blocks|key|281294|text|这个表达式是您可能要研究的内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|281295|%5E(.*?)(\s%2B\/.*)$|code-block|syntax|javascript|281296|在这里，我们有两个捕获组，第一个组收集您想要的输出，第二个组是您不想要的模式，以开始和结束字符为界，只是为了安全起见，如果您想要的话，可以删除：|281297|(.*?)(\s%2B\/.*)|281298|Python测试|281299|#+coding=utf8
#+the+above+tag+defines+encoding+for+this+document+and+is+for+Python+2.x+compatibility

import+re

regex+=+r"%5E(.*?)(\s%2B\/.*)$"

test_str+=+("123+/+some+text+123\n"
++++"anything+else++++/+some+text+123")

subst+=+"\\1"

#+You+can+manually+specify+the+number+of+replacements+by+changing+the+4th+argument
result+=+re.sub(regex,+subst,+test_str,+0,+re.MULTILINE)

if+result:
++++print+(result)

#+Note:+for+Python+2.7+compatibility,+use+ur""+to+prefix+the+regex+and+u""+to+prefix+the+test+string+and+substitution.|281300|JavaScript演示|281301|​|281302|const+regex+=+/%5E(.*?)(\s%2B\/.*)$/gm;
const+str+=+`123+/+some+text+123
anything+else++++/+some+text+123`;
const+subst+=+`\n$1`;

//+The+substituted+value+will+be+contained+in+the+result+variable
const+result+=+str.replace(regex,+subst);

console.log('Substitution+result:+',+result);|281303|281304|RegEx|281305|如果这不是您想要的表达式，则可以修改/更改regex101.com中的表达式。|offset|length|281306|281307|📷|atomic|281308|281309|RegEx电路|281310|您还可以在jex.im中可视化表达式。|281311|281312|281313|281314|空间|281315|对于所需输出之前的空格，我们可以简单地使用负后视添加一个捕获组：|281316|+%5E(\s%2B)?(.*?)(\s%2B\/.*)$|281317|281318|281319|const+regex+=+/%5E(\s%2B)?(.*?)(\s%2B\/.*)$/gm;
const+str+=+`++++++123+/+some+text+123
+++++++++++++anything+else++++/+some+text+123
123+/+some+text+123
anything+else++++/+some+text+123`;
const+subst+=+`$2`;

//+The+substituted+value+will+be+contained+in+the+result+variable
const+result+=+str.replace(regex,+subst);

console.log('Substitution+result:+',+result);|281320|281321|演示|281322|281323|281324|281325|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/Hr3f2n/1/|1|IMAGE|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/78afac69ec85634bf1928ba00e599264.png|imageAlt|2|https://jex.im/regulex/#!flags=&re=%5E(a%7Cb)*?$)|3|https://developer.qcloudimg.com/http-save/yehe-900000/e094e589eeb8af58a7871c65fdc28226.png|4|https://regex101.com/r/NBiUgw/1|5|https://regex101.com/r/85DfUP/1|6|https://developer.qcloudimg.com/http-save/yehe-900000/823d55e01d8e689eef8b4ebe70bb35e2.png^0|0|0|0|0|0|0|0|0|0|0|0|L|C|0|0|0|0|1|1|0|0|0|5|6|2|0|0|0|1|3|0|0|0|L|3|4|0|0|0|0|0|0|0|2|5|0|0|0|1|6|0|0^^$0|@$1|2|3|4|5|6|7|2H|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|2I|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|2J|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|2K|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|2L|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|2M|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|2N|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|2O|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|2P|8|@]|9|@]|A|$E|F]]|$1|U|3|R|5|6|7|2Q|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|2R|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|2S|8|@]|9|@$Z|2T|10|2U|1|2V]]|A|$]]|$1|11|3|R|5|6|7|2W|8|@]|9|@]|A|$]]|$1|12|3|13|5|14|7|2X|8|@]|9|@$Z|2Y|10|2Z|1|30]]|A|$]]|$1|15|3|R|5|6|7|31|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|32|8|@]|9|@]|A|$]]|$1|18|3|19|5|6|7|33|8|@]|9|@$Z|34|10|35|1|36]]|A|$]]|$1|1A|3|R|5|6|7|37|8|@]|9|@]|A|$]]|$1|1B|3|13|5|14|7|38|8|@]|9|@$Z|39|10|3A|1|3B]]|A|$]]|$1|1C|3|R|5|6|7|3C|8|@]|9|@]|A|$]]|$1|1D|3|1E|5|6|7|3D|8|@]|9|@]|A|$]]|$1|1F|3|1G|5|6|7|3E|8|@]|9|@$Z|3F|10|3G|1|3H]]|A|$]]|$1|1H|3|1I|5|D|7|3I|8|@]|9|@]|A|$E|F]]|$1|1J|3|P|5|6|7|3J|8|@]|9|@]|A|$]]|$1|1K|3|R|5|6|7|3K|8|@]|9|@]|A|$]]|$1|1L|3|1M|5|D|7|3L|8|@]|9|@]|A|$E|F]]|$1|1N|3|R|5|6|7|3M|8|@]|9|@]|A|$]]|$1|1O|3|1P|5|6|7|3N|8|@]|9|@$Z|3O|10|3P|1|3Q]]|A|$]]|$1|1Q|3|R|5|6|7|3R|8|@]|9|@]|A|$]]|$1|1R|3|13|5|14|7|3S|8|@]|9|@$Z|3T|10|3U|1|3V]]|A|$]]|$1|1S|3|R|5|6|7|3W|8|@]|9|@]|A|$]]|$1|1T|3|-4|5|6|7|3X|8|@]|9|@]|A|$]]]|1U|$1V|$5|1W|1X|1Y|A|$1Z|20]]|21|$5|22|1X|23|A|$24|25|26|-4]]|27|$5|1W|1X|1Y|A|$1Z|28]]|29|$5|22|1X|23|A|$24|2A|26|-4]]|2B|$5|1W|1X|1Y|A|$1Z|2C]]|2D|$5|1W|1X|1Y|A|$1Z|2E]]|2F|$5|22|1X|23|A|$24|2G|26|-4]]]]

<p>This expression is what you might want to explore: </p>

<pre><code>^(.*?)(\s+\/.*)$
</code></pre>

<p>Here, we have two capturing groups where the first one collects your desired output, and the second one is your undesired pattern, bounded by start and end chars, just to be safe that can be removed if you want: </p>

<pre><code>(.*?)(\s+\/.*)
</code></pre>

<h3>Python Test</h3>

<pre><code># coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"^(.*?)(\s+\/.*)$"

test_str = ("123 / some text 123\n"
    "anything else    / some text 123")

subst = "\\1"

# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0, re.MULTILINE)

if result:
    print (result)

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.
</code></pre>

<h3>JavaScript Demo</h3>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const regex = /^(.*?)(\s+\/.*)$/gm;
const str = `123 / some text 123
anything else    / some text 123`;
const subst = `\n$1`;

// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);

console.log('Substitution result: ', result);</code></pre>
</div>
</div>
</p>

<h3>RegEx</h3>

<p>If this wasn't your desired expression, you can modify/change your expressions in <a href="https://regex101.com/r/Hr3f2n/1/" rel="nofollow noreferrer">regex101.com</a>.</p>

<p><a href="https://i.stack.imgur.com/w1Q3X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w1Q3X.png" alt="enter image description here"></a></p>

<h3>RegEx Circuit</h3>

<p>You can also visualize your expressions in <a href="https://jex.im/regulex/#!flags=&amp;re=%5E(a%7Cb)*%3F%24" rel="nofollow noreferrer">jex.im</a>:</p>

<p><a href="https://i.stack.imgur.com/fBpXt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fBpXt.png" alt="enter image description here"></a></p>

<h3>Spaces</h3>

<p>For spaces before your desired output, we can simply add a capturing group with <a href="https://regex101.com/r/NBiUgw/1" rel="nofollow noreferrer">negative lookbehind</a>: </p>

<pre><code> ^(\s+)?(.*?)(\s+\/.*)$
</code></pre>

<h3>JavaScript Demo</h3>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const regex = /^(\s+)?(.*?)(\s+\/.*)$/gm;
const str = `      123 / some text 123
             anything else    / some text 123
123 / some text 123
anything else    / some text 123`;
const subst = `$2`;

// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);

console.log('Substitution result: ', result);</code></pre>
</div>
</div>
</p>

<h2><a href="https://regex101.com/r/85DfUP/1" rel="nofollow noreferrer">Demo</a></h2>

<p><a href="https://i.stack.imgur.com/KFi00.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KFi00.png" alt="enter image description here"></a></p>


blocks|key|281339|text|这有点棘手。首先从非空白字符开始匹配，然后继续缓慢但肯定地进行匹配，直到紧跟一个可选空格数和斜杠标记的位置：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|281340|\S.*?(?=+*\/)|code-block|syntax|javascript|281341|请参阅现场演示|offset|length|281342|如果斜杠标记可能是输入字符串中的第一个非空格字符，则将\S替换为[%5E\s\/]。|style|CODE|281343|[%5E\s\/].*?(?=+*\/)|281344|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/fR73ON/1^0|0|0|3|4|0|0|R|2|W|7|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@$I|11|J|12|1|13]]|A|$]]|$1|K|3|L|5|6|7|14|8|@$I|15|J|16|M|N]|$I|17|J|18|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

<p>That's a little bit tricky. You first start matching from a non-whitespace character then continue matching slowly but surely up to the position that is immediately followed by an optional number of spaces and a slash mark:</p>

<pre><code>\S.*?(?= *\/)
</code></pre>

<p>See <a href="https://regex101.com/r/fR73ON/1" rel="noreferrer">live demo here</a></p>

<p>If slash mark could be the first non-whitespace character in input string then replace <code>\S</code> with <code>[^\s\/]</code>:</p>

<pre><code>[^\s\/].*?(?= *\/)
</code></pre>


blocks|key|1728232|text|这里有一个可能的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1728233|Regex|offset|length|style|BOLD|1728234|(?<!\/)\S.*\S(?=\s*\/)|code-block|syntax|javascript|1728235|示例|1728236|#+import+regex+#+or+re

string+=+'+123+/+some+text+123'
test+=+regex.search(r'(?<!\/)\S.*\S(?=\s*\/)',+string)
print(test.group(0))
#+prints+'123'

string+=+'a+test+/+some+text+123'
test+=+regex.search(r'(?<!\/)\S.*\S(?=\s*\/)',+string)
print(test.group(0))
#+prints+'a+test'|1728237|短解释|1728238|(?<!\/)说，在可能的匹配之前，不可能有/符号。|unordered-list-item|CODE|1728239|\S.*\S懒散地匹配任何东西(.*)，同时确保它不会以空白(\S)开头或结尾。|1728240|(?=\s*\/)意味着一个可能的匹配必须后面跟着一个/符号或空格%2B一个/。|1728241|entityMap^0|0|0|5|0|0|0|2|0|0|0|3|0|0|7|M|1|0|0|6|G|2|V|2|0|0|9|R|1|10|1|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@$D|14|E|15|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|17|8|@$D|18|E|19|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|S|3|T|5|U|7|1E|8|@$D|1F|E|1G|F|V]|$D|1H|E|1I|F|V]]|9|@]|A|$]]|$1|W|3|X|5|U|7|1J|8|@$D|1K|E|1L|F|V]|$D|1M|E|1N|F|V]|$D|1O|E|1P|F|V]]|9|@]|A|$]]|$1|Y|3|Z|5|U|7|1Q|8|@$D|1R|E|1S|F|V]|$D|1T|E|1U|F|V]|$D|1V|E|1W|F|V]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]]|11|$]]

<p>Here is a possible solution</p>

<p><strong>Regex</strong></p>

<pre><code>(?&lt;!\/)\S.*\S(?=\s*\/)
</code></pre>

<p><strong>Example</strong></p>

<pre><code># import regex # or re

string = ' 123 / some text 123'
test = regex.search(r'(?&lt;!\/)\S.*\S(?=\s*\/)', string)
print(test.group(0))
# prints '123'

string = 'a test / some text 123'
test = regex.search(r'(?&lt;!\/)\S.*\S(?=\s*\/)', string)
print(test.group(0))
# prints 'a test'
</code></pre>

<p><strong>Short explanation</strong></p>

<ul>
<li><code>(?&lt;!\/)</code> says before a possible match there can be no <code>/</code> symbol.</li>
<li><code>\S.*\S</code> matches lazily <em>anything</em> (<code>.*</code>) while making sure it does not start or end with a white space (<code>\S</code>)</li>
<li><code>(?=\s*\/)</code> means a possible match must be followed by a <code>/</code> symbol or by white spaces + a <code>/</code>.</li>
</ul>


blocks|key|1728257|text|你不需要正则表达式就能做到|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1728258|my_string+=+"++++++123+/+some+text+123"
match+=+my_string.split("/")[0].strip()|code-block|syntax|javascript|1728259|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>You could do it without a regex</p>

<pre><code>my_string = "      123 / some text 123"
match = my_string.split("/")[0].strip()
</code></pre>


<p>I'm trying to write a regex to match part of a string that comes before '/' but also ignores any leading or trailing white space within the match.</p>

<p>So far I've got <code>^[^\/]*</code> which matches everything before the '/' but I can't figure out how to ignore the white space.</p>

<pre><code>      123 / some text 123
</code></pre>

<p>should yield</p>

<pre><code>123
</code></pre>

<p>and</p>

<pre><code>     a test / some text 123
</code></pre>

<p>should yield</p>

<pre><code>a test
</code></pre>


Python Regex Match Before Character AND Ignore White Space

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我试图编写一个正则表达式来匹配“/”之前的字符串的一部分，但也忽略了匹配中的任何前导或尾随空格。到目前为止，我已经得到了^[^\/]*，它与'/‘之前的所有内容都匹配，但我不知道如何忽略空白。      123 / some text 123应该屈服123和     a test / some text 123应该屈服...

问Python在字符之前匹配并忽略空白
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