javascript - 使用 AJAX 按值对多个类别进行排序

Question

我正在尝试对使用 ajax get 请求拉取的数据进行排序，我如何获取所有已检查的类别而不是仅按下的一个，如果两者都已检查，则最好同时获得 2/3？

目前我在数据库中的“类别”下插入了一个数字，它们应该通过这些数字进行排序？

如何获取网址中的两个查询？

示例:http://xch07.wi2.sde.dk/sandbox/SQL/

HTML

<form>
    <input class="kategorier" type="checkbox" value="1">Kat 1<br>
    <input class="kategorier" type="checkbox" value="2">Kat 2 <br>
    <input class="kategorier" type="checkbox" value="3">Kat 3
</form>
<br>
<div id="txtHint">Person info will be listed here...
</div>

JS/AJAX

var kategorier = document.getElementsByClassName("kategorier");
var kategorierLength = kategorier.length;

for(let i=0; i < kategorierLength ;i++){
  kategorier[i].addEventListener('click', function(){
    showUser(this.value);
  });
};

function showUser(str) {
    if (str == "") {
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else { 
        if (window.XMLHttpRequest) {
            // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("txtHint").innerHTML = this.responseText;
            }
        };
        xmlhttp.open("GET","index.php?page=getuser&kategori="+str ,true);
        xmlhttp.send();
    }
}

PHP

$kategori = intval($_GET['kategori']);
mysqli_select_db($dbCon,"ajax_demo");
$sql="SELECT * FROM db_test WHERE kategori = $kategori ";
$result = $dbCon->query($sql);

echo "<table>
<tr>
<th>Id</th>
<th>Navn</th>
<th>Efternavn</th>
<th>email</th>
<th>Kategori</th>
</tr>";
while($row = $result->fetch_object()) {
    $id = $row->id;
    $navn = $row->navn;
    $efternavn = $row->efternavn;
    $email = $row->email;
    $kategori = $row->kategori;

    echo "<tr>";
    echo "<td>" . $id . "</td>";
    echo "<td>" . $navn . "</td>";
    echo "<td>" . $efternavn . "</td>";
    echo "<td>" . $email . "</td>";
    echo "<td>" . $kategori . "</td>";
    echo "</tr>";
}
echo "</table>";

Answer 1

<script type="text/javascript">
window.onload = function () {

    var kategorier = document.getElementsByClassName("kategorier");
    var kategorierLength = kategorier.length;

    for(var i=0; i < kategorierLength ;i++){
        kategorier[i].addEventListener('click', function(){

            var checked = '';

            for (var k = 0; k < kategorierLength; k++) {

                if (checked) {
                    var separator = ',';
                } else {
                    var separator = '';
                }

                if (kategorier[k].checked) {
                    checked += separator+kategorier[k].value;
                }
            }
            alert(checked);
            if (checked) {
                showUser(checked);
            }
        });
    }
}

之后，您需要更改查询并使用 IN 语句，如下所示。

$kategori = $_GET['kategori'];
mysqli_select_db($dbCon,"ajax_demo");
$sql="SELECT * FROM db_test WHERE kategori IN ($kategori)";
$result = $dbCon->query($sql);