php，开关案件返回未定义的变量

发布于 2025-01-29 15:52:56 字数 922 浏览 0 评论 0原文

hy，

我在函数时得到了开关案例，但是当我调用它时，我会出现错误的变量，而我不知道为什么（我使用php 8）

private function getIDFromBrand() {
    switch ($this->brand) {
        case "Niky":
            $id = 1;
            break;
        case "Pumo":
            $id = 4;
            break;
        case "Coke":
            if ($this->typecoke== 0) {
                $id = 2;
            } else {
                if ($this->typecoke== 1) {
                    $id = 3;
                }
            }
            break;
        case "Tomato":
            $id = 5;
            break;
        case "Riles":
            $id = 6;
            break;
        case "TEST":
            $id = 7;
            break;
    }

    return $id; // Error Undefined variable $id
}

在我的功能顶部声明$ id时，

$ id = null

或

$ id = 0

开关未更新它，因此它将返回null或0，它将返回声明的值。

原文

Hy,

I got switch case inside function when but when i call it, i got error Undefined Variable and i don't know why (i use PHP 8)

private function getIDFromBrand() {
    switch ($this->brand) {
        case "Niky":
            $id = 1;
            break;
        case "Pumo":
            $id = 4;
            break;
        case "Coke":
            if ($this->typecoke== 0) {
                $id = 2;
            } else {
                if ($this->typecoke== 1) {
                    $id = 3;
                }
            }
            break;
        case "Tomato":
            $id = 5;
            break;
        case "Riles":
            $id = 6;
            break;
        case "TEST":
            $id = 7;
            break;
    }

    return $id; // Error Undefined variable $id
}

When i declare $id at the top of my function, like

$id = null

$id = 0

The switch doesn't update it, so it will return null or 0, it will return the declared value.

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雨的味道风的声音 2025-02-05 15:52:56

您的Switch语句没有默认分支，因此，如果$ this-＆gt; brand是“堆栈溢出”，则不会运行任何语句，并且代码> $ id 永远不会设置。

请参阅 switch> switch> switch> switch语句的PHP手册：

特殊情况是默认情况案例。此案例与其他情况不匹配的任何内容匹配。

同样，如果$ this-＆gt; brand是“可口可乐”，但是$ this- ty- typecoke是42，它将与该分支中的任何条件都不匹配。

switch ($this->brand) {
    case "Niky":
        $id = 1;
        break;
    case "Pumo":
        $id = 4;
        break;
    case "Coke":
        if ($this->typecoke== 0) {
            $id = 2;
        } elseif ($this->typecoke== 1) {
            $id = 3;
        } else {
            $id = -1; // WAS PREVIOUSLY NOT SET
        }
        break;
    case "Tomato":
        $id = 5;
        break;
    case "Riles":
        $id = 6;
        break;
    case "TEST":
        $id = 7;
        break;
    default:
        $id = -1; // WAS PREVIOUSLY NOT SET
        break;
}

Your switch statement has no default branch, so if $this->brand is, say, "Stack Overflow", it will not run any of the statements, and $id will never be set.

See the PHP manual for the switch statement:

A special case is the default case. This case matches anything that wasn't matched by the other cases.

Similarly, if $this->brand is "Coke", but $this->typecoke is, say, 42, it will not match either of the conditions in that branch.

switch ($this->brand) {
    case "Niky":
        $id = 1;
        break;
    case "Pumo":
        $id = 4;
        break;
    case "Coke":
        if ($this->typecoke== 0) {
            $id = 2;
        } elseif ($this->typecoke== 1) {
            $id = 3;
        } else {
            $id = -1; // WAS PREVIOUSLY NOT SET
        }
        break;
    case "Tomato":
        $id = 5;
        break;
    case "Riles":
        $id = 6;
        break;
    case "TEST":
        $id = 7;
        break;
    default:
        $id = -1; // WAS PREVIOUSLY NOT SET
        break;
}

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