如果不在双引号内，则为preg_replace

问题描述

基本上，我想在句子中替换某些单词(例如，单词"tree"和"pizza").限制:当应替换的单词在双引号之间时，不应执行替换.

Basically I want to replace certain words (e.g. the word "tree" with the word "pizza") in sentences. Restriction: When the word that should be replaced is between double quotes, the replace should not be performed.

示例:

The tree is green. -> REPLACE tree WITH pizza
"The" tree is "green". -> REPLACE tree WITH pizza
"The tree" is green. -> DONT REPLACE
"The tree is" green. -> DONT REPLACE
The ""tree is green. -> REPLACE tree WITH pizza

是否可以使用正则表达式执行此操作?我会计算单词前双引号的数量，然后检查它是否为奇数或偶数.但这可以在php中使用preg_replace吗?

Is it possible to do this with regular expressions? I would count the number of double quotes before the word and check if it is odd or even. But is this possible using preg_replace in php?

谢谢！

//

此刻，我的代码如下:

preg_replace("/tree/", "pizza", $sentence)

但是这里的问题是用双引号实现逻辑.我尝试过类似的事情:

But the problem here is to implement the logic with the double quotes. I tried things like:

preg_replace("/[^"]tree/", "pizza", $sentence)

但这是行不通的，因为它仅检查单词前面是否有双引号.但是，在上面的示例中，此检查失败.导入是我只想用正则表达式解决这个问题.

But this does not work, because it checks only if a double quote is in front of the word. But there are examples above where this check fails.Import is that I want to solve that problem with regex only.

推荐答案

正则表达式不是一种可以满足您每项工作所需的工具.您可以在一定程度上使用正则表达式，但是在嵌套引号中的所有情况下，它都会变得越来越复杂.

Regular expression is not a tool that will do what you need for every job. You can use regular expression for this to a certain extent, but for all cases amongst nested quotes, it continues to get more complicated.

您可以在此处使用负向超前.

$text = preg_replace('/\btree\b(?![^"]*"(?:(?:[^"]*"){2})*[^"]*$)/i', 'pizza', $text);

请参见 Working demo

正则表达式:

\b               the boundary between a word char (\w) and not a word char
 tree            'tree'
\b               the boundary between a word char (\w) and not a word char
(?!              look ahead to see if there is not:
 [^"]*           any character except: '"' (0 or more times)
  "              '"'
 (?:             group, but do not capture (0 or more times)
  (?:            group, but do not capture (2 times):
   [^"]*         any character except: '"' (0 or more times)
    "            '"'
  ){2}           end of grouping
 )*              end of grouping
 [^"]*           any character except: '"' (0 or more times)
 $               before an optional \n, and the end of the string
)                end of look-ahead

另一种选择是使用受控回溯，因为您可以在 php

Another option is to use controlled backtracking since your able to do this in php

$text = preg_replace('/"[^"]*"(*SKIP)(*FAIL)|\btree\b/i', 'pizza', $text);

请参见 Working demo

该想法是跳过引号中的内容.我首先匹配引号，后跟除"以外的任何字符，再加上引号，然后使子模式失败，并强制正则表达式引擎不要使用带有(*SKIP)和(*FAIL)回溯控制动词的其他替代方式重试子字符串.

The idea is to skip content in quotations. I first match the quotation followed by any character except " followed by a quotation and then make the subpattern fail and force the regular expression engine to not retry the substring with an other alternative with (*SKIP) and (*FAIL) backtracking control verbs.

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