blocks|key|417147|text|您可以使用.find函数来查找字符.jpg，然后您可以对字符串进行索引，以获得除此之外的所有内容。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417148|string+=+https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC
index+=+string.find(".jpg")
new_string+=+string[:index%2B+4]|code-block|syntax|javascript|417149|您必须添加4，因为这是jpg的长度，所以它不会删除它。|417150|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can use the .find function to find the characters .jpg then you can index the string to get everything but that. Ex:
<pre><code>string = https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC
index = string.find(&quot;.jpg&quot;)
new_string = string[:index+ 4]
</code></pre>
You have to add four because that is the length of jpg so it does not delete that too.

blocks|key|405816|text|如果在给定字符串中找到子字符串，则find()方法返回子字符串的最低索引。如果没有找到，则返回-1。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|405817|str+='https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC'
result+=+str.find('jpg')
print(result)
new_str+=+str[:result]

print(new_str%2B'jpg')|code-block|syntax|javascript|405818|entityMap^0|H|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The <code>find()</code> method returns the lowest index of the substring if it is found in given string. If its is not found then it returns -1.
<pre><code>str ='https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC'
result = str.find('jpg')
print(result)
new_str = str[:result]

print(new_str+'jpg')
</code></pre>

blocks|key|417152|text|你可以使用split+(假设字符串有'jpg'，否则下面的代码将只返回原始url)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417153|string+=+'https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC'
jpg_removed+=+string.split('.jpg')[0]%2B'.jpg'|code-block|syntax|javascript|417154|示例|417155|string+=+'www.google.com'
com_removed+=+string.split('.com')[0]+
#+com_removed+=+'www.google'|417156|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You could use split (assuming the string has 'jpg', otherwise the code below will just return the original url).
<pre><code>string = 'https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC'
jpg_removed = string.split('.jpg')[0]+'.jpg'
</code></pre>
Example
<pre><code>string = 'www.google.com'
com_removed = string.split('.com')[0] 
# com_removed = 'www.google'
</code></pre>

blocks|key|405820|text|请参阅：Extracting+extension+from+filename+in+Python|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|405821|我们提取文件名并添加扩展名，而不是提取扩展名(如果我们知道它总是.jpg，那就好了！)|405822|import+os
filename,+file_extension+=+os.path.splitext('/path/to/somefile.jpg_corruptedpath')
result+=+filename+%2B+'.jpg'|code-block|syntax|javascript|405823|现在，除了最初的问题之外，我认为你最初是如何获得这条信息的可能有问题。必须有一种更好的方法来提取该jpeg，而不会影响路径。遗憾的是，我不能帮助你，因为我是BeautifulSoup的新手。|405824|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/541390/extracting-extension-from-filename-in-python^0|4|18|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

See: <a href="https://stackoverflow.com/questions/541390/extracting-extension-from-filename-in-python">Extracting extension from filename in Python</a>
Instead of extracting the extension, we extract the filename and add the extension (if we know it's always .jpg, it's fine!)
<pre><code>import os
filename, file_extension = os.path.splitext('/path/to/somefile.jpg_corruptedpath')
result = filename + '.jpg'
</code></pre>
Now, outside of the original question, I think there might be something wrong with how you got that piece of information int he first place. There must be a better way to extract that jpeg without messing around with the path. Sadly I can't help you with that since I a novice with BeautifulSoup.

blocks|key|417158|text|您可以使用正则表达式将.jpg之后的所有内容替换为空字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417159|import+re

url++='https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC'

name+=+re.sub(r'(?<=\.jpg).*',"",url)

print(name)
https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg|code-block|syntax|javascript|417160|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|$]]

You could use a regular expression to replace everything after .jpg with an empty string:
<pre><code>import re

url ='https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC'

name = re.sub(r'(?&lt;=\.jpg).*',&quot;&quot;,url)

print(name)
https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg
</code></pre>

blocks|key|405811|text|你可以使用正则表达式。你只想忽略.jpg后面的字符，这样你就可以使用下面这样的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|405812|import+re
new_url=re.findall("(.*\.jpg).*",old_url)[0]|code-block|syntax|javascript|405813|(.*\.jpg)就像一个捕获组，您可以在.jpg之前匹配任意数量的字符。由于.具有特殊的含义，因此您需要使用\转义jpg中的.。.*用于匹配任意数量的字符，但是因为它不在捕获组()中，所以它将被匹配，但不会被提取。|offset|length|style|CODE|405814|entityMap^0|0|0|0|9|13|1|1J|1|1R|1|1T|2|2H|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|N|$]]

You can make use of regular expression. You just want to ignore the characters after .jpg so you can some use of something like this:
<pre><code>import re
new_url=re.findall(&quot;(.*\.jpg).*&quot;,old_url)[0]
</code></pre>
<code>(.*\.jpg)</code> is like a capturing group where you're matching any number of characters before .jpg. Since <code>.</code> has a special meaning you need to escape the <code>.</code> in jpg with a <code>\</code>. <code>.*</code> is used to match any number of character but since this is not inside the capturing group <code>()</code> this will get matched but won't get extracted.

how exactly can I delete characters after .jpg? is there a way to differentiate between the extension I take with python and what follows?
for example I have a link like that
<pre><code>https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DCC3DD9E74DC1D10104F623D7E9BDC
</code></pre>
How can I delete everything after .jpg?
I tried replacing but it didn't work
another way?
Use a forum to count strings or something like ?
I tried to get jpg files with this
<pre><code>for link in links:
 res = requests.get(link).text
 soup = BeautifulSoup(res, 'html.parser')
 img_links = []
 for img in soup.select('a.thumbnail img[src]'):
 print(img[&quot;src&quot;])
 with open('links'+'.csv', 'a', encoding = 'utf-8', newline='') as csv_file:
 file_is_empty = os.stat(self.filename+'.csv').st_size == 0
 fieldname = ['links']
 writer = csv.DictWriter(csv_file, fieldnames = fieldname)
 if file_is_empty:
 writer.writeheader()
 writer.writerow({'links':img[&quot;src&quot;]})

 img_links.append(img[&quot;src&quot;]) 
</code></pre>

Delete all characters that come after a given string

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋 

腾讯云代码助手

CODING DevOps

Cloud Studio

SDK中心

API中心

命令行工具

如何删除.jpg后的字符？有没有办法区分我在python中使用的扩展和后面的扩展？例如，我有一个这样的链接https://s13emagst.akamaized.net/products/29146/29145166/images/res_cd1fa80f252e88faa70ffd465c516741.jpg10DC...

问删除给定字符串后面的所有字符
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问删除给定字符串后面的所有字符EN