blocks|key|82803|text|您可以使用%5E[%5E\(]%2B进行匹配，然后对其进行修剪以删除尾随空格|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|82804|entityMap^0|5|7|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

you would use <code>^[^\(]+</code> to match that and then trim it to remove the trailing space

blocks|key|637494|text|%5E[%5E\(]*|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|637495|[%5E\(]是一个字符类，它匹配除(之外的所有内容，而*是一个贪婪的匹配，它尽可能多次地匹配该类。开头的%5E从字符串的开头开始匹配。|637496|entityMap^0|0|7|0|0|5|G|1|Q|1|1F|1|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$]]

<code>^[^\(]*</code>

<code>[^\(]</code> is a character class, which matches everything except for <code>(</code>, and <code>*</code> is a greedy match, which matches the class as many times as possible. The <code>^</code> at the beginning matches from the beginning of the string.

blocks|key|637514|text|为了避免尾随的空格，可以尝试%5E.*?(?=\s\()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|637515|%5E(.*?)告诉它匹配尽可能少的字符，从字符串的开头开始，(?=\s\()将匹配的另一端锚定到您的paren，而不捕获它或它前面的空格。|637516|entityMap^0|E|C|0|0|6|T|8|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

To avoid the trailing whitespace, try <code>^.*?(?=\s\()</code>.

<code>^(.*?)</code> tells it to match as few characters as possible, from the start of the string, and the <code>(?=\s\()</code> anchors the other end of the match to your paren, without capturing it or the whitespace before it.

blocks|key|3986708|text|您可以在R：*\\(.*中使用这个简单的正则表达式|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3986709|str+<-+"Senior+Software+Engineer+(mountain+view)"

sub("+*\\(.*",+"",+str)
#+[1]+"Senior+Software+Engineer"|code-block|syntax|javascript|3986710|它还避免了尾随的空格。|3986711|entityMap^0|6|6|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You can use this simple regex in R: <code>*\\(.*</code>

<pre><code>str &lt;- "Senior Software Engineer (mountain view)"

sub(" *\\(.*", "", str)
# [1] "Senior Software Engineer"
</code></pre>

It also avoids the trailing whitespace.

blocks|key|4467601|text|一种整齐的方式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4467602|library("stringr")|code-block|syntax|javascript|4467603|4467604|下面的输出，将返回你的列表...|blockquote|4467605|4467606|4467607|str_match_all(str,+"%5E[%5E\\(]%2B")|4467608|如果需要在字符串中输入，则使用|4467609|4467610|4467611|4467612|str_match_all(str,+"%5E[%5E\\(]%2B")+%25>%25+toString()|4467613|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|J|7|15|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|T|3|U|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|V|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|W|$]]

A tidyverse way...

<pre><code>library("stringr")
</code></pre>

<blockquote>
 The output of below, will return you list...
</blockquote>

<pre><code>str_match_all(str, "^[^\\(]+")
</code></pre>

<blockquote>
 If you need in string
</blockquote>

<pre><code>str_match_all(str, "^[^\\(]+") %&gt;% toString()
</code></pre>

blocks|key|637553|text|如果你想匹配括号前后的所有内容，试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|637554|搜索：%5E[%5E$]%2B%7C($).*|offset|length|style|CODE|637555|替换为：\1|637556|如果您想匹配所有括号并将其删除，请尝试以下操作：|637557|搜索：$[%5E(\r\n]*?$%7C$%7C$|637558|替换为：(LEAVE+EMPTY)|637559|entityMap^0|0|3|E|0|4|2|0|0|3|K|0|4|D|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|12|8|@$D|13|E|14|F|G]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|Q|$]]

and if you wanna match everything until and after parenthesis, try this:

SEARCH: <code>^[^$]+|($).*</code>

REPLACE BY: <code>\1</code>

and if you want to match all paranthesis and delete them, try this:

SEARCH: <code>$[^(\r\n]*?$|$|$</code>

REPLACE BY: <code>(LEAVE EMPTY)</code>

blocks|key|3986762|text|package+unglue提供了一个可读的替代方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3986763|#+install.packages("unglue")
library(unglue)

x+<-+"Senior+Software+Engineer+(mountain+view)"
unglue_vec(x,"{job}+({location})",+var+=+"job")
#>+[1]+"Senior+Software+Engineer"|code-block|syntax|javascript|3986764|更多信息：https://github.com/moodymudskipper/unglue/blob/master/README.md|offset|length|3986765|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/moodymudskipper/unglue/blob/master/README.md^0|0|0|5|1R|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

The package unglue offers a readable alternative : 

<pre class="lang-r prettyprint-override"><code># install.packages("unglue")
library(unglue)

x &lt;- "Senior Software Engineer (mountain view)"
unglue_vec(x,"{job} ({location})", var = "job")
#&gt; [1] "Senior Software Engineer"
</code></pre>

More info: <a href="https://github.com/moodymudskipper/unglue/blob/master/README.md" rel="nofollow noreferrer">https://github.com/moodymudskipper/unglue/blob/master/README.md</a>

Given a string <code>str = "Senior Software Engineer (mountain view)"</code>

How can I match everything until I hit the first parenthesis, giving me back "Senior Software Engineer"

match everything until parenthesis

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给定一个字符串str = "Senior Software Engineer (mountain view)"我如何才能匹配所有内容，直到我命中第一个括号，返回“高级软件工程师”

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