PHP MYSQL动态选择框

Question

我正在尝试创建一个搜索框,其中从'box1'中选择的选项填充了'box2'可用的选项.两个框的选项都来自我的MYSQL数据库.我的问题是我不知道如何基于第一个查询执行查询而不刷新页面,这将是乏味和烦人的.

HTML/PHP

<form role="form" action="search.php" method="GET">
          <div class="col-md-3">
              <select class="form-control">
                  <?php
                  $result = mysqli_query($con,"SELECT `name` FROM school");
                  while($row = mysqli_fetch_array($result)) {
                    echo '<option name="'.$row['name'].'">'.$row['name'].' School</option>';
                  }

                  ?>
              </select>
          </div>
          <div class="col-md-3">
              <select class="form-control">
                  <?php
                  $result = mysqli_query($con,"SELECT * FROM products");
                  while($row = mysqli_fetch_array($result)) {
                    echo '<option name="'.$row['product'].'">'.$row['product'].'</option>';
                  }
                  mysqli_close($con);
                  ?>
              </select>
          </div>
          <button type="submit" class="btn btn-info">Search</button>
    </form>

我认为查询会像这样.AJAX可能是这个问题的解决方案,但我不确定如何使用AJAX执行此查询而无需刷新.

SELECT `product` FROM products WHERE `school` = [SCHOOL NAME FROM BOX 1]

提前致谢!

Answer 1

首先使用php创建no1选择菜单,如上所述.然后添加一个'change'eventListener,如:

$('#select1').change(createSelect2);

function createSelect2(){
    var option = $(this).find(':selected').val(),
    dataString = "option="+option;
    if(option != '')
    {
        $.ajax({
            type     : 'GET',
            url      : 'http://www.mitilini-trans.gr/demo/test.php',
            data     : dataString,
            dataType : 'JSON',
            cache: false,
            success  : function(data) {            
                var output = '<option value="">Select Sth</option>';

                $.each(data.data, function(i,s){
                    var newOption = s;

                    output += '<option value="' + newOption + '">' + newOption + '</option>';
                });

                $('#select2').empty().append(output);
            },
            error: function(){
                console.log("Ajax failed");
            }
        }); 
    }
    else
    {
        console.log("You have to select at least sth");
    }
}

现在,no2选择菜单根据选择的1选择选项有新选项.

和php文件:

<?php
header('Content-Type: application/json; charset=utf-8');
header('Access-Control-Allow-Origin: *');

if(isset($_GET['option']))
{
    $option = $_GET['option'];

    if($option == 1)
    {
        $data = array('Arsenal', 'Chelsea', 'Liverpool');
    }
    if($option == 2)
    {
        $data = array('Bayern', 'Dortmund', 'Gladbach');
    }       
    if($option == 3)
    {
        $data = array('Aek', 'Panathinaikos', 'Olympiakos');
    }

    $reply = array('data' => $data, 'error' => false);
}
else
{
    $reply = array('error' => true);
}

$json = json_encode($reply);    
echo $json; 
?>

当然,我使用了一些演示数据,但是您可以使用sql查询填充那里的$ data数组,并将它们作为带有正确标题的json发送.最后在第二个选择菜单中使用一些js:

$('#select2').change(selectSelect2);

function selectSelect2(){
    var option = $(this).find(':selected').val();
    if(option != '')
    {
        alert("You selected: "+option);
    }
    else
    {
        alert("You have to select at least sth");
    }
}

点击这里http://jsfiddle.net/g3Yqq/2/一个工作示例