Passing a variable argument to egrep in a bash script

Question

I have a script, myScript, which is trying to egrep the script argument in a file. Somehow variable expansion isn't working properly with the egrep command. I believe I've isolated the problem in the example as follows: if I write out the argument explicitly in the script, the egrep command works, but if I pass the argument to the script, the egrep command doesn't like the argument I send it.

#!/bin/bash
echo "\def\\$1" > myFile
echo "\def\\$1$1" >> myFile

myVar=\\$1
echo myVar is "$myVar"

grepWorks=$(egrep '\\def\\dog\>' myFile)
echo Without a variable, grep output is $grepWorks

echo Pattern string fed to grep with variable myVar is  "\\def$myVar"
grepFails=$(egrep "\\def$myVar\>" myFile)
echo With a variable, grep output is $grepFails

When I run this script with,

myScript dog

the output is:

myVar is \dog
Without a variable, grep output is \def\dog
Pattern string fed to grep with variable myVar is \def\dog
With a variable, grep output is

Any help would be most appreciated.

Solution

Change the following line:

grepFails=$(egrep "\\def$myVar\>" myFile)

With:

grepFails=$(egrep "\\\\def\\$myVar\>" myFile)

The problem was that you were not escaping the \ properly in the subshell.

To understand, try running eval echo "\\\\". You will notice that the output is \ because of the double evaluation.

Comments

[RobertL]

Another possiblity is: '\\'def"$myVar"'\>' (kind of a tradeoff between backslashes and quotes, take your pick). Also, I'm not sure if you need the \` before the $myVar` (at least I didn't see it in the question).