How to include shell scripts in one big shell script?

worchyld · #1 (**permalink**) 03-17-2009

I am wanting to create one shell script which I run, as I want it to run a series of other shell scripts and PHP scripts in sequence. But, I am not sure how to include shell scripts.

Note: Paths changed for security/aesthetic reasons.

Code:

 
#!/bin/sh  
# FILE: '_final.sh' -- a list of shell scripts and php files to run. 
sh /path/to/delete.sh 
# 
# Just ignore these following lines for the moment 
#sh /path/to/mycronjobs/unzip.sh 
#php /path/to/mycronjobs/ci_cron.php --run=/cron/run/ --show-output 
#sh /path/to/mycronjobs/mysql_import.sh #php /path/to/mycronjobs/ci_cron.php --run=/cron/removeBlankRows/ --show-output

The problem is that whenever I try to run this on the command line it complains that:

Quote:

: No such file or directorynal.sh: line 4: /path/to/mycronjobs/delete.sh

I do not know why it keeps saying "directorynal.sh" for.

I've looked on Google, the most I got was that I am meant to either use the 'source' command or the '.' command, but without explanation of how to use it. I tried both of the following;

Code:

 
#!/bin/sh  
. /path/to/delete.sh 
source /path/to/delete.sh

But I still get the same result/error.

Can someone enlighten me, how do I include the shell scripts? Is there an include function or something?

Thanks.

g.pi · #2 (**permalink**) 03-17-2009

Does delete.sh invoke directorynal.sh? To me, it seems like delete.sh is being called correctly.

worchyld · #3 (**permalink**) 03-17-2009

Quote:

Does delete.sh invoke directorynal.sh? To me, it seems like delete.sh is being called correctly.

I don't think so... I didn't even know you could invoke directorynal.sh? There's nothing in my delete.sh file that suggests this.

But then why does it complain that it can't find the file/directory/contents, even if it exists?

Delete.sh does this;

Code:

find /path/to/inbounddata/data/* -depth -type d ! -name data -ctime 24 -exec /bin/rm -Rf "{}" \;

If I try to write everything in this file, or put anything after the \; it complains that I did not use -exec properly.

I can't even use a "echo 'done'" in this file, so this is part of the reason why I've seperated them all.

g.pi · #4 (**permalink**) 03-17-2009

The find command may be picking up that file. You can try this, to confirm: find /path/to/inbounddata/data/* -depth -type d ! -name data

Check if the list returned contains a file named, directorynal.sh.

If it returns a whole lot of files, redirect the output to a file.

cfajohnson · #5 (**permalink**) 03-17-2009

Did you write the script on a Windows box? If so, you need to remove the carriage returns (use dos2unix).

worchyld · #6 (**permalink**) 03-18-2009

Quote:

Did you write the script on a Windows box? If so, you need to remove the carriage returns (use dos2unix).

I used putty on a windows machine, I don't know if that is what you mean.

Quote:

The find command may be picking up that file. You can try this

I'll try that and report back...

worchyld · #7 (**permalink**) 03-18-2009

Hi there.

I ran:

Code:

find /path/to/inbounddata/data/* -depth -type d ! -name data

It returned:

Quote:

/path/to/inbounddata/data/2009-03-17

Which is correct, as there is only 1 folder right now.

Quote:

If it returns a whole lot of files, redirect the output to a file.

What do you mean by "redirect the output to a file", what does this do and what benefit does this have?

EDIT/UPDATE.

It appears there is an error with my delete.sh file

Code:

# It only works if I set the ctime to 0
find /path/to/inbounddata/data/* -depth -type d ! -name data -ctime 24 -exec /bin/rm -Rf "{}" \;

Quote:

# I always get this output if I set the ctime to anything but 0.
find: missing argument to `-exec'

Is there another way to resolve this problem?

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