当我尝试使用ajax post将值从JavaScript发送到PHP时，URL未加载

Question

I use

if(isset($_POST['category_drop'])){
    var_dump($_POST );
    echo $_POST['category_drop'];
}  {...} to check for the post condition in PHP.

Kindly provide me a solution.

$('.category_filter .dropdown-menu li a ').on('click', function(e) {

    e.preventDefault();
    var category = $(this).text();

    $.ajax({
        type: "POST",
        url: 'page_author.php',
        data: {
            category_drop: category
        },
        success: function(data) {
            // do something;
            alert(category);
        }
    });
});

---

I am getting this below value in an alert box , but my page is not getting loaded with the respective value,

array(1) { ["category_drop"]=> string(12) "Accomodation" }

Accomodation

Answer 1

The commented Code below shows how you may accomplish your goal:

    e.preventDefault();
    var category = $(this).text();

    $.ajax({
        type: "POST",
        url: 'page_author.php',
        data: {
            category_drop: category
        },
        success: function(data) {
            // PERHAPS IT'S NOT category BUT data THAT YOU SHOULD BE INTERESTED IN.
            // AND; YOU BETTER console.log() 
            // TO AVOID THOSE ANNOYING JAVASCRIPT ALERTS...
            // ALTHOUGH, YOU MAY STILL ALERT THE category IF YOU WISHED...
            console.log(data);
            //alert(category);
        }
    });

ON THE PHP SIDE OF THINGS:

<?php

    if(isset($_POST['category_drop'])){
        // ASSIGN THE $_POST['category_drop'] TO A VARIABLE $categoryDrop
        // TO BE USED IN YOUR MySQL DATABASE QUERIES...
        $categoryDrop = $_POST['category_drop'];

        // YOU CAN NOW USE $categoryDrop IN YOUR MySQL QUERIES.
        // AFTERWARDS, IF YOU WANT TO ONLY SEND BACK THE $categoryDrop
        // YOU CAN SIMPLY DO:
        //EITHER:
        die($categoryDrop);

    }