Question

这很简单，但我无法理解逻辑。

有一个简单的在线表格。希望能够在提交后调整变量。

如果有人能够阐明这一点，那将是非常感激的！

$chocolatebar = $HTTP_POST_VARS['chocolatebar'];

if $chocolatebar = "a"  then we want to change the value to "snickers";

if $chocolatebar = "b"  then we want to change the value to "mars";

if $chocolatebar = "c"  then we want to change the value to "cherry ripe";

Answer 1

$chocolatebar = isset($_POST['chocolatebar']) ? $_POST['chocolatebar'] : '';
if ($chocolatebar == "a")  $chocolatebar = "snickers";
elseif ($chocolatebar == "b")  $chocolatebar = "mars";
elseif ($chocolatebar == "c")  $chocolatebar = "cherry ripe";

Answer 2

$chocolatebar = $POST['chocolatebar'];

if($chocolatebar =="a")
{ 
     $chocolatebar="snickers";
}
else if($chocolatebar =="b")
{ 
    $chocolatebar="mars";
}
else
{ 
    $chocolatebar="cherry ripe";
}

Answer 3

首先不要使用$HTTP_POST_VARS，不推荐使用此功能$HTTP_POST_VARS

另外，我猜你应该使用switch而不是if-else。

<?php
$chocolatebar = ($_POST['chocolatebar'] != '') ? $_POST['chocolatebar'] : '';

switch($chocolatebar){
    case 'a':
        $chocolatebar = 'snickers';
        break;

    case 'b':
        $chocolatebar = 'mars';
        break;

    case 'c':
        $chocolatebar = 'cherry ripe';
        break;

    default:
        $chocolatebar = '';
        break;
}

echo $chocolatebar;

Answer 4

在这种情况下，当您有多个选项时，最好使用switch，然后使用多个if-else。

if(isset($_POST['chocolatebar'])){
    $chocolatebar = $_POST['chocolatebar'];
    switch ($chocolatebar) {
        case "a":
            $chocolatebar = "snickers";
            break;
        case "b":
            $chocolatebar = "mars";
            break;
        case "c":
            $chocolatebar = "cherry ripe";
            break;
    }
}

Answer 5

$chocolatebar = $_POST['chocolatebar'];
switch($chocolaterbar){
    case "a": 
        $chocolatebar = "snickers";
    break;

    case "b": 
        $chocolatebar = "mars";
    break;

    case "c": 
        $chocolatebar = "cherry ripe";
    break;

    default:
        $chocolatebar = "unknown";
    break;
}

我的目的是使用开关。或者另一个智能选项是使用数组..

$all_chocolates = array("a"=>"snickers","b"=>"mars","c"=>"cherry ripe");
$chocolatebar = $_POST['chocolatebar'];
$chocolatebar = isset($all_chocolates[$chocolatebar]) ? $all_chocolates[$chocolatebar] : "unknown";

Answer 6

如果您想更改更多值，也可以使用开关案例。

Answer 7

为方便起见，您需要在键值之间进行映射。

$valueMap = array(
    'a' => 'snickers',
    'b' => 'mars',
    'c' => 'cherry ripe'
);

// The '@' sign prevents error if characters other than
// 'a', 'b', 'c' is provided, $value will be NULL in
// those cases.
$value = @$valueMap[$_POST['chocolatebar']];

需要简单的PHP帮助

7 个答案: