UNIX for Dummies Questions & Answers

Hi,

I would like to display the following message from my shell (Korn) script

Copy "old_file.txt" to "new_file.txt"

My code looks as follows

print "Copy "old_file.txt" to "new_file.txt""

However, when I execute the script, I get the following output

Copy old_file.txt to new_file.txt

Can somebody please help me with the syntax to escape the double quotes (") ?

Thanks in advance.

rogers42

Code:

$ print "Copy \"old_file.txt\" to \"new_file.txt\""
Copy "old_file.txt" to "new_file.txt"
$

You could have also gone with ...

Code:

[mddev:/home/cameron]
$ print 'Copy "old_file.txt" to "new_file.txt"'
Copy "old_file.txt" to "new_file.txt"

Thanks for the tips folks. Much appreciated.

rogers42

Hi,

I also need some quoting help. I tracked my problem down in a simple example.
I have a grep that is working if typed at the shell prompt (it is important for me to have a space in the character class):

Code:

grep -E '[a-zA-Z ]{30}:' myFile

But included in a script file (lets say file 'runMe') like the following:

Code:

#!/bin/ksh
typeset pattern='[a-zA-Z ]{30}:'
echo "Command to be executed: grep -E $pattern myFile"
grep -E $pattern myFile

Executing the file 'runMe', I get the following error:
grep: RE error in '[a-zA-Z [ ] imbalance or syntax error

I googled/searched here for the error and for quoting or inserting something else which recognizes a space character, but with no success yet.
Also I tried various quotings in runMe, like "'[a-zA-Z ]{30}:'" or "\"[a-zA-Z ]{30}:\"" or quoting the space character like "[a-zA-Z ]{30}:" but all lead to the same error message. Seems like it is breaking at the space?

Curious for me:
If I execute the script and copy the line which is printed by the 'echo' statement and paste this line at the command prompt, that works! It just doesn't work within the script...

Maybe this not the adequate topic, because its more a grep issue than a quoting issue. If so, please be patient with me

FYI:
I'm relatively new to unix.
In 'myFile', I want to (simplified for this issue) match lines, that have the first 30 characters being (a letter or a space), followed by a colon.
I've aliased grep to /usr/xpg4/bin/grep (and also tried /usr/xpg4/bin/grep manually without alias).

thanks in advance for any hints, Thinkpositiv

Well, I got it myself. I missed the double quotes when using the variable (not a declaration time of the variable)!

Code:

#!/bin/ksh
typeset pattern="[a-zA-Z ]{30}:"
echo "Command to be executed: grep -E $pattern myFile"
grep -E "$pattern" myFile

Hope this helps someone

Hi,

I'm back with a similar issue
Basically I'm grep'ing a file for a pattern. The pattern contains " (double quotes) and a space. So I escaped the quotes. But how do I use this variable in the grep call?

Code:

typeset searchString="path=\"asdf\" mode=\"jkl\""
typeset found=`grep "$searchString" $cs_rte_configtransform` #(a)
typeset found=`grep '$searchString' $cs_rte_configtransform` #(b)

Neither (a) nor (b) works. From my understanding I'd have to use (a), but searchString should contain path=\"asdf\", so I need to declare it like this?

Code:

typeset searchString="path=\\\"asdf\\\" mode=\\\"jkl\\\""

Still no results, the variable found is always empty, although the searchString manually pasted into the grep call works.

Has anyone an idea/hint for me?
Thanks, regards, Thinkpositiv