Question

我想从php变量中的“list_cust_name”中选择项目，通过在WHERE子句中传递该php变量，通过该sql查询获取另一个下拉列表“list_cust_city”中的值。这是我的代码..请帮助我.. < / p>

<td width="228">
    <label style="color:#000">Name </label>
    <?php
        $query_name = "SELECT DISTINCT cust_name FROM customer_db ORDER BY cust_name"; //Write a query
        $data_name = mysql_query($query_name);  //Execute the query
    ?>
    <select id="list_cust_name" name="list_cust_name">
        <?php
            while($fetch_options_name = mysql_fetch_assoc($data_name)) { //Loop all the options retrieved from the query
        ?> 
        <option value="<?php echo $fetch_options_name['cust_name']; ?>"><?php echo $fetch_options_name['cust_name']; ?></option>
        <?php
            }
        ?>
    </select>
</td>
<td width="250"> 
    <label style="color:#000">City </label>
    <?php
        $query_city = "SELECT DISTINCT cust_city FROM customer_db ORDER BY cust_city"; //Write a query
        $data_city = mysql_query($query_city);  //Execute the query
    ?>
    <select id="list_cust_city" name="list_cust_city">
        <?php
            while($fetch_options_city = mysql_fetch_assoc($data_city)) { //Loop all the options retrieved from the query
        ?> 
        <option value="<?php echo $fetch_options_city['cust_city']; ?>"><?php echo $fetch_options_city['cust_city']; ?></option>
        <?php
            }
        ?>
    </select>
</td>

Answer 1

您需要AJAX来电。

首先将JQuery脚本包含在<head>标记内的页面中，如：

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>

然后你需要一个事件监听器来更改你的第一个下拉列表，如下所示：

<script>
$('#list_cust_name').change(function(){
    $.ajax({
        url:'city.php',
        data:{cust_name:$( this ).val()},
        success: function( data ){
            $('#list_cust_city').html( data );
        }
    });
});
</script>

将上述脚本放在<body>标记下方。

现在，您要创建一个名为city.php的页面，如上面的ajax调用所示。您可以根据需要决定名称。

然后在该页面中，您可以获取传递的值cust_name，然后您可以执行查询。新的下拉选项可以生成如下。

<强> city.php

<?php
    $cust_name=$_GET['cust_name'];      //passed value of cust_name
    $query_city = "your query with the cust_name"; //Write a query
    $data_city = mysql_query($query_city);  //Execute the query
    while($fetch_options_city = mysql_fetch_assoc($data_city)) { //Loop all the options retrieved from the query
    ?> 
        <option value="<?php echo $fetch_options_city['cust_city']; ?>"><?php echo $fetch_options_city['cust_city']; ?></option>
    <?php
    }   
?>

完成此脚本后，数据将传递回AJAX调用，并根据我们的代码放在第二个下拉列表中。试一试

Answer 2

就像Jokey说的那样，你需要在这里执行ajax ......

Upto选择客户名称没有问题。

你必须像这样在jquery（ajax）中捕获list_cust_name的更改事件

JS文件：

$('#list_cust_name').change(function(){
    var cust_name = $('#list_cust_name').val();
    $.post( 
             "./php/getCityNames.php",
             { cust_name: cust_name },
             function(data) {
                $('#list_cust_city').html(data);
             }

          );
 });

PHP文件（名为getCityNames.php）：

<?php
$cust_name =  $_POST["cust_name"];  //getting the customer name sent from JS file
$query_city = "SELECT DISTINCT cust_city FROM customer_db WHERE cust_name = '$cust_name' ORDER BY cust_city"; //added where as per your requirement
    $data_city = mysql_query($query_city);  //Execute the query

foreach($categories as $category){  //Its my style
        echo "<option>";
        echo $data_city['cust_city'];
        echo "</option>";
    }

如何在php变量中获取选定的下拉项

2 个答案: