CODEWITHSUNDEEP
bashawk
Tim Eagle
Tim Eagle

Reputation: 51

whats going wrong? Awk output from a Temporary variable into BASH array

I declare a temporary file

    TEMPFILE="$(mktemp)"

Then I have an awk statement fill it with an output... Next step, I have another AWK statement take out particular field to put into an array... this is 'crapping out'

DATES = ($(awk -F'/' '{print $2}' '${TEMPFILE}'))

I'm also parsing out into a separate array using a CUT (not sure if it's working either)

IPS = ($(cut -f2 $(TEMPFILE)))

I'm getting the error:

Script12.sh: line 36: syntax error near unexpected token `('

Script12.sh: line 36: `DATES = ($(awk -F'/' '{print $2}' '${TEMPFILE}'))'*

Upvotes: 0

Views: 453

Answers (3)

Mingliang Liu
Mingliang Liu

Reputation: 5787

Don't use blanks for variable assignment in BASH. There should be no blanks around the = sign

Upvotes: 3

Jens
Jens

Reputation: 72717

The shell is space-sensitive. The shell syntax for assignment is

var1=[word1] ...

where the [] indicates an optional part, and ... repetition.

Note: no blanks around the = sign.

Then, there is no parameter expansion (replacing $var with its value) inside single quotes. Use double quotes:

DATES=($(awk -F/ '{print $2}' "${TEMPFILE}"))

Upvotes: 3

Ed Morton
Ed Morton

Reputation: 204229

You can't put spaces around assignments in shell and you need to change single to double quotes around your shell variable when used in your awk command line, i.e. "$TEMPFILE" not '${TEMPFILE}' and then not try to execute that variable on your cut command line (think about what $(TEMPFILE) means).

Upvotes: 2

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