英文:
Finding a word in a file using Shell in Terminal on macOS
问题
我有一个文本文件，其中有一行，如下所示：
Submitting job with ID: 50fd2805-6fcc-4bc6-91a2-638528180ab6
（这是完整的一行，之前和之后还有其他行。）
我需要找到并提取ID到一个变量中。我想出了一个正则表达式模式：
[\da-f]{8}(-[\da-f]{4}){3}-[\da-f]{12}
这个模式应该有效，但似乎在Shell中无论我尝试什么都不起作用：grep、sed等。
我的一次尝试是这样的：
JOB_ID=$(grep '[\da-f]{8}(-[\da-f]{4}){3}-[\da-f]{12}' file)
明确一下，我希望JOB_ID包含字符串"50fd2805-6fcc-4bc6-91a2-638528180ab6"（不带引号）。
对于这个看似初学者的问题，我感到抱歉，但我在互联网上找不到任何答案：我已经阅读了数十个类似的问题在SO和SE上，还搜索了几个教程，但仍然无法找到答案。
感谢任何指导！
英文:
I have a text file with a line among others:
Submitting job with ID: 50fd2805-6fcc-4bc6-91a2-638528180ab6
(It's a full line, and it has other lines before and after.)
I need to find and extract to a variable the ID. I came up with a RexEx pattern:
[\da-f]{8}(-[\da-f]{4}){3}-[\da-f]{12}
The patterns should work, but it doesn't seem to work in Shell with anything I tried: grep, sed – whatever.
One of my attempts was this:
JOB_ID=$(grep '[\da-f]{8}(-[\da-f]{4}){3}-[\da-f]{12}' file)
To be clear, as a result I need JOB_ID to contain string "50fd2805-6fcc-4bc6-91a2-638528180ab6" (without quotes.)
I'm sorry for such a seemingly noob question, but I couldn't find anything on the internet: I've read dozens of similar questions on SO and SE, and googled several tutorials, and still couldn't figure out the answer.
Any guidance is appreciated!
答案1
得分: 1
你可以使用 awk 来打印以 Submitting 开头的行中的最后一个字段，如下所示：
JOB_ID=$(awk '/^Submitting/ {print $NF}' file)
英文:
You could use awk to print the last field on a line starting Submitting like this:
JOB_ID=$(awk '/^Submitting/ {print $NF}' file)
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