Question

在此 PHP代码中，我想将静态值定义为多个变量 而将字符串连接到其父变量，如下面的代码

$Alpha = "Alpha-";
$Beta = "Beta-";
$Gamma = "Gamma-";
function func($vars, $number, $letter) {
    switch ($number) {
        case 1:
            foreach ($vars as $var) {
                $var .= "1";
            }
                break;
        case 2:
            foreach ($vars as $var) {
                $var .= "2";
            }
                break;
        case 3:
            foreach ($vars as $var) {
                $var .= "3";
            }
    }
    switch ($letter) {
        case "a":
            foreach ($vars as $var) {
                $var .= "A";
            }
                break;
        case "b":
            foreach ($vars as $var) {
                $var .= "B";
            }
                break;
        case "c":
            foreach ($vars as $var) {
                $var .= "C";
            }
    }
    return $var;
}
func([$Alpha, $Beta, $Gamma], 1, "a");
$Alpha .= "-haEND";
$Beta .= "-taEND";
$Gamma .= "-maEND";

echo $Alpha;
echo '<br>';
echo $Beta;
echo '<br>';
echo $Gamma;

我希望结果是

α-1A-haEND

β-1A-taEND

伽玛-1A-maEND

但我没有得到 1A

Answer 1

您正在尝试更新变量副本的值（在每个foreach中）。要更新该值，您必须在循环中通过引用传递它们。

然后，您没有使用返回值。

你可以尝试这样的事情。请注意foreach中的&，返回的使用以及list()从数组中提取数据。

$Alpha = "Alpha-";
$Beta = "Beta-";
$Gamma = "Gamma-";
function func($vars, $number, $letter) {
    switch ($number) {
        case 1:
            foreach ($vars as &$var) {$var .= "1"; }
                break;
        case 2:
            foreach ($vars as &$var) {$var .= "2"; }
                break;
        case 3:
            foreach ($vars as &$var) {$var .= "3"; } 
    }

    switch ($letter) {
        case "a":
            foreach ($vars as &$var) {$var .= "A"; }
                break;
        case "b":
            foreach ($vars as &$var) {$var .= "B"; }
                break;
        case "c":
            foreach ($vars as &$var) {$var .= "C"; }
    }
    return $vars;
}

$arr = func([$Alpha, $Beta, $Gamma], 1, "a");
list($Alpha, $Beta, $Gamma) = $arr ;

$Alpha .= "-haEND";
$Beta .= "-taEND";
$Gamma .= "-maEND";

echo $Alpha;
echo '<br>';
echo $Beta;
echo '<br>';
echo $Gamma;

输出：

Alpha-1A-haEND
Beta-1A-taEND
Gamma-1A-maEND

Answer 2

你在这里有很多重复的代码 - 没有必要使用六个单独的循环来为一些字符串添加数字和字母。

PHP的array_walk方法允许您将一个用户定义的回调应用于数组的每个成员，这可以大大减少这一点，并且如果您需要，还可以不需要添加其他的case语句展开支持的数字/字母集：

$Alpha = "Alpha-";
$Beta = "Beta-";
$Gamma = "Gamma-";

$values = [&$Alpha, &$Beta, &$Gamma];
$number = 1;
$letter = 'a';

array_walk($values, function (&$item) use ($number, $letter) {
    $item .= $number . strtoupper($letter);
});

$Alpha .= "-haEND";
$Beta .= "-taEND";
$Gamma .= "-maEND";

正如@Syscall所提到的，重要的区别在于元素通过引用传递给函数 - 您的原始代码是修改副本，而不是原件。

请参阅https://eval.in/949087

用等值和连接定义多个变量

2 个答案: