如何在Unix Shell脚本中找到文件中字符串的位置

Question

Can you please help me solve this puzzle? I am trying to print the location of a string (ie, line #) in a file, first to the std output, and then capture that value in a variable to be used later. The string is “my string”, the file name is “myFile” which is defined as follows:

this is first line
this is second line
this is my string on the third line
this is fourth line
the end

Now, when I use this command directly at the command prompt:

% awk ‘s=index($0, “my string”) { print “line=” NR, “position= ” s}’ myFile

I get exactly the result I want:

% line= 3, position= 9

My question is: if I define a variable VAR=”my string”, why can't I get the same result when I do this:

% awk ‘s=index($0, $VAR) { print “line=” NR, “position= ” s}’ myFile

It just won't work!! I even tried putting the $VAR in quotation marks, to no avail? I tried using VAR (without the $ sign), no luck. I tried everything I could possibly think of ... Am I missing something?

Answer 1

awk variables are not the same as shell variables. You need to define them with the -v flag

For example:

$ awk -v var="..." '$0~var{print NR}' file

will print the line number(s) of pattern matches. Or for your case with the index

$ awk -v var="$Var" 'p=index($0,var){print NR,p}' file

using all uppercase may not be good convention since you may accidentally overwrite other variables.

to capture the output into a shell variable

$ info=$(awk ...)

for multi line output assignment to shell array, you can do

$ values=( $(awk ...) ); echo ${values[0]}

however, if the output contains more than one field, it will be assigned it's own array index. You can change it with setting the IFS variable, such as

$ IFS=$(echo -en "\n\b"); values=( $(awk ...) )

which will capture the complete lines as the array values.