根据选择的项目数回显答案

【问题标题】：Echo based on number of items selected根据选择的项目数回显
【发布时间】：2020-03-07 14:58:38
【问题描述】：

<form action="exercise2.php" method="post"> 
    Type your name: <input type="text" name="name2" ><br>

    Toyota<input name="vehicle[]" type="checkbox" value="Toyota"><br>
    Bmw<input name="vehicle[]" type="checkbox" value="Bmw"><br>
    Audi<input name="vehicle[]" type="checkbox" value="Audi"><br>
    Subaru<input name="vehicle[]" type="checkbox" value="Subaru"><br>

    <input type="submit" name="submit">
</form>

** 事情是这样的；当我不选择任何车辆时，程序崩溃，当它应该出现在案例 1 中时。 **

我想呼应车辆，例如检查两三个选项时。

<?php

$name = $_POST['name2'];
$vehicles = $_POST['vehicle'];
$cont=0;

foreach($vehicles as $i){
    $cont++;
}

switch($cont){
    case 0:
        echo "You should choose some favs";
        break;
    case 1:
        echo "$name, you should look for more fav cars";
        break;

    case 4:
        echo "$name, I think you have too many fav cars";
        break;
}?>

【问题讨论】：

你应该输出你的$_POST变量的内容（例如var_dump），以便在你不检查任何东西时了解其中的内容。
$vehicles = $_POST['vehicle'];应该是 $vehicles = !empty($_POST['vehicle']) ？ $_POST['vehicle']:[0];
复选框中未选择任何内容时，程序崩溃。有什么想法吗？

标签： javascript php html switch-statement case

【解决方案1】：

您正在尝试迭代一个不存在的数组。未选中时，表单不会在复选框上发送任何内容，因此您需要更改 php：

<?php
# If you are echoing this back, it's good to sanitize it a bit
$name = htmlspecialchars($_POST['name2']);
# Make the default be an empty array
$vehicles = (!empty($_POST['vehicle']))? $_POST['vehicle'] : [];
$cont=0;
# This won't fail because it can iterate an empty array
foreach($vehicles as $i){
    $cont++;
}
# 0 should now work
switch($cont){
    case 0:
        echo "You should choose some favs";
        break;
    case 1:
        echo "$name, you should look for more fav cars";
        break;

    case 4:
        echo "$name, I think you have too many fav cars";
        break;
}

同样打开错误，它会告诉你它不能迭代empty。

【讨论】：