CODEWITHSUNDEEP
phpajaxwordpress
Kumar
Kumar

Reputation: 60

AJAX Using PHP File in Wordpress Plugin

I am following a tutorial on W3Schools for AJAX PHP. https://www.w3schools.com/php/php_ajax_php.asp

Here is the twist: I am doing this in wordpress. The following is a method that DOES work, but it is less than ideal.

1) Create the following gethint.php in the root directory.

<?php
// Array with names
$a[] = "Anna";
$a[] = "Brittany";
...
$a[] = "Vicky";

// get the q parameter from URL
$q = $_REQUEST["q"];

$hint = "";

// lookup all hints from array if $q is different from "" 
if ($q !== "") {
    $q = strtolower($q);
    $len=strlen($q);
    foreach($a as $name) {
        if (stristr($q, substr($name, 0, $len))) {
            if ($hint === "") {
                $hint = $name;
            } else {
                $hint .= ", $name";
            }
        }
    }
}

// Output "no suggestion" if no hint was found or output correct values 
echo $hint === "" ? "no suggestion" : $hint;

2) Using the CSS & Javascript toolbox plugin, add this code to the header:

<script>
function showHint(str) {
    if (str.length == 0) { 
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else {
        var xmlhttp = new XMLHttpRequest();
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("txtHint").innerHTML = this.responseText;
            }
        };
        xmlhttp.open("GET", "/gethint.php?q=" + str, true);
        xmlhttp.send();
    }
}
</script>

3) Create a page with the following code (in plain text):

<p><b>Start typing a name in the input field below:</b></p>
<form> 
First name: <input type="text" onkeyup="showHint(this.value)">
</form>
<p>Suggestions: <span id="txtHint"></span></p>

While this works, having to create a php file and adding to the root directory seems like bad practice. It would be better to have this php file stored in the plugins directory. However that causes this line of the header script to fail as 404:

xmlhttp.open("GET", "/gethint.php?q=" + str, true);

Simply changing the relative path won't work, because theoretically, different users can have their plugin folder in different locations.

I figure I should be using the wp_ajax_ and wp_ajax_nopriv_ hooks, but my attempts I have failed, so I am probably doing it wrong. Please help.

Upvotes: 1

Views: 1617

Answers (1)

Dexter Bengil
Dexter Bengil

Reputation: 6625

Doing ajax in WordPress should all be sent to /wp-admin/admin-ajax.php, to do that, in your plugin's main file or the index.php file, register your ajax action like this:

// let's do the ajax thing here
add_action( 'wp_ajax_theNameOfMyCustomAjax', 'theFunctionThatMyAjaxWillCall' );

function theFunctionThatMyAjaxWillCall() {
  // include your ajax file here, in this case
  // I assumed that we placed the gethint.php in 
  // /wp-content/plugins/yourpluginname/gethint.php
  include( plugin_dir_path( __FILE__ ).'gethint.php' );

  // don't forget to add "die;" every time you return a response to your ajax
  //Example: echo $hint ?? "no suggestion"; die;

  // or you can add the termination code right here
  die; // this will prevent the ajax response to have 0 in the end.

}

Now, in your javascript, instead of calling the filename of your ajax file, you can now use the global ajaxurl javascript variable like this:

xmlhttp.open("GET", ajaxurl+"?action=theNameOfMyCustomAjax&q=" + str, true);

Upvotes: 1

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