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How do I print a string with variables?
Trying this
awk -F ',' '{printf /p/${3}_abc/xyz/${5}_abc_def/}' file
Need this at output
/p/APPLE_abc/xyz/MANGO_abc_def/
where ${3} = APPLE
and ${5} = MANGO
487
To print a blank line, use print "" , where "" is the empty string. To print a fixed piece of text, use a string constant, such as "Don't Panic" , as one item. If you forget to use the double-quote characters, your text is taken as an awk expression, and you will probably get an error.
If you notice awk 'print $1' prints first word of each line. If you use $3, it will print 3rd word of each line.
awk '{ print $2; }' prints the second field of each line. This field happens to be the process ID from the ps aux output.
printf allows interpolation of variables. With this as the test file:
$ cat file
a,b,APPLE,d,MANGO,f
We can use printf to achieve the output you want as follows:
$ awk -F, '{printf "/p/%s_abc/xyz/%s_abc_def/\n",$3,$5;}' file
/p/APPLE_abc/xyz/MANGO_abc_def/
In printf, the string %s means insert-a-variable-here-as-a-string. We have two occurrences of %s, one for $3 and one for $5.
130
Not as readable, but the printf isn't necessary here. Awk can insert the variables directly into the strings if you quote the string portion.
$ cat file.txt
1,2,APPLE,4,MANGO,6,7,8
$ awk -F, '{print "/p/" $3 "_abc/xyz/" $5 "_abc_def/"}' file.txt
/p/APPLE_abc/xyz/MANGO_abc_def/
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