building array with awk and/or shell

Question

I used the following line to create an array of fields from a string:

var=$(echo $FieldName | awk -F";" '{print $1,$2,$3,$4,$5,$6,$7,$8,$9}')

$FieldName is a string having fields separated by semicolon.

When $FieldName contains string without whitespace it behaves properly. The problem is when it contains any whitespace character embedded then the script treats the whitespace as newline.

e.g.:

$FieldName = aa;bcd;cal;da;ea;fa;ga;ha;ia

gives [aa,bcd,cal,da,ea,fa,ga,ha,ia] which is as expected.

But $FieldName = aa;bcd;cal;da;ea;fa;ga;ha;ia <= ia2

gives [aa,bcd,cal,da,ea,fa,ga,ha,ia] , [<=, , , , , , , , ] and [ia2, , , , , , , , ]

Any ideas?

that doesn't change anything. still the whitespace creates the same problem. — Abinash Koirala
– Abinash Koirala, Commented Sep 10, 2012 at 10:57
Like @chepner also implies, the fundamental problem is probably that you are using echo $value instead of echo "$value" with proper double quotes. Always quote your variable interpolations, unless you specifically require the shell to perform word splitting and wildcard expansion on the value. — tripleee
– tripleee, Commented Sep 10, 2012 at 13:27
I think the problem is what you do with $var -- what do you do with $var? Is $var quoted when you use it later in your script? Are you using printf? — glenn jackman
– glenn jackman, Commented Sep 10, 2012 at 13:40

perreal · Accepted Answer · 2012-09-10 10:18:08Z

2

You don't need awk:

IFS=\;
set $FieldName
var=$@

answered Sep 10, 2012 at 10:18

perreal

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4 Comments

fork0 Over a year ago

Is set -- $FieldName any better?

Abinash Koirala Over a year ago

this solution is more elegant but it still can't solve the problem created by the whitespace

chepner Over a year ago

@abhayK: $9 will contain "ia <= ia2", whitespace and all. But when you expand the parameter, you need to quote it to protect that whitespace.

Thor Over a year ago

To join the string with , use this, i.e. IFS=, and var="$*".

Thor · Accepted Answer · 2012-09-10 12:41:01Z

1

You seem to be trying to change the delimiter of $FieldName, in that case use awk like this:

echo "$FieldName" | awk -v FS=',' -v OFS=';' 1

The 1 executes the default block: { print $0 }.

Or tr:

echo "$FieldName" | tr ';' ','

Output:

aa,bcd,cal,da,ea,fa,ga,ha,ia <= ia2

answered Sep 10, 2012 at 12:41

Thor

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