blocks|key|490203|text|在python3中，这对我起了作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|490204|for+line+in+texfile:
++++x+=+re.search("(.*?)(text)",+line)
++++try:
++++++++print+(x.group(1))
++++except:
++++++++print(line)|code-block|syntax|javascript|490205|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|$]]

This worked for me in python3:

<pre><code>for line in texfile:
 x = re.search("(.*?)(text)", line)
 try:
 print (x.group(1))
 except:
 print(line)
</code></pre>

blocks|key|52863|text|如果您想检测文件的一部分在几行上扩展，如果文件不是太大，那么通过一次检查一行来限制regexes的功能并不是一种特别好的方法。当文件可以被读取并完全放在RAM中时，最好用正则表达式来分析它，将文本作为一个独特的整体来研究。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|52864|请注意，如果没有使用标志re.MULTILINE，则'%5E'具有“字符串的开始”的意义，如果使用此标志，则具有“行的开始”。|offset|length|style|CODE|52865|顺便说一句，如果您使用match()方法，您不需要在RE模式的开头添加"%5E"，因为match()试图从字符串的一开始就进行匹配。|52866|因此，下面是一种分析整个文本的方法，正如您似乎想要的那样(我使用splitlines(True)获取字符串ss中的行列表，这个列表模拟文件°：|BOLD|52867|import+re

ss+=+"""+++first+line
++++bli+bli+hhhhhhhh+TEXT+bla+blajjjjjjjjj
hhhhhhhh++VVVVV
ZZZZZZ
++++tttt
bolo+bolo+TEXTrumunu+and+badad
yyyyyyyyyyyyyyyy
kkkkkkkkkkk
jjjjjjjjjjjjjjj
+++nnnn++++uytr
++++++poiurrr
ahahahah+bobobo
++ppppp+TEXT+aaaabbbbb+cccccg++++
++++++kmsms
TEXT+fedex+redex+bidex
pududadi
A

no-whitespace-before-that
+++hhrhezipo"""

regx+=+re.compile('TEXT+*(.%2B(?<!+)(?<!\r)(?:\n[%5E+]%2B(?<!\n))?)')

for+fnd+in+regx.findall(ss):
++++print+'\n'.join(map(repr,fnd.splitlines(True)))
++++print+'---------------------------------'|code-block|syntax|javascript|52868|结果：|52869|'bla+blajjjjjjjjj\n'
'hhhhhhhh'
---------------------------------
'rumunu+and+badad\n'
'yyyyyyyyyyyyyyyy\n'
'kkkkkkkkkkk\n'
'jjjjjjjjjjjjjjj'
---------------------------------
'aaaabbbbb+cccccg'
---------------------------------
'fedex+redex+bidex\n'
'pududadi\n'
'A\n'
'\n'
'no-whitespace-before-that'
---------------------------------|52870|。|52871|如果文件是巨大的，不能在RAM中的一个块中充电，您可以这样做：|52872|import+re

ss+=+"""+++first+line
++++bli+bli+hhhhhhhh+TEXT+bla+blajjjjjjjjj
hhhhhhhh++VVVVV
ZZZZZZ
++++tttt
bolo+bolo+TEXTrumunu+and+badad
yyyyyyyyyyyyyyyy
kkkkkkkkkkk
jjjjjjjjjjjjjjj
+++nnnn++++uytr
++++++poiurrr
ahahahah+bobobo
++ppppp+TEXT+aaaabbbbb+cccccg++++
++++++kmsms
TEXT+fedex+redex+bidex
pududadi
A

no-whitespace-before-that
+++hhrhezipo"""

rigx+=+re.compile('TEXT+*(.%2B\n?)')
li+=+[]
for+line+in+ss.splitlines(True):
++++mat+=+rigx.search(line)
++++if+'TEXT'+in+line:
++++++++li.append(mat.group(1))
++++elif+'+'+in+line+and+li:
++++++++if+not+line.startswith('+'):
++++++++++++li.append(line.split('+')[0])
++++++++li[-1]+=+li[-1].rstrip('+\r\n')
++++++++print+'\n'.join(map(repr,li))
++++++++print+'====================='
++++++++li+=+[]
++++elif+li:
++++++++li.append(line)|52873|这段代码给出的结果与前一个相同。你看，这不那么简单。那是因为大文件更有问题。|52874|entityMap^0|0|C|C|Q|3|0|B|7|Z|3|15|7|0|W|G|1H|2|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@$D|17|E|18|F|G]|$D|19|E|1A|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1B|8|@$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|1I|8|@$D|1J|E|1K|F|G]|$D|1L|E|1M|F|L]]|9|@]|A|$]]|$1|M|3|N|5|O|7|1N|8|@]|9|@]|A|$P|Q]]|$1|R|3|S|5|6|7|1O|8|@]|9|@]|A|$]]|$1|T|3|U|5|O|7|1P|8|@]|9|@]|A|$P|Q]]|$1|V|3|W|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1R|8|@]|9|@]|A|$]]|$1|Z|3|10|5|O|7|1S|8|@]|9|@]|A|$P|Q]]|$1|11|3|12|5|6|7|1T|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]]|14|$]]

If you want to detect a part of a file extending on several lines, and if the file isn't too much gigantic, it's not a particularly good method to limit the power of regexes by examining one line at a time.
When the file can be read and put entirely in RAM, it is better to analyze it with a regex exploring the text as a unique whole.

Note that <code>'^'</code> has significance "beginning of the string" if flag <code>re.MULTILINE</code> isn't used, and "beginning of a line" if this flag is used.

By the way, if you employ the <code>match()</code> method, you don't need to add <code>"^"</code> at the start of the RE pattern, since <code>match()</code> try to match from the very beginning of the string.

So, here's a manner to analyze the whole text as you seem to want it (I use <code>splitlines(True)</code> to obtain a list of the lines in the string ss , this list simulates a file° :

<pre><code>import re

ss = """ first line
 bli bli hhhhhhhh TEXT bla blajjjjjjjjj
hhhhhhhh VVVVV
ZZZZZZ
 tttt
bolo bolo TEXTrumunu and badad
yyyyyyyyyyyyyyyy
kkkkkkkkkkk
jjjjjjjjjjjjjjj
 nnnn uytr
 poiurrr
ahahahah bobobo
 ppppp TEXT aaaabbbbb cccccg 
 kmsms
TEXT fedex redex bidex
pududadi
A

no-whitespace-before-that
 hhrhezipo"""

regx = re.compile('TEXT *(.+(?&lt;! )(?&lt;!\r)(?:\n[^ ]+(?&lt;!\n))?)')

for fnd in regx.findall(ss):
 print '\n'.join(map(repr,fnd.splitlines(True)))
 print '---------------------------------'
</code></pre>

result:

<pre><code>'bla blajjjjjjjjj\n'
'hhhhhhhh'
---------------------------------
'rumunu and badad\n'
'yyyyyyyyyyyyyyyy\n'
'kkkkkkkkkkk\n'
'jjjjjjjjjjjjjjj'
---------------------------------
'aaaabbbbb cccccg'
---------------------------------
'fedex redex bidex\n'
'pududadi\n'
'A\n'
'\n'
'no-whitespace-before-that'
---------------------------------
</code></pre>

.

If the file is gigantic and can't be charged in only one chunk in RAM, you can do:

<pre><code>import re

ss = """ first line
 bli bli hhhhhhhh TEXT bla blajjjjjjjjj
hhhhhhhh VVVVV
ZZZZZZ
 tttt
bolo bolo TEXTrumunu and badad
yyyyyyyyyyyyyyyy
kkkkkkkkkkk
jjjjjjjjjjjjjjj
 nnnn uytr
 poiurrr
ahahahah bobobo
 ppppp TEXT aaaabbbbb cccccg 
 kmsms
TEXT fedex redex bidex
pududadi
A

no-whitespace-before-that
 hhrhezipo"""

rigx = re.compile('TEXT *(.+\n?)')
li = []
for line in ss.splitlines(True):
 mat = rigx.search(line)
 if 'TEXT' in line:
 li.append(mat.group(1))
 elif ' ' in line and li:
 if not line.startswith(' '):
 li.append(line.split(' ')[0])
 li[-1] = li[-1].rstrip(' \r\n')
 print '\n'.join(map(repr,li))
 print '====================='
 li = []
 elif li:
 li.append(line)
</code></pre>

This code gives the same result as the former one.
You see it's less simple. That's because big-big files are more problematic.

blocks|key|52940|text|如果你不是被迫使用雷克斯，你可以用这个.|type|unstyled|depth|inlineStyleRanges|entityRanges|data|52941|将文件加载为列表：|52942|with+open("texte.txt",+"r")+as+fileInput:
++++listLines+=+fileInput.readlines()|code-block|syntax|javascript|52943|获取包含关键字的行的索引，如果它不止一次存在，您可能无法得到预期的结果，但这是一个简单的修复：|52944|listIndex+=+[i+for+i,+item+in+enumerate(listLines)+if+"text"+in+item]|52945|这是在关键字后面没有空行的行，方法是对列表进行切片：|52946|lines+=+[line+for+line+in+listLines[listIndex[0]:]+if+line]|52947|您可能希望在关键字后面使用以下任何文本：|52948|lineMatched+=+listLines[listIndex].split("text")[1].strip()|52949|并打印结果：|52950|print+"\n".join([lineMatched]+%2B+lines+if+lineMatched+else+lines)|52951|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

if you're not forced into using regexes you could use this...

Load the file as a list:

<pre><code>with open("texte.txt", "r") as fileInput:
 listLines = fileInput.readlines()
</code></pre>

Get the index of the line that contains your keyword, if it exists more than once you might not get the expected result, but it's an easy fix:

<pre><code>listIndex = [i for i, item in enumerate(listLines) if "text" in item]
</code></pre>

This are the lines that follow your keyword without blank lines, by slicing the list:

<pre><code>lines = [line for line in listLines[listIndex[0]:] if line]
</code></pre>

You might want to get any text following you keyword with:

<pre><code>lineMatched = listLines[listIndex].split("text")[1].strip()
</code></pre>

&amp; print the result:

<pre><code>print "\n".join([lineMatched] + lines if lineMatched else lines)
</code></pre>

Hi I'm looking for a way to extract a part of a text file with Python using a Regex:
here is my code:

<pre><code> texfile=open("texte.txt", "r")
 for line in texfile:
 if re.match("^text(.*)", line):
 print line,
</code></pre>

I'm seaching for the text following the word text until it reaches the end of paragraph or when it reach a white space,but my code return just the words which follow the word "text" on 1 line.

For Example :

<pre><code>bla bla hhhhhhhh text bla blajjjjjjjjjjjjjjjjjjjjj
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
ffff
</code></pre>

Must return :

<pre><code>bla blajjjjjjjjjjjjjjjjjjjjj
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
ffff
</code></pre>

<hr>

thank you I tried all your codes but no one works as I want:I want now to render it simple,extract the stream which follow a certain "text" until it reach a blank line: 

<pre><code> text
 sssssssssssssssss
 ssssssss
 kkkk
 lllmmm

 kkkk

 ;must return 
 sssssssssssssssss
 ssssssss
 kkkk
 lllmmm
 ;because of the blank line
</code></pre>

How can I extract text from a file using regex?

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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嗨，我正在寻找一种使用Python使用Regex提取文本文件一部分的方法:下面是我的代码：    texfile=open("texte.txt", "r")        for line in texfile:            if re.match("^text(.*)", line):          ...

问如何使用regex从文件中提取文本？
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问如何使用regex从文件中提取文本？EN