使用Bash脚本查找文件中的字符串行数

Question

How can I use a bash script to find the line number where a string occurs?

For example if a file looked like this,

Hello I am Isaiah
This is a line of text.
This is another line of text.

and I ran the script to look for the string "line" it would output the number 2, as it is the first occurance.

Answer 1

Given that your example only prints the line number of the first occurrence of the string, perhaps you are looking for:

awk '/line/{ print NR; exit }' input-file

If you actually want all occurrences (eg, if the desired output of your example is actually "2\\n3\\n"), omit the exit .

Answer 2

I like Siddhartha's comment on the OP. Why he didn't post it as an answer escapes me.

I usually just want the line number of the first line that shows what I'm looking for.

lineNum="$(grep -n "needle" haystack.txt | head -n 1 | cut -d: -f1)"

Explained: after the grep, grab just the first line ( num:line ), cut by the colon d elimiter and grab the first f ield