使用preg_replace进行可变长度的遮罩 [英] variable length masking with preg_replace
问题描述
我正在使用preg_replace_callback()屏蔽字符串中单引号(包括两端)之间的所有字符.但我只想在可能的情况下使用preg_replace()，但无法弄清楚.任何帮助将不胜感激.
这是我使用的preg_replace_callback()所产生的正确输出的结果:
function maskCallback( $matches ) {
return str_repeat( '-', strlen( $matches[0] ) );
}
function maskString( $str ) {
return preg_replace_callback( "('.*?')", 'maskCallback', $str );
}
$str = "TEST 'replace''me' ok 'me too'";
echo $str,"\n";
echo $maskString( $str ),"\n";
输出为:
TEST 'replace''me' ok 'me too'
TEST ------------- ok --------
我尝试使用:
preg_replace( "/('.*?')/", '-', $str );
但是破折号被消耗掉了，例如:
TEST -- ok -
我尝试过的所有其他方法也不起作用. (我显然不是正则表达式专家.)这可能吗?如果可以，怎么办?
是的，您可以做到(假设引号是平衡的)示例:
$str = "TEST 'replace''me' ok 'me too'";
$pattern = "~[^'](?=[^']*(?:'[^']*'[^']*)*+'[^']*\z)|'~";
$result = preg_replace($pattern, '-', $str);
这个想法是:如果字符是一个引号或后面跟一个奇数的引号，则可以替换该字符.
不带引号:
$pattern = "~(?:(?!\A)\G|(?:(?!\G)|\A)'\K)[^']~";
$result = preg_replace($pattern, '-', $str);
仅当与前一个匹配连续的字符时(换句话说，当它在最后一个匹配之后紧接时)，或者当它的前面带有与该前一个匹配不连续的引号时，该模式才匹配一个字符. /p>
\G是最后一场比赛之后的位置(开始时是字符串的开头)
图案细节:
~ # pattern delimiter
(?: # non capturing group: describe the two possibilities
# before the target character
(?!\A)\G # at the position in the string after the last match
# the negative lookbehind ensure that this is not the start
# of the string
| # OR
(?: # (to ensure that the quote is a not a closing quote)
(?!\G) # not contiguous to a precedent match
| # OR
\A # at the start of the string
)
' # the opening quote
\K # remove all precedent characters from the match result
# (only one quote here)
)
[^'] # a character that is not a quote
~
请注意，由于模式不匹配右引号，因此以下非引号的字符将无法匹配，因为没有先例匹配.
(*SKIP)(*FAIL)方式:
您可以使用回溯控制动词(*SKIP)和(*FAIL)(可以是缩短为(*F)).
$pattern = "~(?:(?!\A)\G|')(?:'(*SKIP)(*F)|\K[^'])~";
$result = preg_replace($pattern, '-', $str);
由于该模式在每个右引号上均失败，因此以下字符将不匹配，直到下一个右引号.
像这样写的模式可能更有效:
$pattern = "~(?:\G(?!\A)(?:'(*SKIP)(*F))?|'\K)[^']~";
(您也可以使用(*PRUNE)代替(*SKIP).)
I am masking all characters between single quotes (inclusively) within a string using preg_replace_callback(). But I would like to only use preg_replace() if possible, but haven't been able to figure it out. Any help would be appreciated.
This is what I have using preg_replace_callback() which produces the correct output:
function maskCallback( $matches ) {
return str_repeat( '-', strlen( $matches[0] ) );
}
function maskString( $str ) {
return preg_replace_callback( "('.*?')", 'maskCallback', $str );
}
$str = "TEST 'replace''me' ok 'me too'";
echo $str,"\n";
echo $maskString( $str ),"\n";
Output is:
TEST 'replace''me' ok 'me too'
TEST ------------- ok --------
I have tried using:
preg_replace( "/('.*?')/", '-', $str );
but the dashes get consumed, e.g.:
TEST -- ok -
Everything else I have tried doesn't work either. (I'm obviously not a regex expert.) Is this possible to do? If so, how?
Yes you can do it, (assuming that quotes are balanced) example:
$str = "TEST 'replace''me' ok 'me too'";
$pattern = "~[^'](?=[^']*(?:'[^']*'[^']*)*+'[^']*\z)|'~";
$result = preg_replace($pattern, '-', $str);
The idea is: you can replace a character if it is a quote or if it is followed by an odd number of quotes.
Without quotes:
$pattern = "~(?:(?!\A)\G|(?:(?!\G)|\A)'\K)[^']~";
$result = preg_replace($pattern, '-', $str);
The pattern will match a character only when it is contiguous to a precedent match (In other words, when it is immediately after the last match) or when it is preceded by a quote that is not contiguous to the precedent match.
\G is the position after the last match (at the beginning it is the start of the string)
pattern details:
~ # pattern delimiter
(?: # non capturing group: describe the two possibilities
# before the target character
(?!\A)\G # at the position in the string after the last match
# the negative lookbehind ensure that this is not the start
# of the string
| # OR
(?: # (to ensure that the quote is a not a closing quote)
(?!\G) # not contiguous to a precedent match
| # OR
\A # at the start of the string
)
' # the opening quote
\K # remove all precedent characters from the match result
# (only one quote here)
)
[^'] # a character that is not a quote
~
Note that since the closing quote is not matched by the pattern, the following characters that are not quotes can't be matched because there is no precedent match.
EDIT:
The (*SKIP)(*FAIL) way:
Instead of testing if a single quote is not a closing quote with (?:(?!\G)|\A)' like in the precedent pattern, you can break the match contiguity on closing quotes using the backtracking control verbs (*SKIP) and (*FAIL) (That can be shorten to (*F)).
$pattern = "~(?:(?!\A)\G|')(?:'(*SKIP)(*F)|\K[^'])~";
$result = preg_replace($pattern, '-', $str);
Since the pattern fails on each closing quotes, the following characters will not be matched until the next opening quote.
The pattern may be more efficient written like this:
$pattern = "~(?:\G(?!\A)(?:'(*SKIP)(*F))?|'\K)[^']~";
(You can also use (*PRUNE) in place of (*SKIP).)
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