Unfortunately, despite having tried to learn regex at least one time a year for as many years as I can remember, I always forget as I use them so infrequently. This year my new year's resolution is to not try and learn regex again - So this year to save me from tears I'll give it to Stack Overflow. (Last Christmas remix).
不幸的是，尽管我试着每年至少有一年的时间学习正则表达式，但我总是忘记，因为我很少使用正则表达式。今年我的新年决心是不要再尝试和学习regex——所以今年要把我从眼泪中拯救出来，我将把它交给Stack Overflow。(去年圣诞节混音)。
I want to pass in a string in this format {getThis}, and be returned the string getThis. Could anyone be of assistance in helping to stick to my new year's resolution?
我想在这个格式{getThis}中传递一个字符串，并返回字符串getThis。有人能帮助我坚持我的新年决心吗?
Related questions on Stack Overflow:
Stack Overflow相关问题:
- How can one turn regular quotes (i.e. ', ") into LaTeX/TeX quotes (i.e. `', ``'')
- 一个人如何能转换规则引号(即?“，”)进入乳胶/TeX报价(即”,“”)
- Regex: To pull out a sub-string between two tags in a string
- 在一个字符串中从两个标签之间抽出一个子串。
- Regex to replace all \n in a String, but no those inside [code] [/code] tag
- Regex将所有的\n替换为一个字符串，但不替换在[代码][/code]标签内。
12 个解决方案
#1
44
If your string will always be of that format, a regex is overkill:
如果您的字符串始终是这种格式，那么regex将被过度使用:
>>> var g='{getThis}';
>>> g.substring(1,g.length-1)
"getThis"
#2
223
Try
试一试
/{(.*?)}/
That means, match any character between { and }, but don't be greedy - match the shortest string which ends with } (the ? stops * being greedy). The parentheses let you extract the matched portion.
这意味着，匹配{和}之间的任何字符，但不要贪心-匹配以}(?停止*贪婪)。圆括号让您提取匹配的部分。
Another way would be
另一个方法是
/{([^}]*)}/
This matches any character except a } char (another way of not being greedy)
这与任何字符(不贪婪的另一种方法)匹配
#3
133
/\{([^}]+)\}/
/ - delimiter
\{ - opening literal brace escaped because it is a special character used for quantifiers eg {2,3}
( - start capturing
[^}] - character class consisting of
^ - not
} - a closing brace (no escaping necessary because special characters in a character class are different)
+ - one or more of the character class
) - end capturing
\} - the closing literal brace
/ - delimiter
#4
15
This one works in Textmate and it matches everything in a CSS file between the curly brackets.
这个在Textmate中工作，它与CSS文件中花括号之间的所有内容相匹配。
\{(\s*?.*?)*?\}
selector {. . matches here including white space. . .}
选择器{。。这里的匹配包括空格。
If you want to further be able to return the content, then wrap it all in one more set of parentheses like so:
如果您希望进一步能够返回内容，则将其全部封装在一组括号中，如下所示:
\{((\s*?.*?)*?)\}
and you can access the contents via $1.
你可以通过$1访问内容。
This also works for functions, but I haven't tested it with nested curly brackets.
这也适用于函数，但我还没有使用嵌套的花括号来测试它。
#5
11
Here's a simple solution using javascript replace
这里有一个使用javascript替换的简单解决方案。
var st = '{getThis}';
st = st.replace(/\{|\}/gi,''); // "getThis"
As the accepted answer above points out the original problem is easily solved with substring, but using replace can solve the more complicated use cases
正如上面所提到的，用子字符串可以很容易地解决原来的问题，但是使用replace可以解决更复杂的用例。
If you have a string like "randomstring999[fieldname]" You use a slightly different pattern to get fieldname
如果您有一个类似于“randomstring999[fieldname]”的字符串，您可以使用稍微不同的模式来获得fieldname。
var nameAttr = "randomstring999[fieldname]";
var justName = nameAttr.replace(/.*\[|\]/gi,''); // "fieldname"
#6
8
You want to use regex lookahead and lookbehind. This will give you only what is inside the curly braces:
您希望使用regex lookahead和lookbehind。这只会给你在花括号里面的东西:
(?<=\{)(.*?)(?=\})
#7
7
Try this:
试试这个:
/[^{\}]+(?=})/g
For example
例如
Welcome to RegExr v2.1 by #{gskinner.com}, #{ssd.sd} hosted by Media Temple!
will return gskinner.com, ssd.sd.
将返回gskinner.com ssd.sd。
#8
3
var re = /{(.*)}/;
var m = "{helloworld}".match(re);
if (m != null)
console.log(m[0].replace(re, '$1'));
The simpler .replace(/.*{(.*)}.*/, '$1') unfortunately returns the entire string if the regex does not match. The above code snippet can more easily detect a match.
更简单的.replace(/ * {(. *)}。*/，“$1”)不幸的是，如果regex不匹配，则返回整个字符串。上面的代码片段可以更容易地检测到匹配。
#9
3
Try this one, according to http://www.regextester.com it works for js normaly.
试试这个吧，根据http://www.regextester.com，它适用于js规范。
([^{]*?)(?=\})
([^ {]* ?)(? = \ })
#10
2
Regex for getting arrays of string with curly braces enclosed occurs in string, rather than just finding first occurrence.
Regex用于获取带有花括号的字符串数组，而不是只查找第一次出现。
/\{([^}]+)\}/gm
#11
2
i have looked into the other answers, and a vital logic seems to be missing from them . ie, select everything between two CONSECUTIVE brackets,but NOT the brackets
我已经研究了其他的答案，其中一个重要的逻辑似乎缺失了。ie，在两个连续的方括号之间选择所有内容，但不要选择括号。
so, here is my answer
这就是我的答案。
\{([^{}]+)\}
#12
0
You can use this regex recursion to match everythin between, even another {} (like a JSON text) :
您可以使用这个regex递归来匹配每个薄层，甚至另一个{}(如JSON文本):
\{([^()]|())*\}
#1
44
If your string will always be of that format, a regex is overkill:
如果您的字符串始终是这种格式，那么regex将被过度使用:
>>> var g='{getThis}';
>>> g.substring(1,g.length-1)
"getThis"
#2
223
Try
试一试
/{(.*?)}/
That means, match any character between { and }, but don't be greedy - match the shortest string which ends with } (the ? stops * being greedy). The parentheses let you extract the matched portion.
这意味着，匹配{和}之间的任何字符，但不要贪心-匹配以}(?停止*贪婪)。圆括号让您提取匹配的部分。
Another way would be
另一个方法是
/{([^}]*)}/
This matches any character except a } char (another way of not being greedy)
这与任何字符(不贪婪的另一种方法)匹配
#3
133
/\{([^}]+)\}/
/ - delimiter
\{ - opening literal brace escaped because it is a special character used for quantifiers eg {2,3}
( - start capturing
[^}] - character class consisting of
^ - not
} - a closing brace (no escaping necessary because special characters in a character class are different)
+ - one or more of the character class
) - end capturing
\} - the closing literal brace
/ - delimiter
#4
15
This one works in Textmate and it matches everything in a CSS file between the curly brackets.
这个在Textmate中工作，它与CSS文件中花括号之间的所有内容相匹配。
\{(\s*?.*?)*?\}
selector {. . matches here including white space. . .}
选择器{。。这里的匹配包括空格。
If you want to further be able to return the content, then wrap it all in one more set of parentheses like so:
如果您希望进一步能够返回内容，则将其全部封装在一组括号中，如下所示:
\{((\s*?.*?)*?)\}
and you can access the contents via $1.
你可以通过$1访问内容。
This also works for functions, but I haven't tested it with nested curly brackets.
这也适用于函数，但我还没有使用嵌套的花括号来测试它。
#5
11
Here's a simple solution using javascript replace
这里有一个使用javascript替换的简单解决方案。
var st = '{getThis}';
st = st.replace(/\{|\}/gi,''); // "getThis"
As the accepted answer above points out the original problem is easily solved with substring, but using replace can solve the more complicated use cases
正如上面所提到的，用子字符串可以很容易地解决原来的问题，但是使用replace可以解决更复杂的用例。
If you have a string like "randomstring999[fieldname]" You use a slightly different pattern to get fieldname
如果您有一个类似于“randomstring999[fieldname]”的字符串，您可以使用稍微不同的模式来获得fieldname。
var nameAttr = "randomstring999[fieldname]";
var justName = nameAttr.replace(/.*\[|\]/gi,''); // "fieldname"
#6
8
You want to use regex lookahead and lookbehind. This will give you only what is inside the curly braces:
您希望使用regex lookahead和lookbehind。这只会给你在花括号里面的东西:
(?<=\{)(.*?)(?=\})
#7
7
Try this:
试试这个:
/[^{\}]+(?=})/g
For example
例如
Welcome to RegExr v2.1 by #{gskinner.com}, #{ssd.sd} hosted by Media Temple!
will return gskinner.com, ssd.sd.
将返回gskinner.com ssd.sd。
#8
3
var re = /{(.*)}/;
var m = "{helloworld}".match(re);
if (m != null)
console.log(m[0].replace(re, '$1'));
The simpler .replace(/.*{(.*)}.*/, '$1') unfortunately returns the entire string if the regex does not match. The above code snippet can more easily detect a match.
更简单的.replace(/ * {(. *)}。*/，“$1”)不幸的是，如果regex不匹配，则返回整个字符串。上面的代码片段可以更容易地检测到匹配。
#9
3
Try this one, according to http://www.regextester.com it works for js normaly.
试试这个吧，根据http://www.regextester.com，它适用于js规范。
([^{]*?)(?=\})
([^ {]* ?)(? = \ })
#10
2
Regex for getting arrays of string with curly braces enclosed occurs in string, rather than just finding first occurrence.
Regex用于获取带有花括号的字符串数组，而不是只查找第一次出现。
/\{([^}]+)\}/gm
#11
2
i have looked into the other answers, and a vital logic seems to be missing from them . ie, select everything between two CONSECUTIVE brackets,but NOT the brackets
我已经研究了其他的答案，其中一个重要的逻辑似乎缺失了。ie，在两个连续的方括号之间选择所有内容，但不要选择括号。
so, here is my answer
这就是我的答案。
\{([^{}]+)\}
#12
0
You can use this regex recursion to match everythin between, even another {} (like a JSON text) :
您可以使用这个regex递归来匹配每个薄层，甚至另一个{}(如JSON文本):
\{([^()]|())*\}