Shell Programming and Scripting

Hello,
I am trying to print a formatted test using awk command and its not working for me:

for ex: there is one file matching the pattern test as shown below

Code:

$ ls -ltr test* | awk '{print $9}'
test.dat

now, I want to print to a file the test file_name: test.dat using awk like

Code:

$  FLAG=file_name
$ls -ltr test* | awk '{print "$FLAG",$9}'
$FLAG test.dat

In the above awk command, I want to print the value of variable FLAG which will have some value (set above to file_name, this value can change at runtime) but I am not able to print it as it prints the variable name and not its value.

Any help is appreciated. Txs.

Please use code tags, not icode tags for console output and commands.

Try either one:

Code:

FLAG=file_name
ls -ltr test* | awk -v FLAG=$FLAG '{print FLAG,$9}'
ls -ltr test* | awk -v FLAG=$FLAG '/FLAG/ {print $9}'

hth

Hello calredd,

Following may help you in same.

Code:

FLAG=file_name
ls -l test* | awk -vflag="$FLAG" '{print flag,$9}'

EDIT:
Hello Sea,

There is no need of using ls -ltr, in spite ls -l can be used

Thanks,
R. Singh

It worked. Thanks both, really appreciate it.

Why not printf "${FLAG} %s\n" test* to avoid the ls, awk, and pipe. ${FLAG} cannot contain any % characters of course.

Hello Mr Singh,
I need to print two variables this time, like:

Code:

 FLAG=file_name
 TIME=file_time
awk -v flag="$FLAG" '{print flag,$9}' test*

In the above code, I need to print TIME also after $9 but its not working. How can we print multiple variables using awk?
Thanks.

Shell variables are not awk variables. It won't see TIME if you don't do -v TIME="$TIME"

ls file* | awk '{ print $9 }' is something VERY different from awk '{ print $9 }' file* by the way. The second actually reads the contents of the files in question. Don't swap them.