正则表达式以获取特定模式

Question

I need to fetch the specific rule from the filter list that how many rules are there in with this category.

I have tried to fetch this type of rule from the filter list. The rule pattern is as follows

" /example.com $script,domain=example.com "

and the second exception rule is

"@@/example.com $script,domain=example.com "

third rule with domain anchor is

"||example.com

whereas fourth rule with anchor and domain tag is

"||jizz.best^$popup,domain=vivo.sx

and the fifth one is

"@@||pagead2.googlesyndication.com/pagead/js/adsbygoogle.js$script,domain=quebeccoupongratuit.com

6th one with domain restriction is as follows

"example.com###examplebanner

7th without domain restriction is

"###examplebanner

8th is exeptional with element hidding

example.com#@##examplebanner

These are different categories of rules I have to fetch separately

a=open('1-19-16anti-adblock-killer-filters.txt','r')
text=a.read()
 line_starts_with_2pipes_no_domain = 0
  line_starts_with_2pipes_with_domain = 0
 line_starts_with_2ats_with_domain = 0
 line_with_domain = 0

 for line in text.split("\n"): 
   if line.startswith("||"):
    if ",domain" in line: 
        line_starts_with_2pipes_with_domain += 1
    else:
        line_starts_with_2pipes_no_domain += 1
  elif line.startswith("@@") and ",domain" in line:
    line_starts_with_2ats_with_domain += 1
   elif ",domain" in line: 
    line_with_domain += 1
   elif line.strip(): 
      print(f"No idea what to do with :{line}")

print("2pipes_no_group", line_starts_with_2pipes_no_domain ) 
print("2pipes_with_group", line_starts_with_2pipes_with_domain ) 
 print("2@_with_group", line_starts_with_2ats_with_domain ) 
 print("line_with_domain", line_with_domain)

i am trying now to fetch 5th , 6th ,7th and 8th rule . Your response will be appreciated, thanks.

Answer 1

Your regex does not fit the , before domain:

"\/[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+domain="
#                  ^^^^^^^^^^^^ no , allowed

You can also simplify this a lot:

with open("easylist.txt") as f: 
    print('There are total Rule With Domain tag are =', f.read().count(",domain="))

should give you your answer of how often ',domain=' occures. If your file is big, you can also count linewise:

domain_rule_count = 0
with open("easylist.txt") as f:
    for line in f:
        domain_rule_count += 1 if ",domain=" in line else 0

---

Edit after question in comment: You simply test for what you want:

text = """ some text
/example.com $script,domain=example.com
@@/example.com $script,domain=example.com 
||example.com
||jizz.best^$popup,domain=vivo.sx
"""

line_starts_with_2pipes_no_domain = 0
line_starts_with_2pipes_with_domain = 0
line_starts_with_2ats_with_domain = 0
line_with_domain = 0

for line in text.split("\n"): 
    if line.startswith("||"):
        if ",domain" in line: 
            line_starts_with_2pipes_with_domain += 1
        else:
            line_starts_with_2pipes_no_domain += 1
    elif line.startswith("@@") and ",domain" in line:
        line_starts_with_2ats_with_domain += 1
    elif ",domain" in line: 
        line_with_domain += 1
    elif line.strip(): 
        print(f"No idea what to do with '{line}'")

print("2pipes_no_group", line_starts_with_2pipes_no_domain ) 
print("2pipes_with_group", line_starts_with_2pipes_with_domain ) 
print("2@_with_group", line_starts_with_2ats_with_domain ) 
print("line_with_domain", line_with_domain)

Output:

No idea what to do with ' some text'
2pipes_no_group 1
2pipes_with_group 1
2@_with_group 1
line_with_domain 1