insert();
}
2014/03/29