PHP Code to Post Variable to PHP Without Form or Controls

PHP Code [2] Ways to Send Variables to PHP Without Having to Have or Form or Any Input Controls on the Web Page

This first method of doing a php post without a form involves using FormData() constructor which essentially sends post data to php without having to have a form or any controls on the web page. Its an effective way to submit data to php tricking php into thinking its form data.

///////// JAVASCRIPT CODE ////////////////////// let fData = new FormData(); // use the append method of the FormData object // to add variables you want to send to php fData.append("anyNameOfVariableYouWish1", varVariableX); fData.append("anyNameOfVariableYouWish2", varVariableY); // open ajax connection and send variables to php xhttp.open("POST", "url-of-php-page.php", true); xhttp.send(fData); ///////// PHP CODE ////////////////////// if($_SERVER["REQUEST_METHOD"] == "POST") { if(isset($_POST["anyNameOfVariableYouWish1"])) { $var1 = $_POST["anyNameOfVariableYouWish1"]; } if(isset($_POST["anyNameOfVariableYouWish2"])) { $var2 = $_POST["anyNameOfVariableYouWish2"]; } }

This second method of sending post data/javascript variables to php is sort like a hybrid between a POST and a GET request, even though it is really a POST request.

///////// JAVASCRIPT CODE ////////////////////// // the string you send php is constucted like a get request let postData = "anyNameOfVariableYouWish1=varVariableX&anyNameOfVariableYouWish2=varVariableY&"; // open ajax connection and send variables to php xhttp.open("POST", "url-of-php-page.php", true); // were letting php know to treat what is coming is form data, even though its not xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); xhttp.send(fData); ///////// PHP CODE ////////////////////// if($_SERVER["REQUEST_METHOD"] == "POST") { if(isset($_POST["anyNameOfVariableYouWish1"])) { $var1 = $_POST["anyNameOfVariableYouWish1"]; } if(isset($_POST["anyNameOfVariableYouWish2"])) { $var2 = $_POST["anyNameOfVariableYouWish2"]; } }

Of course dont forget to include the rest of the required ajax request Javascript code ...

var xhttp = new XMLHttpRequest(); xhttp.onreadystatechange = function() { if (this.readyState == 4 && this.status == 200) { console.log(this.response); } };