grep - searching for a specific string

Shell Programming and Scripting
1

ppl,

this is my "file" with fields

orderno orderdate orderdesc telno street city
1 01/04/2006 abc 123 100 tampa
2 01/04/2006 abc 123 100 tampa
3 01/04/2006 abc 123 100 tampa
4 01/04/2006 abc 123 100 tampa
5 01/04/2006 abc 123 100 tampa
6 01/04/2006 abc 123 100 tampa

searching for lines with ordernos 3,4 and 5 in a loop as:

ll=3
ul=5

while [ $ll -le $ul ]
do
grep '^$ll\>' file
ll=$((ll+1))
done

gives me null output.

pls help with the correct expression to retrieve the required lines. UNDERLINE.. the key to search is only the first field.. all other fields are same and can't be used. (p.s. however the expression is working with substituted values of $ll (if i hard code $ll with 3, 4 or 5.. i get correct results)

Thanks a lot in advance,
Sirisha

2

From man sh

       Enclosing  characters  in  single quotes preserves the literal value of
       each character within the quotes.  A single quote may not occur between
       single quotes, even when preceded by a backslash.

       Enclosing  characters  in  double quotes preserves the literal value of
       all characters within the quotes, with the exception of $,  ?,  and  \.
       The  characters  $  and  ?  retain  their special meaning within double
       quotes.  The backslash retains its special meaning only  when  followed
       by one of the following characters: $, ?, ", \, or <newline>.  A double
       quote may be quoted within double quotes by preceding it with  a  back-
       slash.

Replace the ' quotes with "

grep "^$ll\>" file

3

Thanks a lot Vino.. that was of really helpful!

regards
sirisha