Shell Programming and Scripting

I have question about awk if someone can help me here. I want to find a string (file has just one row) and then get the position of the column.

let say i have 4 columns

a b c d

Look for b and somehow get the column position in this case b is at $2 script doesn't know what is the position of the column. So if awk or something else can help me to get position of the column.

Thanks in Advance.

It is necesary to do it with awk?

Because a script can do it

.........
cont=0
for i in `cat (the file)`
do
if [ `echo $i` = (the name you would like to find) ]
then
break;
else
(( cont = cont + 1 ))
done
echo "the position is -$cont-"
.........

Bye

Code:

ruby -e 'p gets.split.index("bar").succ' myfile

Code:

newlisp -e '(+ 1(find "foo" (parse (read-line))))' <myfile

In awk:

Code:

$ cat file 
a b c d
e c g h
a b j c
$ awk -v var=c '{for (i=0;i<=NF;i++){if(var == $i){ print "Var "var" found in column "i" line "NR }}}' file
Var c found in column 3 line 1
Var c found in column 2 line 2
Var c found in column 4 line 3

Try...

Code:

echo 'a b c d' | awk 'BEGIN{RS=FS}/b/{print NR}'

Thanks for everybody's input couldn't accomplish my mission yet

Output of $* (a b c d)

typeset -u P=`echo $*|awk 'BEGIN{RS=FS}/b/{print "$"NR}'`

Rather then giving me b it gives me $2.

Thanks in advance.

Let me explain little bit what i am trying to accomplish i want to find a string in arugments that have been passed to script then get the column number and based on column number get the value of that arugment.

Thanks