从ajax

Question

今天刚刚学习ajax并且在某处。尝试链接我的PHP链接以使用ajax，以便显示具有该链接ID的数据而不刷新页面。以下是我尝试过的代码。

的index.php

    <div class="page">
    <ul>
    <?php
    $sql = <<<EOF
    SELECT * FROM addcategory ;
    EOF;
    $ret = $db->query($sql); 
    While ($row = $ret->fetchArray(SQLITE3_ASSOC)) {
            $catname = $row['catname'];
            $catid = $row['catid'];

            echo "<li><a href='index.php?category_id=$catid'>$catname</a></li>";
    }
    ?>
    </ul>
    </div>
    <div class="cat_question">response displays here</div>

php脚本

    require_once ("db.php");
    $db = new MyDb();

    $catid = (int)$_GET['category_id'];
    $csql = <<<EOF
    SELECT * FROM questions WHERE category_id = {$catid};

    $cret = $db->query($csql);
    While ($crow = $cret->fetchArray(SQLITE3_ASSOC)) {
    $catquestion = $crow['question'];
    $catans = $crow['answer'];

    echo "<div>$catquestion</div><p>$catans</p>";
   }

的JavaScript

   $('.page a).click(function(e) {
            e.preventDefault();
            If (window.XMLHttpRequest) {
            var xhttp = new XMLHTtpRequest();
            } else {
                    xhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xhttp.onreadystatechange = function() {
                    If (xhttp.readyState == 4 && xhttp.status == 200) {
            var response = xhttp.responseText;
            $('.cat_question').html(response);
                    }
            };
            xhttp.open("GET", "category.php?category_id=$catid", true);
            xhttp.send();
    }); 

我知道如果我将xhttp URL更改为category.php？category_id = 1或任何其工作的ID，但这意味着必须为每个标签编写ajax请求。请问解决这个问题的方法是什么。提前谢谢。