[英]Get string inside parentheses, removing parentheses, with regex
I have these two strings... 我有这两个字符串......
var str1 = "this is (1) test";
var str2 = "this is (2) test";
And want to write a RegEx to extract what is INSIDE the parentheses as in "1"
and "2"
to produce a string like below. 并且想要编写一个RegEx来提取INSIDE括号中的内容,如
"1"
和"2"
以生成如下所示的字符串。
var str3 = "12";
right now I have this regex which is returning the parentheses too... 现在我有这个正则表达式也返回括号...
var re = (/\((.*?)\)/g);
var str1 = str1.match(/\((.*?)\)/g);
var str2 = str2.match(/\((.*?)\)/g);
var str3 = str1+str2; //produces "(1)(2)"
Like this 像这样
Javascript 使用Javascript
var str1 = "this is (1) test",
str2 = "this is (2) test",
str3 = str1.match(/\((.*?)\)/)[1] + str2.match(/\((.*?)\)/)[1];
alert(str3);
See MDN RegExp 请参阅MDN RegExp
(x) Matches x and remembers the match.
(x)匹配x并记住匹配。 These are called capturing parentheses.
这些被称为捕获括号。
For example, /(foo)/ matches and remembers 'foo' in "foo bar."
例如,/(foo)/匹配并记住“foo bar”中的'foo'。 The matched substring can be recalled from the resulting array's elements 1 , ..., [n] or from the predefined RegExp object's properties $1, ..., $9.
可以从结果数组的元素1 ,...,[n]或预定义的RegExp对象的属性$ 1,...,$ 9中调用匹配的子字符串。
Capturing groups have a performance penalty.
捕获组会降低性能。 If you don't need the matched substring to be recalled, prefer non-capturing parentheses (see below).
如果您不需要调用匹配的子字符串,则更喜欢非捕获括号(参见下文)。
Try: 尝试:
/[a-zA-Z0-9\s]+\((\d+)\)[a-zA-Z0-9\s]+/
This will capture any digit of one or more inside the parentheses. 这将捕获括号内的一个或多个数字。
Example: http://regexr.com?365uj 示例: http : //regexr.com?365uj
EDIT: In the example you will see the replace field has only the numbers, and not the parentheses--this is because the capturing group $1
is only capturing the digits themselves. 编辑:在示例中,您将看到替换字段只有数字,而不是括号 - 这是因为捕获组
$1
仅捕获数字本身。
EDIT 2: 编辑2:
Try something like this: 尝试这样的事情:
var str1 = "this is (1) test";
var str2 = "this is (2) test";
var re = /[a-zA-Z0-9\s]+\((\d+)\)[a-zA-Z0-9\s]+/;
str1 = str1.replace(re, "$1");
str2 = str2.replace(re, "$1");
console.log(str1, str2);
try this 试试这个
var re = (/\((.*?)\)/g);
var str1 = str1.match(/\((.*?)\)/g);
var new_str1=str1.substr(1,str1.length-1);
var str2 = str2.match(/\((.*?)\)/g);
var new_str2=str2.substr(1,str2.length-1);
var str3 = new_str1+new_str2; //produces "12"
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