On Tuesday 24 May 2005 01:19, Paul M Foster wrote:
> 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?
Ummmmm....BLUSH !!! Yes, I copied it and modified it to work ?? for me.
Made the suggested changes by you and Eben and still doesn't work. The
copied script is below.
>
> I'll go over it, and try to explain the prior comments.
>
...........<snip some stuff>.............
OK, Thanks
> >###### # 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.
OK, understood and removed.
>
> > 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.
OK, understood and learning, (hopefully)
>
> 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 $1. That way the script would tell you what file it
> was processing for each loop.
OK, changed that.
>
> If my count is right, I don't know why your script is telling you
> this line is an unknown command. I suspect there is some alteration
> between the running script and what you posted. Otherwise, it doesn't
> make sense.
>
> > if [ ! -e "$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;
> > -------------------------------------------------------------------
.....................<snip more>...............
>
> 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
Sean, running SuSE 9.2 Pro. Yes, my convert is at /usr/bin/convert also.
Kwan, would like to try your suggested changes but if they are not
necessary I'd like to get this to work first.
Eben, You said, "Is this /home/bob/commands/ImageMagickConvert ?"
Yes, that is where the script is and how I run it.
Following is the script after Paul's and your suggestions:
------------------------------------------------------------------------------------
#!/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
cd $FROM_DIR
for i in *.$FROM; do
echo $i
if [[ ! -e "$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;
---------------------------------------------------------------------------------------
Bob S.
-----------------------------------------------------------------------
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