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Passing a variable to sed command

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# 1  
Old 10-07-2009
Passing a variable to sed command
Hi guys,

I wanted to pass a variable to the sed command which tells which line to be deleted.

Code:
a=2;
echo $a;
sed '$ad' c.out

it is throwing an error.
Code:
sed: 0602-403 "$a"d is not a recognized function.

I even tried "$a" and \$a.. but it is of no use.

Can you please correct me where i am going wrong.

Thanks,
Magesh.
# 2  
Old 10-07-2009
Hi.

You should enclose the variable in braces: ${a} - and use double quotes for your sed.

Cheers

Code:
wc Test
      20      68     452 Test

a=2
sed "${a}d" Test | wc
      19      63     431

# 3  
Old 10-07-2009
Code:
 export a=2;sed "${a}d" file

# 4  
Old 10-07-2009
cool man.. i missed the double quotes.. But usually sed will use only single quotes, please correct me if i am wrong, then when we should use double quotes??
# 5  
Old 10-07-2009
Quote:
Originally Posted by mac4rfree
cool man.. i missed the double quotes.. But usually sed will use only single quotes, please correct me if i am wrong, then when we should use double quotes??
Yes, you're wrong Smilie

You can use either, but if you're using variables, or anything that the shell needs to expand you should use double quotes.
# 6  
Old 10-07-2009
Quote:
Originally Posted by mac4rfree
Code:
sed '$ad' c.out

Code:
sed "/${a}/d" c.out

# 7  
Old 10-07-2009
Thanks.. for the explanation..
finally the code which worked is

Code:
sed "${a}d" c.out

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