Using a variable with gsub & sub

[Peter Vanderhaden]

In the following statement, I'm replacing ,, with ,NULL,.

$inrec = $inrec.sub(',,\\',',NULL,')

What I want to do is something like this:

$inrec = $inrec.sub(',,\\',',$variable,')

Unfortunately, I haven't been able to figure out the syntax for using a
variable with sub & gsub. Can someone help me out?
Thanks.

 

[yermej]

In the following statement, I'm replacing ,, with ,NULL,.

$inrec = $inrec.sub(',,\\',',NULL,')

What I want to do is something like this:

$inrec = $inrec.sub(',,\\',',$variable,')

Unfortunately, I haven't been able to figure out the syntax for using a
variable with sub & gsub. Can someone help me out?

$inrec.gsub! ',,\\', ",#{$variable},"

Though I'm not sure why you have the \\ in there, unless it just
didn't end up on the newsgroup correctly.

 

[Harry Kakueki]

What I want to do is something like this:

$inrec = $inrec.sub(',,\\',',$variable,')

Unfortunately, I haven't been able to figure out the syntax for using a
variable with sub & gsub. Can someone help me out?

Try this.

",#{var},"

Harry

 

[Peter Vanderhaden]

yermej,
Your solution was exactly what I needed, thanks. As for the \\, that is
part of the first value that is being replaced. The first backslash is
an escape character for the second backshash.
Thanks again,
PETERV

 

[Justin Collins]

yermej,
Your solution was exactly what I needed, thanks. As for the \\, that is
part of the first value that is being replaced. The first backslash is
an escape character for the second backshash.
Thanks again,
PETERV

If you really want to use global variables ($var) then you can also do

$inrec.gsub!(',,\\', ",#$variable,")