如何有效率地替換字串中的佔位符變數？

最佳化「替換字串中的佔位符變數」

人們可能會遇到各種場景，其中識別和替換字符字串中的佔位符變數至關重要。此程式碼片段示範了一個動態替換函數dynStr，它可以定位大括號({}) 括起來的鍵值對並相應地更新它們：

function dynStr($str,$vars) {
    preg_match_all("/\{[A-Z0-9_]+\}+/", $str, $matches);
    foreach($matches as $match_group) {
        foreach($match_group as $match) {
            $match = str_replace("}", "", $match);
            $match = str_replace("{", "", $match);
            $match = strtolower($match);
            $allowed = array_keys($vars);
            $match_up = strtoupper($match);
            $str = (in_array($match, $allowed)) ? str_replace("{".$match_up."}", $vars[$match], $str) : str_replace("{".$match_up."}", '', $str);
        }
    }
    return $str;
}

$variables = array("first_name" => "John","last_name" => "Smith","status" => "won");
$string = 'Dear {FIRST_NAME} {LAST_NAME}, we wanted to tell you that you {STATUS} the competition.';
echo dynStr($string,$variables); // Would output: 'Dear John Smith, we wanted to tell you that you won the competition.'

但是，正如作者強調的那樣，此實作存在冗餘數組存取並且顯得計算密集。我們可以透過消除正規表示式的使用並根據「vars」數組中的鍵值對實現簡單的字串替換來簡化這一過程：

$variables = array("first_name" => "John","last_name" => "Smith","status" => "won");
$string = 'Dear {FIRST_NAME} {LAST_NAME}, we wanted to tell you that you {STATUS} the competition.';

foreach($variables as $key => $value){
    $string = str_replace('{'.strtoupper($key).'}', $value, $string);
}

echo $string; // Dear John Smith, we wanted to tell you that you won the competition.

這種方法刪除了不必要的嵌套數組遍歷，並且簡化替換過程，從而產生更乾淨、更有效率的程式碼。

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