如何檢查網址中是否存在參數？

Question

每當URL包含以p2開頭的任何參數時，我想輸出一條消息，例如在以下所有實例中：

example.com/?p2=hello

example.com/?p2foo=hello

example.com/?p2

example.com/?p2=

我試過了：

if (!empty($GET['p2'])) {
    echo "a parameter that starts with p2 , is showing in your url address";

} else {
    echo "not showing";
}

Answer 1

這應該涵蓋您所有的情況

$filtered = array_filter(array_keys($_GET), function($k) {
    return strpos($k, 'p2') === 0;
});

if ( !empty($filtered) ) {
    echo 'a paramater that starts with p2 , is showing in your url address';
}
else {
    echo 'not showing';
}

Answer 2

只需遍歷$_GET數組，並為匹配操作所需的鍵添加鍵p2開頭的條件。

foreach($_GET as $key=>$value){
    if (substr($key, 0, 2) === "p2"){
        // do your thing
        print $value;
    }
}

substr($key,0,2)從字符串中獲取前兩個字符

Answer 3

嘗試

if (isset($GET['p2'])) {
echo "a paramater that starts with p2 , is showing in your url address";

} else {
echo "not showing";
}

Answer 4

最快的方法是

if(preg_match("/(^|\|)p2/",implode("|",array_keys($_GET)))){
    //do stuff
}