[Solved] Python Regex starting with hashtag

by


Just split by a newline and get the first element:

test_str = "# <init> (Lorg/cyanogenmod/audiofx/ActivityMusic;)V\n         , 0 iput-object v1, v0, Lorg/cyanogenmod/audiofx/ActivityMusic$11;->this$0 Lorg/cyanogenmod/audiofx/ActivityMusic;\n         , 4 invoke-direct v0, Ljava/lang/Object;-><init>()V\n         , a return-void "
print(test_str.split('\n')[0])

See demo

Output: # <init> (Lorg/cyanogenmod/audiofx/ActivityMusic;)V.

If it is not the first line:

A non-regex way:

test_str = "some string\n# <init> (Lorg/cyanogenmod/audiofx/ActivityMusic;)V\n         , 0 iput-object v1, v0, Lorg/cyanogenmod/audiofx/ActivityMusic$11;->this$0 Lorg/cyanogenmod/audiofx/ActivityMusic;\n         , 4 invoke-direct v0, Ljava/lang/Object;-><init>()V\n         , a return-void\n# <init> (Lorg/cyanogenmod/audiofx/ActivityMusic;)V "
ss = test_str.split('\n')
for s in ss:
    if s.strip()[0] == "#":
        print s
        break

For a regex way, see this:

p = re.compile(r'^#.*', re.M)
print p.search(test_str).group()

The main point in the regex approach is

  • Use re.M multiline flag
  • Use re.search that will return just 1 match object
  • The # must be the first character (or add \s* – optional whitespace) so that the line starting with it could be matched.

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solved Python Regex starting with hashtag