如何替换字符串的多个子出现次数

huangapple
go评论128阅读模式
英文:

How to replace multiple sub occurrences of string

问题

我有一个使用情况，其中我可以拥有一个包含任何内容的文本字符串。我想要实现的目标是替换给定字符串中的某个特定模式。

假设我有一个给定的字符串：

  1. :es1
  2. es2
  3. aaes1aa
  4. :es3,
  5. es1:
  6. ees1,
  7. ees1 
  9. {
  10. "es1 :

我在这里尝试的是，假设我必须替换这个字符串中的所有es1，但有一个条件。它必须以[, | ; | : | " | ' | \\ | \s]结尾或开头。:es1，es1,，es1:等都是可以接受的，但eees1sss不是。

我尝试了类似于这样的正则表达式：([, | ; | : | " | ' | \\ | \s])(es1)([, | ; | : | " | ' | , | \s])，但我不认为这是我需要的。

Go程序：

  1. match := regexp.MustCompile(`([, | ; | : | " | ' | \\ | \s])(es1)([, | ; | : | " | ' | , | \s])`)
  2. test := `:es1
  3. es2
  4. aaes1aa
  5. :es3,
  6. es1:
  7. ees1,
  8. ees1 
  10. {
  11. "es1 :`
  13. fmt.Println(match.ReplaceAllString(test, "$1es4$3"))

输出：

  1. es2
  2. aaes1aa
  3. :es3,
  4. :
  5. ees1,
  6. ees1 
  8. {
  9.  :

我原本期望的输出更像是：

  1. :es4
  2. es2
  3. aaes1aa
  4. :es3,
  5. es4:
  6. ees1,
  7. ees1 
  9. {
  10. "es4 :
英文:

I have one use case where I can have a text string which can contain anything. What I want to achieve is to replace a certain pattern within that given string.

Let's say I have given string as

  1. :es1
  2. es2
  3. aaes1aa
  4. :es3,
  5. es1:
  6. ees1,
  7. ees1 
  9. {
  10. "es1 :

What I am trying to do is here is suppose I have to replace all es1 in this string but with one condition. It has to be either end or start with [, | ; | : | " | ' | \\ | \s]. :es1, es1,, es1: and so on are accepted but eees1sss is not.

I tried ([, | ; | : | " | ' | \\ | \s])(es1)([, | ; | : | " | ' | , | \s]) something like this but I don't think it's what I need.

Go program:

  1. match := regexp.MustCompile(`([, | ; | : | " | ' | \\ | \s])(es1)([, | ; | : | " | ' | , | \s])`)
  2. 	test := `:es1
  3. 	es2
  4. 	aaes1aa
  5. 	:es3,
  6. 	es1:
  7. 	ees1,
  8. 	ees1 
  9. 	
  10. 	{
  11. 	"es1 :`
  12. 	
  13. 	fmt.Println(match.ReplaceAllString(test, "$1es4$3"))

output:

  1.        es2
  2.         aaes1aa
  3.         :es3,
  4. :
  5.         ees1,
  6.         ees1 
  8.         {
  9.          :

I was expecting my output to be more like

  1. :es4
  2.     es2
  3.     aaes1aa
  4.     :es3,
  5.     es4:
  6.     ees1,
  7.     ees1 
  9.     {
  10.     "es4 :

答案1

得分: 2

下面提供的解决方案没有经过全面测试，但似乎可以工作。

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. func main() {
  9. 	match := regexp.MustCompile(`([, | ; | : | " | ' | \\ | \s])es1|^es1([, | ; | : | " | ' | , | \s])`)
  10. 	test := `:es1
  11.     es2
  12.     aaes1aa
  13.     :es3,
  14.     es1:
  15.     ees1,
  16.     ees1 
  17.     
  18.     {
  19.     "es1 :`
  21. 	fmt.Println(match.ReplaceAllString(test, "es4"))
  22. }

点击此处查看代码。

英文:

the solution provided below is not well tested against all possibilities, but it seems to be working.

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. func main() {
  9. 	match := regexp.MustCompile(`([, | ; | : | " | ' | \\ | \s])es1|^es1([, | ; | : | " | ' | , | \s])`)
  10. 	test := `:es1
  11.     es2
  12.     aaes1aa
  13.     :es3,
  14.     es1:
  15.     ees1,
  16.     ees1 
  17.     
  18.     {
  19.     "es1 :`
  21. 	fmt.Println(match.ReplaceAllString(test, "es4"))
  22. }

https://play.golang.org/p/E8lb9vmM_Sa

答案2

得分: 1

你可以使用以下代码：

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. func main() {
  9. 	match := regexp.MustCompile(`([,;:'\s])es1\b|\bes1([,;:'\s])`)
  10. 	test := ":es1\n    es2\n    aaes1aa\n    :es3,\n    es1:\n    ees1,\n    ees1 \n    \n    {\n    \"es1 :"
  11. 	fmt.Println(match.ReplaceAllString(test, "es4$2"))
  12. }

请参考Go演示和正则表达式演示。请注意，方括号内的空格和|字符是有意义的，并且会与这些字符完全匹配，因此您需要从模式中删除它们。

该正则表达式匹配：

  • ([,;:'\s])es1\b - 第1组：逗号、分号、冒号、双引号、单引号、反斜杠或空格；然后是作为一个完整单词的es1（\b是一个单词边界）

  • | - 或者

  • \bes1 - 一个完整的单词es1

  • ([,;:'\s]) - 第2组：逗号、分号、冒号、双引号、单引号、反斜杠或空格

英文:

You can use

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. func main() {
  9. 	match := regexp.MustCompile(`([,;:"'\\\s])es1\b|\bes1([,;:"'\\\s])`)
  10. 	test := ":es1\n    es2\n    aaes1aa\n    :es3,\n    es1:\n    ees1,\n    ees1 \n    \n    {\n    \"es1 :"
  11. 	fmt.Println(match.ReplaceAllString(test, "${1}es4$2"))
  12. }

See the Go demo and the regex demo. Note that the spaces and | chars inside square brackets are meaningful and match these chars literally, thus, you need to remove them all from your pattern.

The regex matches:

  • ([,;:"'\\\s])es1\b - Group 1: a comma, or a semi-colon, colon, double or single quotation mark, backslash or whitespace; then es1 as a whole word (\b is a word boundary)
  • | - or
  • \bes1 - a whole word es1
  • ([,;:"'\\\s]) - Group 2: a comma, or a semi-colon, colon, double or single quotation mark, backslash or whitespace

huangapple
  • 本文由 发表于 2021年9月2日 16:03:46
  • 转载请务必保留本文链接：https://go.coder-hub.com/69026377.html
  • go
  • regex
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