更换里面的PHP引号字符出现的所有(Replace all occurences of char i

我怎么能转换是这样的：

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

为此：

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

基本上，我想逃离“只”是引号里的括号。

编辑

我对新的正则表达式，因此，我尝试这样做：

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

但是，这只会逃避的第一次出现（。

编辑2

也许我应该提及我的字符串可以有这些括号已经逃脱了，在这种情况下，我不希望他们再次逃脱做。

顺便说一句，我需要工作两个双人和单引号，但我想我能做到这一点，只要我对他们的一个工作示例。

Answer 1:

这应该对单引号和双引号做到这一点：

$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;

产量

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

它是PHP> = 5.3。 如果你有一个较低的版本（> = 5）你有一个单独的函数来代替回调的匿名函数。

Answer 2:

您可以使用preg_replace_callback此;

// 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)');

什么已逃走串;

// 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:

鉴于串

$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";

但如果你是递归的粉丝

 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";

结果：

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

Answer 4:

这里是你如何可以用做preg_replace_callback()函数。

$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);

文章来源: Replace all occurences of char inside quotes on PHP