Quoting Paul M Foster <paulf@quillandmouse.com>:
> On Thu, May 19, 2005 at 11:00:02PM -0400, Bob Stia wrote:
>
> > Hello Sluggers,
> >
> > What is wrong with this script??
>
> Bob, by putting out a script like this without any explanation, we
> assumed you wrote it, or knew how it worked. It looks, instead, like you
> copied it from somewhere else. Right?
>
> I'll go over it, and try to explain the prior comments.
>
> > --------------------------------------------------
> > #!/bin/sh
> >
> > #########################################################################
> > # A simple script to create smaller and lower quality images from
> > # one directory to another
> > #########################################################################
> >
> > # The file format you wish to change
> > FROM=jpg
> >
> > # The end result file format
> > TO=jpg
> >
> > # Options for "convert"
> > CONVERT_OPTIONS='-quality 70 -geometry 800x600'
> >
> > # Pull the directory name for the title
> > FROM_DIR=/home/bob/WebPages/PorschePilots/fog/temp
> > TO_DIR=/home/bob/Webpages/PorschePilots/fog/carlouelimages
> >
> > #########################################################################
> > # The actual conversion stuff
> > # You can edit stuff below here to change how this works, or just stick
> > # with the variables above (which should be good for most people
> > #########################################################################
> >
> > echo Running $0
> > echo
> >
> > COUNT=0
>
> Someone asked if this is an unused variable. You've said here you want
> to create a variable called COUNT, and you want to assign it a 0 value.
> That's fine except that it's not referenced anywhere else in the script.
> So it appears you're just assigning a variable a number for no reason,
> thus an "unused variable". You can eliminate this line if you like.
>
> >
> > cd $FROM_DIR
>
> Someone else mentioned $VARNAME versus ${VARNAME}. No, you won't find
> this exact thing in your script, but the above is an example of what the
> poster meant. When he said "VARNAME", he meant "variable name". Anything
> preceded by a $ in a bash script is a variable. In this case, the
> variable name is $FROM_DIR. In bash, you can also say ${FROM_DIR}, and
> it means mostly the same thing. In most cases, it's a matter of
> preference which form you use. In some cases you need to use ${FROM_DIR}
> instead. But I don't think you have any of that kind of thing in your
> script.
>
> You'll note that earlier, you created the COUNT variable and assigned it
> a 0 value. Notice you didn't have a $ in front of it. When you first
> create a variable and assign a value to it, you don't put the $ there.
> But later if you wanted to _use_ the COUNT variable, like:
>
> echo $COUNT
>
> you'd have to put the $ in front of it, or bash would choke. Um, I
> should also say that "echo" tells bash to print out everything else
> that's on that line. So the line above would tell bash to print out the
> value of the COUNT variable.
>
> > for i in *.$FROM; do
> >
> > echo $1
>
> The above appears to by line 35 by my count. Terms like $0, $1, $2
> indicate the parameters with which you call the program. $0 is always
> the name of the script. So for example, if you said:
>
> bobs_script joe sam
>
> then those variables would be as follows:
>
> $0 = bobs_script
> $1 = joe
> $2 = sam
>
> You can see from this how it would progress; the next thing on the
> command line corresponds to the next number.
>
> In this case, I don't think you're actually passing the script any
> parameters, so the $1 is meaningless. I suspect this should be
>
> echo $i
>
> instead of echo $1. That way the script would tell you what file it was
> processing for each loop.
>
> If my count is right, I don't know why your script is telling you this
> line is an unknown command. I suspect there is some alteration between
> the running script and what you posted. Otherwise, it doesn't make
> sense.
>
> > if [ ! -e "$TO_DIR/$i" ] # Process only files not already done
> > then
> > echo Converting $i... with $CONVERT_OPTIONS
> > convert $CONVERT_OPTIONS $FROM_DIR/$i $TO_DIR/${i%%$FROM}$TO
> > echo
> > fi
> >
> > done;
> >
> > exit;
> > ----------------------------------------------------------------------
> > This is what I get:
> >
> > /home/bob/commands/ImageMagickConvert: line 35: : command not found
> > /home/bob/commands/ImageMagickConvert: line 36: : command not found
> > /home/bob/commands/ImageMagickConvert: line 37: : command not found
> > /home/bob/commands/ImageMagickConvert: line 38: : command not found
> > /home/bob/commands/ImageMagickConvert: line 39: : command not found
> > /home/bob/commands/ImageMagickConvert: line 40: : command not found
> > /home/bob/commands/ImageMagickConvert: line 41: : command not found
> > bob@EasyStreet:/>
>
> Make the modification(s) I suggested above, and try the script again. If
> still no joy and the error is the same, try to cut and paste the script
> into an email back to the list (so there are no alterations). We can
> look at it again.
>
> Paul
Could it be that the script cannot find the convert command on this box? That
might explain why it is giving the command not found.
Try doing a which convert from the command line and see what it tells you. If it
does not come back with the full path theres your answer. On my system (debian
unstable/experimental) it is located at /usr/bin/convert.
HTH
Sean
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