替換字符出現的所有引號內的PHP

Question

我怎麼能轉換是這樣的：

"hi (text here) and (other text)" come (again) 

要這樣：

"hi \(text here\) and \(other text\)" come (again) 

基本上，我想逃離「只」是括號裏面的引號。

編輯

我對新的正則表達式，因此，我嘗試這樣做：

$params = preg_replace('/(\'[^\(]*)[\(]+/', '$1\\\($2', $string); 

但這隻會逃避的第一次出現（

EDIT 2。

也許我應該提一下，我的字符串可能有這些括號al隨時可以逃脫，在這種情況下，我不想再逃脫他們。順便說一句，我需要它適用於雙引號和單引號，但我認爲只要我有其中一個例子的工作示例，我就可以做到這一點。

Answer 1

這應爲單引號和雙引號做到這一點：

$str = '"hi \(text here)" and (other text) come \'(again)\''; 

$str = preg_replace_callback('`("|\').*?\1`', function ($matches) { 
    return preg_replace('`(?<!\\\)[()]`', '\\\$0', $matches[0]); 
}, $str); 

echo $str; 

輸出

"hi \(text here\)" and (other text) come '\(again\)' 

這是PHP> = 5.3。如果您的版本較低（> = 5），則必須使用單獨的函數替換回調中的匿名函數。

Answer 2

您可以使用preg_replace_callback這個;

// outputs: hi \(text here\) and \(other text\) come (again) 
print preg_replace_callback('~"(.*?)"~', function($m) { 
    return '"'. preg_replace('~([\(\)])~', '\\\$1', $m[1]) .'"'; 
}, '"hi (text here) and (other text)" come (again)'); 

那麼已經逃過的字符串呢？

// outputs: hi \(text here\) and \(other text\) come (again) 
print preg_replace_callback('~"(.*?)"~', function($m) { 
    return '"'. preg_replace('~(?:\\\?)([\(\)])~', '\\\$1', $m[1]) .'"'; 
}, '"hi \(text here\) and (other text)" come (again)'); 

Answer 3

鑑於串

$str = '"hi (text here) and (other text)" come (again) "maybe (to)morrow?" (yes)'; 

迭代方法

for ($i=$q=0,$res='' ; $i<strlen($str) ; $i++) { 
    if ($str[$i] == '"') $q ^= 1; 
    elseif ($q && ($str[$i]=='(' || $str[$i]==')')) $res .= '\\'; 
    $res .= $str[$i]; 
} 

echo "$res\n"; 

但如果你是遞歸的粉絲

function rec($i, $n, $q) { 
    global $str; 
    if ($i >= $n) return ''; 
    $c = $str[$i]; 
    if ($c == '"') $q ^= 1; 
    elseif ($q && ($c == '(' || $c == ')')) $c = '\\' . $c; 
    return $c . rec($i+1, $n, $q); 
} 

echo rec(0, strlen($str), 0) . "\n"; 

結果：

"hi \(text here\) and \(other text\)" come (again) "maybe \(to\)morrow?" (yes) 

Answer 4

以下說明如何使用preg_replace_callback()函數執行此操作。

$str = '"hi (text here) and (other text)" come (again)'; 
$escaped = preg_replace_callback('~(["\']).*?\1~','normalizeParens',$str); 
// my original suggestion was '~(?<=").*?(?=")~' and I had to change it 
// due to your 2nd edit in your question. But there's still a chance that 
// both single and double quotes might exist in your string. 

function normalizeParens($m) { 
    return preg_replace('~(?<!\\\)[()]~','\\\$0',$m[0]); 
    // replace parens without preceding backshashes 
} 
var_dump($str); 
var_dump($escaped);