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ksh using `` or $()

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# 1  
Old 11-04-2010
ksh using `` or $()
In csh I use

Code:
opt=` echo $arg | awk 'BEGIN {FS="="} {print $1}' `

I am wondering what I should use when using ksh

Code:
opt=$(print -R $arg | awk 'BEGIN {FS="="} {print $1}')

or

Code:
opt=` print -R $arg | awk 'BEGIN {FS="="} {print $1}' `

# 2  
Old 11-04-2010
The $() are nicer, as they can nest and vi will match them up for you as long as you don't have '()' unbalanced lying around, like if you use old 'case' syntax.
# 3  
Old 11-04-2010
Thanks.

I am also getting syntax error at line 299: `(' unexpected, see below

Code:
  if [ ($optfstmod -eq 1) && ($optfrestore -eq 1) ] ; then
    msg="Can't set both -fstmod and -frestore"
    print ""
    print "ERROR: "$msg
    print "      either -fstmod or -frestore"
    print ""
    exit 1
  fi

What is the difference between these two statements??

Code:
if [ $optfstmod -eq 1 ] ; then 

if [[ $optfstmod -eq 1 ]] ; then

# 4  
Old 11-04-2010
Code:
if [ $optfstmod -eq 1 ] && [ $optfrestore -eq 1 ] ; then

---------- Post updated at 20:34 ---------- Previous update was at 20:28 ----------

[ $optfstmod -eq 1 ] is a test statement that will work in any bourne type shell and is POSIX compliant. [[ $optfstmod -eq 1 ]] is a conditional expression that is specific to ksh and bash and offers more possibilities and is a bit faster, but it is not POSIX compliant and as as such will likely not work in other Bourne type shells.
# 5  
Old 11-04-2010
Rather than running two tests you could also use -a (for and) test option, Note the -eq is for numeric comparison where = is for string.
Code:
if [ $optfstmod -eq 1 -a $optfrestore -eq 1 ] ; then

&& works fine within the [[ with the bash/ksh version, also note == and -eq are the same so no need to worry about alpha values in your vars here.
Code:
if [[ $optfstmod == 1 && $optfrestore == 1 ]] ; then

# 6  
Old 11-04-2010
Code:
if (( allows++ == (( $qqq += 3 ) % 17 ) && $xxx > 7 ))

# 7  
Old 11-05-2010
Quote:
Originally Posted by Chubler_XL
Rather than running two tests you could also use -a (for and) test option, Note the -eq is for numeric comparison where = is for string.
Code:
if [ $optfstmod -eq 1 -a $optfrestore -eq 1 ] ; then

Note: this will still work, but it is considered deprecated. See test
Quote:
&& works fine within the [[ with the bash/ksh version, also note == and -eq are the same so no need to worry about alpha values in your vars here.
== and -eq are definitely not the same. They happen to have the same result in this case. But this would also be the case with a regular test statement.
A nice thing about Conditional expressions is that they are much less picky about quoting and white space and can do pattern matching..
Quote:
Originally Posted by DGPickett
Code:
if (( allows++ == (( $qqq += 3 ) % 17 ) && $xxx > 7 ))

In bash/ksh arithmetic conditions, there is even no need for the dollar signs ( $qqq += 3 won't work anyway )
Code:
if (( allows++ == (( qqq += 3 ) % 17 ) && xxx > 7 ))
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