不能發送數據從ajax調用到php獲取數據庫值

Question

我有一個問題，從數據庫檢索我的數據。

在我的Ajax調用中，我測試了我的表單值，並且它工作正常。

$(document).ready(function(){ 
$("form#form").submit(function(event) { 
    event.preventDefault(); 
    var color = $('#Color').val(); 
    var radio = $('input[name="filter_opties"]:checked').val(); 
    filter(color, radio); 
    $.ajax({ 
     type: "POST", 
     url: "db_querys.php", 
     data: {'color' : color, 'radio' : radio}, 
     success: function(data) { 
      alert(data); 
     } 
    }); 
}); 

但在我的db_querys.php中，我無法獲取顏色和無線電的值。

<?php 
$gekozenGemeente = $_POST['color']; 
$gekozenCategorie = $_POST['radio']; 

if($gekozenGemeente != null) 
    { 
    echo $gekozenGemeente . $gekozenCategorie; 
    } 
else 
    { 
    echo "<br> Values are null"; 
    } 
?> 

這是我的形式：

<form id = "form" action="#" method="post" > 
    <!----- Select Option Fields Starts Here -----> 
    <label class="heading">Selecteer uw gemeente:</label> 
    <br> 
    <select name="Color" id="Color"> 
     <option value="heemstede">Heemstede</option> 
     <option value="bloemendaal">Bloemendaal</option> 
    </select> 
    <br> 
    <!---- Radio Button Starts Here -----> 
    <label class="heading">Radio Buttons :</label><br> 
    <input type="radio" id="radio1" name="filter_opties" value="Betaald"><label for="radio1">Betaald</label><br/> 
    <input type="radio" id="radio2" name="filter_opties" value="Vergunning"><label for="radio2">Vergunning</label><br/> 
    <input type="radio" id="radio3" name="filter_opties" value="Blauwe zone"><label for="radio3">Blauwe zone</label><br/> 
    <br> 
    <input id= "submit" name="submit" type="submit" value="Get Selected Values" onclick="filter()"> 
</form> 

你們能解釋我缺少的是什麼？

編輯：增加了過濾功能。

function filter(color, radio){ 

    var locations = <?= json_encode($markers_json); ?>; 
    var locations2 = JSON.parse(locations); 


    var polygons = <?=json_encode($polygons_json);?>; 
    //var polygons2 = JSON.parse(polygons); 
    //document.getElementById("demo").innerHTML = polygons2; 


    initialize(locations2,polygons) 

     } 

Answer 1

$(document).ready(function(){ 
$("#form").submit(function(event) { 
    event.preventDefault(); 
    var color = $('#Color').val(); 
    var radio = $('input[name="filter_opties"]:checked').val(); 
    filter(color, radio); 
    $.ajax({ 
     type: "POST", 
     url: "db_querys.php", 
     data: {'color' : color, 'radio' : radio}, 
     success: function(data) { 
      alert(data); 
     } 
    }); 
});