按国家筛选mysql中的数据

Question

我想从下拉菜单中按国家筛选数据：

<form name="filter_form" method="POST" action="display_data.php">
Select a country:
<select name="value">
    <option name="country" value="AU">Austria</option>
    <option name="country" value="BE">Belgium</option>
    <option name="country" value="BU">Bulgaria</option>
</select>
<input type="submit" name="btn_submit" value="Submit Filter" />
<?php
if($_POST['country'] == 'BE') {  

    $query = mysql_query("SELECT * FROM think WHERE Country='Belgium'");  
}  
elseif($_POST['country'] == 'AU') { 
 $query = mysql_query("SELECT * FROM think WHERE Country='Austria'");  
} else {  
    $query = mysql_query("SELECT * FROM think");  
}  
?>

代码不过滤任何数据。如果有人能帮忙的话，谢谢！

Answer 1

避免编写多余的代码。用以下代码更改您的代码：

<form name="filter_form" method="POST" action="display_data.php">
Select a country:
<select name="country">
    <option value="AU">Austria</option>
    <option value="BE">Belgium</option>
    <option value="BU">Bulgaria</option>
</select>
<input type="submit" name="btn_submit" value="Submit Filter" />

<?php
if(isset($_POST['country']))
{
    switch($_POST['country'])
    {
    case 'BE' : $countryName = 'Belgium';
                break;
    case 'AU' : $countryName = 'Austria';
                break;
    default : $countryName = '';
              break;
    }
    $where = '';
    if($countryName != '')
    {
        $where = "WHERE Country='".$countryName."'";
    }   
    $query = mysql_query("SELECT * FROM think ".$where."");  
}
?>

Answer 2

使用select标记时，服务器页将引用select标记的名称，而不是选项。

更改代码如下：

<select name="country">
    <option value="AU">Austria</option>
    <option value="BE">Belgium</option>
    <option value="BU">Bulgaria</option>
</select>

Answer 3

if($_POST['country'] == 'BE') {  

应该是

if($_POST['value'] == 'BE') {  

等等等等！