英文:
Go lang get matching substring from string
问题
我正在尝试从字符串中提取所有在引号之间的单词。
这是我的当前代码：
func StrExtract(word string) []string {
r, _ := regexp.Compile(`".*"`)
result := r.FindAllString(word, -1)
RemoveDuplicates(&result)
return result
}
在这里测试代码
输入如下：
`Hi guys, this is a "test" and a "demo" ok?`
我得到的输出是：
["test" and a "demo"]
但我想要得到的是：
[test demo]
请帮我修复这个问题，或者提供更好的解决方案。
英文:
I'm trying to extract all words from a string which are between quotes.
Here's my current code:
func StrExtract(word string) []string {
r, _ := regexp.Compile(`".*"`)
result := r.FindAllString(word, -1)
RemoveDuplicates(&result)
return (result)
}
Test the code here
With an input like:
`Hi guys, this is a "test" and a "demo" ok?`
I get the output:
["test" and a "demo"]
But I'd like to get:
[test demo]
Please help me fix this, or suggest better alternatives.
答案1
得分: 2
你可以简单地添加一个懒惰量词.*?，即正则表达式".*?"，如果你想保持简单。你之所以得到"test" and a "demo"，是因为.*是贪婪的，它会尽可能匹配尽量多的文本（因此，它实际上匹配了test之前和demo之后的"，忽略了它们之间的其他引号）。
通常，更好但在某些方面稍微复杂一些的方法是使用字符类"[^"]*"，禁止匹配引号之间的引号。这也可能导致一些其他行为，比如包括换行符（在这种情况下，你也可以禁用它们[^"\n]，或者你实际上想要这种情况）。
由于你还想去掉引号，需要进行一些额外的操作。你可以使用回顾后发断言：(?<=")[^"]*(?=")，或者使用捕获组："(.*?)"和"([^"]*)"。如果你选择使用捕获组的方法，你必须使用捕获组，而不是整个匹配。
英文:
You can just add a lazy quantifier .*?, ".*?" being the regex, if you want to keep it simple. The reason you are getting "test" and a "demo" is because just .* is greedy and matches as much text as possible (therefore, it actually matches the " before test and after demo, ignoring the fact that there are other quotes in between).
Normally a better but in some ways slightly more complicated way to do this is using character classes "[^"]*", disabling matching quotes in between. This can also cause some other behaviors like including newlines (in which case you can also disable them [^"\n], or perhaps you actually want such a case)
Since you want to also not have the quotes some additional things need to be done. You can do that with either lookarounds: (?<=")[^"]*(?="), or with capture groups: "(.*?)" and "([^"]*)". If you choose the capture group route, you have to use the capture group, not whole matches.
答案2
得分: 2
正则表达式：
"(.*?)"
这是一个在线演示：
https://regex101.com/r/sI4tA9/1
现在你需要做的就是将匹配项连接起来。不幸的是，我对go不太了解，所以在这种情况下无法帮助你。
英文:
Regex:
"(.*?)"
Here is an online demo:
https://regex101.com/r/sI4tA9/1
All you have to do now is to join matches. Unfortunately I'm not so into go that's why I can't help you in that case
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