javascript-更改Ajax Post参数并基于交替依赖下拉列表返回HTML

我有3个下拉菜单,其中包含在页面加载时填充的值

<select class='form-control' id='make' placeholder='Make:'>
<select class='form-control' id='model' placeholder='Model:'>
<select class='form-control' id='version' placeholder='Version:'>

我有一个函数,可以根据单击的下拉列表的值来更新未单击的“其他”下拉列表中的值-但对于每个下拉列表,我都会重复此函数3次

$('#model').change(function(){
    let selectedModel = $(this).val();
    $.ajax({
        url: 'PHP/dropdown.PHP',
        type: 'POST',
        data: {model: selectedModel},
        success:function(data)
        {  $('#make').html('');
           $('#version').html('');
            let makeJSON = JSON.parse(data)[0];
            let versionjsON = JSON.parse(data)[2];

            for (let i = 0; i < makeJSON.length; i++) {
                if (makeJSON[i].mMake!= '' && makeJSON[i].mMake!= null) {
                    $('#make').html($('#make').html() + '<option value="' + makeJSON[i].mMake + '">' + makeJSON[i].mMake + '</option>');
                }
            }

            for (let i = 0; i < versionjsON.length; i++) {
                if (versionjsON[i].mVersion != '' && versionjsON[i].mVersion != null) {
                    $('#version').html($('#version').html() + '<option value="' + versionjsON[i].mVersion + '">' + versionjsON[i].mVersion + '</option>');
                }
            }

        }
    });
});

PHP看起来像这样：

$model = $_REQUEST['model'];

$sqlupdateModel = "SELECT disTINCT mMake, mVersion FROM Cars WHERE mModel = '$model';
$stmtModel = sqlsrv_query( $conn, $sqlupdateModel);
if( $stmtModel === false)
{
    die( print_r( sqlsrv_errors(), true));
}
$updateModel = [];
while( $row = sqlsrv_fetch_array( $stmtModel, sqlSRV_FETCH_ASSOC)){
    $updateModel[] = $row;
}
echo json_encode(array($updateMake, $updateModel, $updateVersion));

…这一切都很好,
基本上,我正在寻找一种更简单的解决方案来重用该函数(包括JS和PHP),而不是重写3次！

根据我的尝试,

$('#make, #model, #version').change(function(){
let columnValue = $(this).val();
.......
data: {model: columnValue},
success:function(data)
{$(this).html(''); //this doesn't work obvIoUsly!

在这之后我被斯诺克

解决方法:

这应该适用于JS端,您将必须检查映射功能以获得响应

$('#make, #model, #version').change(function(ev){
    let selected = $(this).val();
    let id = ev.target.id;
    let data = {};
    data[id] = selected;
    $.ajax({
        url: 'PHP/dropdown.PHP',
        type: 'POST',
        data: data,
        success:function(data)
        {  
           let options = ['make', 'model', 'version']
           const response = {
             make: JSON.parse(data[0].map(make => make.mMake)),
             model: JSON.parse(data[1].map(make => make.mModel)),
             version: JSON.parse(data[2].map(make => make.mVersion))
           }

           options.filter(option => option !== id).forEach(option => setDropdown(option, response[option]));
        }
    });
});
function setDropdown(id, data) {
    const id = `#${id}`
    $(id).html('');
    for (let i = 0; i < data.length; i++) {
        if (data[i] != '' && data[i] != null) {
                $(id).html($(id).html() + '<option value="' + data[i] + '">' + data[i] + '</option>');
        }
    }
}