Question

如果在表单中设置了两个变量，我试图显示特定数据。

$retired = "";
$stolen = "";

if(isset($_POST['submit'])){
    $retired = $_GET['showretired'];
    $stolen = $_GET['showstolen'];
}

我试过了吼叫：

<?php }elseif(isset($_GET['showretired']) && (isset($_GET['showstolen'])){ ?>

哪个不起作用

<?php }elseif(isset($retired, $stolen)){

未设置时有效

<?php }elseif(isset($_GET['showretired'], $_GET['showstolen'])) {

这只显示退休部分。

我不确定最好的方法。

以下是表格：

    <form>
        <label>Show Retired Column </label><input type="checkbox" name="showretired"> 
        <label class="space">Show Stolen Column </label><input type="checkbox" name="showstolen">
        <input class="space" type="submit" name="submit" value="Refine">
    </form>

这就是我在同一个文件中使用变量的方式。

              <?php }elseif(isset($_GET['showretired'], $_GET['showstolen'])) {
                    ?>
                <tr>

                    <th>Retired</th>
                    <th>Stolen</th>
                </tr>
                <?php }else{ ?>

Answer 1

<?php

if(isset($_POST['submit'])){
    if(isset($_POST['showretired'])) {
        $retired = $_POST['showretired'];
    }
    if(isset($_POST['showstolen'])) {
        $stolen = $_POST['showstolen'];
    }

    if(isset($retired)){
    echo $retired;
}
if(isset($stolen)) {
    echo $stolen;
}
}



?>

<form action="" method="POST">
    <label>Show Retired Column </label><input type="checkbox" name="showretired" value="The retired field"> 
    <label class="space">Show Stolen Column </label><input type="checkbox" name="showstolen" value="The showstolen field">
    <input class="space" type="submit" name="submit" value="Refine">
</form>

Answer 2

表单上没有方法意味着它将默认为GET。更改条件并选中复选框。

<?php

if(isset($_GET['submit'])){
    $foo = isset($_GET['foo']) ? true : false;
    $bar = isset($_GET['bar']) ? true : false;
    var_dump($foo);
    var_dump($bar);
}

?>

<form>
    <input type="checkbox" name="foo">
    <input type="checkbox" name="bar">
    <input type="submit" name="submit">
</form>

Answer 3

if(isset($_GET['submit'])){

        if(isset($_GET['showretired']) &&  $_GET['showretired'] !="" && isset($_GET['showstolen']) && $_GET['showstolen'] !="" )
        {

            echo $retired."   ".$stolen;
        }


    }

和表格

<form>
    <label>Show Retired Column </label><input type="checkbox" name="showretired" value="showretired" checked="checked"> 
    <label class="space">Show Stolen Column </label><input type="checkbox" name="showstolen" value="showstolen" checked="checked">
    <input class="space" type="submit" name="submit" value="Refine">
</form>

希望它会有所帮助

无法获取多个Isset变量

3 个答案: