避免在替换字符串中找到替换占位符

Question

假设我有这个字符串：

"my string ? other string ?"

• 我想替换第一个“？”使用"first param ?"（注意文本中的占位符？）
• 和第二个"second param".

如果我做一个 preg_replace 我得到这个：

my string first param second param other string ?
          ^^^^^^^^^^^^^^^^^^^^^^^^              ^
                WRONG                    NOT REPLACED

基本上，因为第一个替换也有占位符，preg_replace 愚蠢到替换那个占位符，而不是最后的真正的第二个。

使用preg_replace 编码：

$search = ["?", "?"];
$params = ["first param ?", "second param"];
$query ="first text ? other text ?";

//> Marker in the query are ?, so I create the array to preg_replace
$search = array_fill(0,count($params),'/\?/');
$query = preg_replace(
    $search,                // a list of ?
    $params,                // escaped values
    $query,                 // from query
    1                       // replace only 1 time
);
//output: first text first param second param other text ?

关于如何避免在替换中搜索占位符的任何提示？

带有preg_replace的实时代码：http://sandbox.onlinephpfunctions.com/code/e705ba454d030103344bc826e0fe0bf42d5b7b90

不适用于str_replace

$search = ["?", "?"];
$params = ["first param ?", "second param"];
$query ="first text ? other text ?";

$query = str_replace ($search, $params, $query);
echo $query;

// output: first text first param second param other text first param second param

带有 str_replace 的实时代码： http://sandbox.onlinephpfunctions.com/code/dc259325411ee42de759f145eac78b339f329f74

异常输出

给定：

$search = ["?", "?"];
$params = ["first param ?", "second param"];
$query ="first text ? other text ?";

预期的输出是：

first text first param ? other text second param
           ^^^^^^^^^^^^^            ^^^^^^^^^^^^
         first placeholder         second placeholder

带有 3 个参数的异常输出

$search = ["?", "?", "?"];
$params = ["first param", "second param ?", "third param"];
$query ="first text ? other text ? other chunk ?";

预期的输出是：

first text first param other text  second param ? other chunk third param
           ^^^^^^^^^^^^^            ^^^^^^^^^^^^              ^^^^^^^^^
         first placeholder         second placeholder         third placeholder

我的自定义解决方案

我想出了一个使用preg_split 的可能解决方案，但老实说，这太老套了，肯定有更好的解决方案：

 $parts = preg_split('/(\?)/', $query, -1, PREG_SPLIT_DELIM_CAPTURE);

 foreach($parts as $k=>&$v) {
        // if is odd, then it's a placeholder
        if ($k%2 == 1)
            $v = $params[$k/2];  // replace placeholder with a param
 }

 $query = implode('',$parts);

Answer 1

任何自定义替换逻辑都应该用preg_replace_callback实现，例如：

$params = ["first param", "second param ?", "third param"];
$query ="first text ? other text ? other chunk ?";

echo preg_replace_callback('/\?/', function($m) use (&$params) {
    return array_shift($params);
}, $query);

直播代码：http://sandbox.onlinephpfunctions.com/code/33f4804b49103e54e8070e8d9959ec9642930857

Answer 2

这仅适用于 2 个占位符。

$search = [
    "/^([^?]*)\?/",     # matches everything that is not a question mark in group 1, then a question mark
    "/^(.*)\?/"         # matches everything until last question mark in group 1, then a question mark
];
$params = ["$1first param ?", "$1second param"];

$query = "first text ? other text ?";

$query = preg_replace($search, $params, $query);

echo $query;

输出：

first text first param ? other text second param

Answer 3

这可以使用while() 循环并检查strpos() 来完成，但explode() 是解决此问题的另一种方法：

$params = ['first param ?', 'second param'];
$query = 'my string ? other string ?';
$str = '';

foreach(explode('?', $query, count($params) + 1) as $token) {
    $str .= $token . array_shift($params);
}

echo $str;

见live demo here

Answer 4

试试：

$s = "my string ? other string ?";
$s = preg_replace('/\?/', "first ?", $s, 1);
$s = preg_replace('/((?:^.*?\?)?.*?)\?/', "$1second text", $s);
echo $s;

基于How to skip first regex match?