Passing Multiple shell variables to awk script doesn't work

ripudaman.singh · 08-27-2014, 04:42 AM

echo $jdbc_url
echo $v_Password
echo $jdbc_url

awk -F"=" -v awk_v_SchemaName=$v_SchemaName -v awk_v_Password=$v_Password -v awk_jdbc_url=$jdbc_url '{
if ( $1 == "jdbc.url" ) {
print awk_jdbc_url
}
else if ( $1 == "jdbc.username" ) {
print awk_v_SchemaName
}
else if ( $1 == "jdbc.password" ) {
print awk_v_Password
}
else {
print $0
}
}' file_name >> file_name_modified

why does it print blank for awk_v_SchemaName, awk_v_Password? though echo statements on top of awk command prints assigned values

Any alternative do I have? This is hell irritating with awk, something work, something doesn't... lot many times I waste time on it and finally apt other approaches

Please help me if possible

Ygrex · 08-27-2014, 05:03 AM

there is an ENVIRON special variable available in awk:

Code:

$ awk 'BEGIN{print ENVIRON["USER"]}' < /dev/null
ygrex
$ env USER=fake-user-name awk 'BEGIN{print ENVIRON["USER"]}' < /dev/null
fake-user-name

should be suitable for your use-case

NevemTeve · 08-27-2014, 05:12 AM

If you used [code] and [/code] tags, we could read your program.

PS: here as an example:

Code:

awk -v var1='first val' \
    -v var2='second val' \
    'BEGIN { print var1; print var2; }' </dev/null

grail · 08-27-2014, 07:30 AM

What version of awk?

What derivation of awk? (ie gawk, nawk, mawk??)

Although it should be obvious from the awk script, may we have some examples of what the file contents look like? (obviously change any sensitive information values)

And as above, please use code tags.

As far as awk_v_SchemaName goes, according to your bash echoes, this variable is set to nothing as there was no bash variable called v_SchemaName