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shell script -- awk variable substitution

Hi,

I have a script in which I'm reading a file line by line and I'm finding certain position value using awk index as

Code:

# 2 spaces before open parens

pos=`echo | awk -v Line="$Line" '{ print index("'"$Line"'","  (")}'`

When reading the file line by line and if the line is like this and when the script tries to executes the line:

Quote:

CALL ECHO ( BUILD ("person id: " , REQUEST -> PERSONID ))

Then the awk functionality is failing and giving me errors as

Code:

awk: { print index("CALL ECHO ( BUILD ("person id: " ,  REQUEST -> PERSONID ))","  (")}

awk:                                            ^ syntax error

awk: fatal: 0 is invalid as number of arguments for index

Please do let me know how to resolve this.

Thanks

Code:

Line1='CALL ECHO ( BUILD ("person id: " , REQUEST -> PERSONID ))'

Line2='CALL ECHO  ( BUILD ("person id: " , REQUEST -> PERSONID ))'

Line3='CALL ECHO ( BUILD  ("person id: " , REQUEST -> PERSONID ))'

echo "$Line1"|awk -v Substr="  (" '{ print index($0,Substr)}'

0

echo "$Line2"|awk -v Substr="  (" '{ print index($0,Substr)}'

10

echo "$Line3"|awk -v Substr="  (" '{ print index($0,Substr)}'

18

So it seems, if I understand correctly what you want to do, that your code should be :

Code:

# 2 spaces before open parens

pos=$(echo "$Line"|awk -v Substr="  (" '{ print index($0,Substr)}')

At least it seems to give what you expect.

why are you using "-v Line=$Line" when you are not using it in your awk command ?
What is $Line?

Code:

awk -v Line="$Line" '{ print index(Line, ...) }'

and you should use $() with bash , not `..`