如何提取PAT1与第3次出现的PAT2之间的文本行？

Hey there! Let's figure out how to get exactly the content you need. Your input is a single line where PAT1 appears once, and PAT2 shows up multiple times—you want everything from PAT1 up to (and including) the 3rd PAT2. The basic awk '/PAT1/PAT2/' command stops at the first PAT2, so we need a smarter approach.

Solution for Single-Line Input (Your Case) #

Since your content is all in one line, we can use a non-greedy regular expression in awk to target the exact substring:

awk '{
    match($0, /PAT1.*?PAT2.*?PAT2.*?PAT2/);
    print substr($0, RSTART, RLENGTH)
}' your_file.txt

Let's break this down:

• match($0, /PAT1.*?PAT2.*?PAT2.*?PAT2/): This regex looks for PAT1, then the shortest possible text until the first PAT2, repeats that twice more to reach the 3rd PAT2. The .*? is key here—it's non-greedy, so it won't jump to the last PAT2 in the line.
• substr($0, RSTART, RLENGTH): After match() finds the target substring, it sets RSTART to the start position and RLENGTH to the length of the match. We use these to extract and print the exact part we want.

When you run this on your sample input:

1 2 PAT1 3 4 PAT2 5 6 PAT2 7 PAT2 8 9 PAT2 10 

You'll get your desired output:

PAT1 3 4 PAT2 5 6 PAT2 7 PAT2

Bonus: Solution for Multi-Line Input #

If your content was spread across multiple lines (e.g., PAT1 on one line, PAT2 on separate lines), we'd use a flag-and-count approach instead:

awk '/PAT1/{flag=1} flag{print; if(/PAT2/) count++} count==3{exit}' your_file.txt

How this works:

• /PAT1/{flag=1}: Flip a flag to start capturing content when we hit PAT1.
• flag{print; if(/PAT2/) count++}: Print every line while the flag is set, and increment a counter each time we see PAT2.
• count==3{exit}: Stop processing as soon as we hit the 3rd PAT2.

内容的提问来源于stack exchange，提问作者dbar