Shell Programming and Scripting

Hi Gurus

I am writing a script to capture conents of a file starting from a regular expression till end of the file. When I execute sed -n '/2010/,$p' filename
I can get correct output. Now I wanted date part so I did this

sed -n '/2010\/04\/27/,$p' filename

Everything is fine till here. But as soon as I try to put the date in variable and use with sed it doest work. The date variable will not have characters escaped. My question is how can I put date in a variable and use sed to capture content of file starting from the date till end of the file

for example var1=2010/04/27

now i want to print content of my file starting $var1 till end of file ..please help

Thanks in advance
perl

This should work:

Code:

var='2010\/04\/27'
sed -n '/'$var'/,$p' filename

Afaik following should also work:

Code:

var1="2010/04/27"
sed -n ''/$var1/,$p'' filename

This can't work because $var1 contains the regexp delimiter.

That's right! Thank you for your hint. I just was too fast and did not test it, like I normally use to.

Thanks guys but i still have a problem. The value of variable var1 in the given example is not known and can differ with each execution. This variable will have a date value which will be something like var1="2010/04/27"

Therefore I can not use escape charcter since value will not be known.
Is there any way i can use delimeter other than / in sed ? i tried with may characters but it doesnt work.


thanks in advance

I read about delimiteres prior to posting here and also tried but unfortunately it doesnt work. My problem looks simple but i dont have solution

I have a variable containing date as "2010/04/27" and I want to print all lines from a file where date apprears till the end of file , and I can not hardcode the value of variable since it keeps on changing


any ideas how to achive this ?

will appriciate your help on this

Thanks