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Old 05-24-2012, 05:52 AM   #1
newbiecolonopenparens
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Registered: May 2012
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Problem with analyzing command line arguments: bash shell script


So far, my code looks like this:
Code:
commands=$@
for arg in commands
do
    # echo $arg
    if [ $arg == "-print" ] then
        # do something
    if [ $arg == "-name" ] then
        # do something
    ...
done
The problem I have is that for some reason, the if statements get skipped when I type in a prompt like: ./program -name newb -print
I double checked to make sure that the correct arg was being passed in as well. Any suggestions?
 
Old 05-24-2012, 05:57 AM   #2
pan64
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you can add the following line before this loop:
set -xv
to see what's happening
 
Old 05-24-2012, 05:59 AM   #3
TobiSGD
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You forgot the fi to end the if statements:
Code:
commands=$@
for arg in commands
do
    # echo $arg
    if [ $arg == "-print" ] then
        # do something
    fi
    if [ $arg == "-name" ] then
        # do something
    fi
    ...
done
However, for test like this you should have a look at the "case" statement.
 
Old 05-24-2012, 10:19 AM   #4
grail
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I would query the purpose of assigning $@ to a variable and not just using it as is?

Also, if used directly you either need to quote it or remove it all together, like:
Code:
for arg
do
    ...
 
Old 05-24-2012, 02:25 PM   #5
newbiecolonopenparens
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Registered: May 2012
Posts: 8

Original Poster
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Well, I did include the fi's to close the if statements. This was posted before I slept, so sorry if I left those out. I did commands=$@ instead of directly doing for arg in $@ because I thought I was losing words when I used shift 1 inside the if statements
 
Old 05-24-2012, 05:56 PM   #6
newbiecolonopenparens
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Registered: May 2012
Posts: 8

Original Poster
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So, is it wrong of me to shift through my arguments, even though i keep track of it myself?
I was doing:
find
Code:
for arg in $@
do
    if [ $arg == "name" ]
    then
        shift 1
        echo "$1"
    fi
    if [ $arg == "type" ]
    then
        shift 1
        if [ -f $1 ] then
        #echo all directories out. Is this where it's wrong? how do i check the next argument in the loop?
        #maybe something like string=($arg+1)
        #so that i can do if [ -f $string ] ?
        fi
    fi
done
 
Old 05-24-2012, 08:03 PM   #7
chrism01
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1. 'shift 1'; '1' is redundant ie just 'shift' is enough/default
2. Try using getopts instead of rolling your own code http://wiki.bash-hackers.org/howto/getopts_tutorial

http://rute.2038bug.com/index.html.gz
http://tldp.org/LDP/Bash-Beginners-G...tml/index.html
http://www.tldp.org/LDP/abs/html/
 
Old 05-24-2012, 08:29 PM   #8
David the H.
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You need to first take the time to understand exactly how the shell uses arguments, and the "@" form of parameter expansion in particular.

To start with, read these three links:

http://mywiki.wooledge.org/Arguments
http://mywiki.wooledge.org/WordSplitting
http://mywiki.wooledge.org/Quotes


Now, $@ outputs all the parameter arguments as separate entities, but only if you quote it first. Otherwise the shell simply word-splits it back out again. So always double quote it. The "${array[@]}" array expansions work the same way.

Your first attempt:
Code:
commands=$@
for arg in $commands ; do
This is bad. First it puts the entire contents of $@ into a single scalar variable, then it relies on word-splitting that variable back into pieces for the for loop to iterate over.


Now if you really want to back-up/duplicate the $@ parameters, set them into an array.

Code:
commands=( "$@" )
for arg in "${commands[@]}"; do
Now the strings are never left unprotected.

Do note however that while the $@ parameters start with "1", array indexes start with "0", so you'll have to do a bit of math if you need them to match up. Or else start with a dummy 0 array entry before you populate it.

Code:
commands[0]=""		#start with an empty array 0
commands+=( "$@" )	#+= appends entries to an existing array
unset commands[0]	#unset 0, so command 1 will be the first one in the array

for arg in "${commands[@]}"; do
This way "$1" and "${commands[1]}", etc. should be identical.


How can I use array variables?
http://mywiki.wooledge.org/BashFAQ/005



Now for a few more points:


When using bash or ksh, it's recommended to use ((..)) for numerical tests, and [[..]] for string/file tests and complex expressions. Avoid using the old [..] test unless you specifically need POSIX-style portability.

http://mywiki.wooledge.org/ArithmeticExpression
http://mywiki.wooledge.org/BashFAQ/031
http://wiki.bash-hackers.org/commands/classictest
http://wiki.bash-hackers.org/syntax/...nal_expression


When testing the contents of a single variable for multiple values, you should use a single case statement, instead of multiple if..elif..thens.

http://wiki.bash-hackers.org/syntax/ccmd/case


Finally, since you appear to be trying to create an option parser of some kind, I recommend you take a look at the built-in getopts instead. No need to reinvent the wheel if you don't have to.

http://wiki.bash-hackers.org/howto/getopts_tutorial
 
  


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