搜索包含数字的字符串python

Question

So I have the following code:

cus_str = "show configuration " #intiate router command string variable

with open(var) as config_file: #open file
    for line in config_file: #line by line reading of file
        if '"xe-" + #anynumber + "/" #anynumber + "/" + #anynumber' in line:
            line = line.replace('\n' , '').replace('"' , '').replace(']' , '')     #removes all extra characters
            i = "| except " + (line.split("cust ", 1)[1]) + " " #split the line and save last index (cust name)
        cus_str+=i #append to string
config_file.close()

The line: if '"xe-" + #anynumber + "/" #anynumber + "/" + #anynumber' in line: Is what I'm struggling with syntax-wise.

I am looking to see if a line in the file contains the following string: "xe- number / number / number " (ex: "xe-6/1/10"). It will always have this format, but the numbers will change. What sort of syntax would I use to do this most efficiently.

Thanks!

Answer 1

You can use regular expressions for this. Regular expressions allow you to specify a pattern of text (though not all patterns can be expressed as a regular expression). We can then compile that expression into a Pattern object, and use that object to search strings for the pattern.

import re

pattern = re.compile(r'"xe-\d+/\d+/\d+"')  # \d+ is "one or more digits".  
                                           # Everything else is literal

with open(var) as config_file:
    for line in config_file:
        if pattern.search(line):  # will return a Match object if found, else None
            ...

Answer 2

Sounds like a job for the Regular Expressions library!

Is this number a date? You can restrict the amount of digits.

from re import search, compile #Import the re library with search
pattern = compile(r"xe-\d{1,2}\/\d{1,2}\/\d{1,2}") #Use this pattern to find lines

cus_str = "show configuration " #intiate router command string variable

with open(var) as config_file: #open file
    for line in config_file: #line by line reading of file
        if search(pattern, line): #Returns None if match is not found
            line = line.replace('\n' , '').replace('"' , '').replace(']' , '')     #removes all extra characters
            i = "| except " + (line.split("cust ", 1)[1]) + " " #split the line and save last index (cust name)
        cus_str+=i #append to string

The regular expression: xe-\\d{1,2}\\/\\d{1,2}\\/\\d{1,2} matches "xe-" followed by 3 groups of digit pairs with a slash separator. Each group of digits can be 1 or 2 characters in length.

Sidenotes

• You don't need to close config_file since the with statement does that for you when you exit the block.

• I'm not sure what you were trying to accomplish with cus_str += i when there is no match for the pattern. As it stands, it will just repeat the same i from the previous line, unless you indent that line by 1 level. Or give you an error if the first line doesn't contain the pattern.

Answer 3

You can test a regex with a simple program like with some online tools like https://pythex.org/ , and then make your program, a little sample to point you in the right direction would be:

import re

config_file = ["xe-6/1/10", "xf-5/5/542", "xe-4/53/32" ]
rex = re.compile(r'xe-[\d]+/[\d]+/[\d]+')
for line in config_file: #line by line reading of file
    if rex.search(line):
      print(line)

xe-6/1/10
xe-4/53/32