Question

我从数据库中检索数据时遇到了问题。

在我的Ajax调用中，我测试了我的表单值并且它有效。

$(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，我无法获得颜色和广播的值。

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

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

这是我的表格：

<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>

你们能解释我错过的东西吗？

编辑：添加了过滤功能。

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);
        }
    });
});

无法从ajax调用发送数据到php来获取数据库值

1 个答案: