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

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

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);

    }  

PHP:

 $result = array(
 'success' => false,
 );
 if(isset($_POST['category_drop'])){
    $result['category'] = $_POST['category_drop'];
    $result['success'] = true;
    // you can pass in result array whatever you need
 }
echo json_encode($result);
exit;

Js:

$('.category_filter .dropdown-menu li a ').on('click', function(e) {
e.preventDefault();
var category = $(this).text();
$.ajax({
    type: "POST",
    url: 'page_author.php',
    dataType: 'application/json',
    data: {
        category_drop: category
    },
    success: function(data) {
       if(data.success){
        //do something here   
        console.log(data.category);
        console.log(data);
        }
    }
 });
});

Always use good practices :)

As per my understanding of your question you may follow this.

PHP

     $result = array(
        'category' => null,
        'success' => false
    );
    if (isset($_POST['category_drop'])) {
        $result['category'] = $_POST['category_drop'];
        $result['success'] = true;
    }
    echo json_encode($result);
    exit;

JS

        $('.category_filter .dropdown-menu li a ').on('click', function (e) {
            e.preventDefault();
            var category = $(this).text();
            $.ajax({
                type: "POST",
                url: 'page_author.php',
                dataType: 'application/json',
                data: {
                    category_drop: category
                },
                success: function (data) {
                    if (data.success) {
                       // append here the category value
                       // to your respective TAG
                       $("YOUR-RESPECTIVE TAG").text(data.category);

                       //OR

                       $("YOUR-RESPECTIVE TAG").text(data['category']);   

                    }
                }
            });
        });