如何使用选择和提交按钮重定向

问题描述：

我使用PHP来填充选项列表，然后想要使用提交并将我带到另一个页面...如何检查在下拉列表中选择了什么，提交并把我带到那个页面？如何使用选择和提交按钮重定向

form action="/test.php" method="post"> 
    <select> 
    <option value hidden>All</option> 
    <?php 

    $connection = mysqli_connect("127.0.0.1","root",""); 
    mysqli_select_db($connection,"test"); 
    $result = mysqli_query($connection,"select description from categories"); 
    while($row = mysqli_fetch_array($result)) 
    { 
    print "<option value = 
    ".$row['description'].".php".">".$row['description']."</option>"; 
    } 

    ?> 
    </select> 
    <button type = "submit" form= "menu" value="Submit">Filter</button>`

你需要给你的'

答

巧妙之处在于，当你通过$_POST发送选项，你需要给<select>元素一个name属性（未每个<option>）：

<form action="/test.php" method="post"> 
    <select name="description"> 
    <option value hidden>All</option> 
    <?php 
    $connection = mysqli_connect("127.0.0.1","root",""); 
    mysqli_select_db($connection,"test"); 
    $result = mysqli_query($connection,"select description from categories"); 
    while($row = mysqli_fetch_array($result)) { 
     print "<option name='some_Name' value = 
     ".$row['description'].".php".">".$row['description']."</option>"; 
    } 
    ?> 
    </select> 
    <button type="submit" form="menu" value="Submit">Filter</button> 
</form>

然后你可以检索与所选<option>：

$_POST['description']

希望这有助于！ :)

答

首先你需要给你的选项元素一个名字。

form action="/test.php" method="post"> 
    <select> 
    <option value hidden>All</option> 
    <?php 

    $connection = mysqli_connect("127.0.0.1","root",""); 
    mysqli_select_db($connection,"test"); 
    $result = mysqli_query($connection,"select description from categories"); 
    while($row = mysqli_fetch_array($result)) 
    { 
    print "<option name='some_Name' value = 
    ".$row['description'].".php".">".$row['description']."</option>"; 
    } 

    ?> 
    </select> 
    <button type = "submit" form= "menu" value="Submit">Filter</button>`

然后在test.php的文件，你可以使用访问选择：

$selectOption = $_POST['some_Name'];

如何使用选择和提交按钮重定向

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