I have a silly question, how do I get the post value (
$.ajax({
type: "POST",
url: "b.php",
data: function (data) {
data.firstName = $('#firstName').val();
data.lastName = $('#lastName').val();
}
});
I guess it's not $_POST['firstName'] or $_POST['data']...
Thanks for helping
=.=" It's working now.
ajax:
$.ajax({
type: "POST",
url: "b.php",
data: {
firstName = $('#firstName').val(),
lastName = $('#lastName').val()
}
});
php
echo $_POST['firstName'] . "+" . $_POST['lastName'];
Your $_POST object will contain an array with the name 'data', as you are passing a JavaScript object.
I would recommend creating a JavaScript object and then using JSON.stringify() to pass to PHP. Try something like this:
let ajaxData = {
firstname: $('#firstName').val(),
lastName: $('#lastName').val()
};
let ajaxString = JSON.stringify(ajaxData);
$.ajax({
type: "POST",
url: "b.php",
data: { data: ajaxString }
});
Then, in your PHP controller, you would do something like this:
$ajaxData = $_POST['data'];
$userData = json_decode($ajaxData, true); //Decodes JSON data and creates associative array.
You can check the structure using vardump($userData);
$ajaxData and set it equal to $_POST['data']. This is a JSON-encoded string. Then, you would call the native PHP function json_decode() and pass in the variable, like this: $userData = json_decode($ajaxData, true); This creates an associative array with the data you gave it from the POST request.$_POST['firstName']. However, I still recommend encoding your data in JSON as this makes it much easier to work with larger datasets you may be sending to the backend. For small data sets like this, you can send a single object, but you will reach difficulty if you need to send more complex datasets.