Question

我有以下IF条件，但总是给我最后的$ genreid 6000

if ($URL == "/apps.html"
    or "/apps/all-genres.html"
    or "/apps/all-genres/top-paid-apps.html"
    or "/apps/all-genres/top-free-apps.html"
) {
    $genreid = "36";
}

if ($URL == "/apps/business.html"
    or "/apps/business/top-paid-apps.html"
    or "/apps/business/top-free-apps.html"
) {
    $genreid = "6000";
}

有人可以帮我纠正吗？

Answer 1

你的所有人都搞砸了。 OR“foo”将解决为真。

尝试这样的事情：

if (in_array($URL,array('test1','test2','test3') ))
{
    $genreid = "36";
}

Answer 2

非空（和非“0”）字符串的计算结果为“true”。您的代码看起来应该更像这样：

if ($URL == "/apps.html"
    or $URL == "/apps/all-genres.html"
    or $URL == "/apps/all-genres/top-paid-apps.html"
    or $URL == "/apps/all-genres/top-free-apps.html") {
        $genreid = "36";
}

Answer 3

if ($URL == "/apps/business.html"
            or "/apps/business/top-paid-apps.html"
            or "/apps/business/top-free-apps.html"
            )

这将检查以下其中一项是否真实：

$URL是"/apps/business.html"
"/apps/business/top-paid-apps.html"
"/apps/business/top-free-apps.html"

这两个字符串是真的，这不是or运算符的工作方式。依次将$URL与每个进行比较，或使用数组：

if(in_array($URL, ['/apps/business.html', …])) {
    …
}

但更不用说 - 为什么你不首先使用elseif？

IF子句仅给出最后IF条件的值

3 个答案: