I have bash script and try run command inside it
That's ok
echo ${something:="zip -r -q $TAG -P $PASS $LOCPATH"}
>zip -r -q evolution -P evolution ~/.gconf/apps/evolution
That's ok too
zip -r -q evolution -P evolution ~/.gconf/apps/evolution
But here order have been changed only when passed values and added strange . -i
zip -r -q $TAG -P $PASS $LOCPATH
>zip error: Nothing to do! (try: zip -r -q -P evolution evolution . -i ~/.gconf/apps/evolution
Thanks for any advice.
BASH FAQ entry #50: "I'm trying to put a command in a variable, but the complex cases always fail!"
something=(zip -r -q "$TAG" -P "$PASS" "$LOCPATH")
"${something[#]}"
Try doing type zip, seems it's aliased to something.
Maybe put the full path of zip to override this ,something like :
/usr/bin/zip
Related
I have the following code on cron job, it runs but the code does not really do what it supposed to. It does not create the directory plus is does not do anything in the code. Please help check if the way I pointed to the directory is wrong.
#!/bin/bash
NAMEDATE=`date +%F_%H-%M`_`whoami`
NAMEDATE2=`date `
mkdir ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE -m 0755
mysqldump -u u3811*****_boss -p"*******" u3811*****_data | gzip ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE/db.sql.gz
echo "This is the database backup for website.com on $NAMEDATE2" |
mailx -a ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE/db.sql.gz -s "website.com Database attached" -- mail#gmail.com
chmod -R 0644 ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE/*
exit 0
Your NAMEDATE variable needs to be modified a bit, as shown below, for more information about variables in bash see this link
NAMEDATE=$(date +%F_%H-%M"_"$(whoami))
When you issue the mkdir command you will need to pass the -p option to create the complete directory structure if it doesn't exists.
mkdir -p ~/home/u3811numbers/domains/website.com/public_html/cron/backup/files/$NAMEDATE -m 0755
Also, the ~ character on Linux based distributions is used as a shortcut for the home directory of the user that invokes it so, in the line below the result is /home//home/u3811*****/domains/website.com/public_html/cron/backup/files/2020-09-04_23-13_ you can read more about it in here
In you last command before the exit, you might need to pass a wildcard (*) to avoid removing the executable bit on the directory, see below
chmod -R 0644 ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE/
The final version of your script will look something like this.
#!/bin/bash
NAMEDATE=$(date +%F_%H-%M"_"$(whoami))
NAMEDATE2=date
mkdir -p ~/home/u3811******/domains/website.com/public_html/cron/backup/files/$NAMEDATE -m 0755
mysqldump -u u3811*****_boss -p"******" u3811*****_data | gzip > ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE/db.sql.gz
echo "This is the database backup for website.com on $NAMEDATE2" | mailx -a ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE/db.sql.gz -s "website.com Database attached" -- mail#gmail.com
chmod -R 0644 ~/home/u3811*****/domains/website.com/public_html/cron/backup/files/$NAMEDATE/*
To debug a bash script you can always pass the -x flag for more information take a look at this article
I am trying to automate a procedure where the system will fetch the contents of a file (1 Url per line), use wget to grab the files from the site (https folder) and then remove the line from the file.
I have made several tries but the sed part (at the end) cannot understand the string (I tried escaping characters) and remove it from that file!
cat File
https://something.net/xxx/data/Folder1/
https://something.net/xxx/data/Folder2/
https://something.net/xxx/data/Folder3/
My line of code is:
cat File | xargs -n1 -I # bash -c 'wget -r -nd -l 1 -c -A rar,zip,7z,txt,jpg,iso,sfv,md5,pdf --no-parent --restrict-file-names=nocontrol --user=test --password=pass --no-check-certificate "#" -P /mnt/USB/ && sed -e 's|#||g' File'
It works up until the sed -e 's|#||g' File part..
Thanks in advance!
Dont use cat if it's posible. It's bad practice and can be problem with big files... You can change
cat File | xargs -n1 -I # bash -c
to
for siteUrl in $( < "File" ); do
It's be more correct and be simpler to use sed with double quotes... My variant:
scriptDir=$( dirname -- "$0" )
for siteUrl in $( < "$scriptDir/File.txt" )
do
if [[ -z "$siteUrl" ]]; then break; fi # break line if him empty
wget -r -nd -l 1 -c -A rar,zip,7z,txt,jpg,iso,sfv,md5,pdf --no-parent --restrict-file-names=nocontrol --user=test --password=pass --no-check-certificate "$siteUrl" -P /mnt/USB/ && sed -i "s|$siteUrl||g" "$scriptDir/File.txt"
done
#beliy answers looks good!
If you want a one-liner, you can do:
while read -r line; do \
wget -r -nd -l 1 -c -A rar,zip,7z,txt,jpg,iso,sfv,md5,pdf \
--no-parent --restrict-file-names=nocontrol --user=test \
--password=pass --no-check-certificate "$line" -P /mnt/USB/ \
&& sed -i -e '\|'"$line"'|d' "File.txt"; \
done < File.txt
EDIT:
You need to add a \ in front of the first pipe
I believe you just need to use double quotes after sed -e. Instead of:
'...&& sed -e 's|#||g' File'
you would need
'...&& sed -e '"'s|#||g'"' File'
I see what you trying to do, but I dont understand the sed command including pipes. Maybe some fancy format that I dont understand.
Anyway, I think the sed command should look like this...
sed -e 's/#//g'
This command will remove all # from the stream.
I hope this helps!
I am trying to make some workaround to solve a problem.
We have a gtk+ program that call a bash script who calls rdesktop.
In a machine, we discover that the rdesktop call need on extra parameter...
Since i didnt write anything of this code, and i can modify the GTK part of the problem, i can only edit the bash script that make the middle call between the calls.
i have a variable called CMD with something that look like:
rdesktop -x m -r disk:USBDISK=/media -r disk:user=/home/user/ -r printer:HP_Officejet_Pro_8600 -a 16 -u -p -d -g 80% 192.168.0.5
i need to "live edit" this line for when the printer parameter exists, it append ="MS Publisher Imagesetter" after the printer name.
The best i accompplish so far is
ladb#luisdesk ~ $ input="rdesktop -x m -r disk:USBDISK=/media -r disk:user=/home/user/ -r printer:HP_Officejet_Pro_8600 -a 16 -u -p -d -g 80% 192.168.0.5"
ladb#luisdesk ~ $ echo $input | sed s/'printer:.*a /=\"MS Publisher Imagesetter\" '/
Which return me:
rdesktop -x m -r disk:USBDISK=/media -r disk:user=/home/user/ -r ="MS Publisher Imagesetter" 16 -u -p -d -g 80% 192.168.0.5
Almost this, but i need to append the string, not replace it.
help?
Edit: i pasted incomplete exemples. fixed
Edit2:
With the help of who respond, i end up with
echo "$input" | sed 's/\(printer:\)\([^ ]*\)/\1\2="MS Publisher Imagesetter"/'
If you want the output to look like:
rdesktop -x m -r disk:USBDISK=/media -r disk:user=/home/user/ -r printer:"HP_Officejet_Pro_8600 MS Publisher Imagesetter" -a 16 -u -p -d -g 80% 192.168.0.5
This sed will do, it matches the printer: part first then the existing printer name and quotes both, if not you can adjust the replacement
variables to put the quotes/spacing where you want:
input="rdesktop -x m -r disk:USBDISK=/media -r disk:user=/home/user/ -r printer:HP_Officejet_Pro_8600 -a 16 -u -p -d -g 80% 192.168.0.5"
echo "$input" | sed 's/\(printer:\)\([^ ]*\)/\1"\2 MS Publisher Imagesetter"/'
output:
rdesktop -x m -r disk:USBDISK=/media -r disk:user=/home/user/ -r printer:"HP_Officejet_Pro_8600 MS Publisher Imagesetter" -a 16 -u -p -d -g 80% 192.168.0.5
You can use this:
sed 's/printer:[^=]\+=/\0 "MS Publisher Imagesetter"/' <<< "$input"
The \0 in the replacement pattern outputs the match itself.
I'm using a script that downloads images. I am having issues while running the wget command with a variable that contains "{" and "}" characters. It transforms into "%7B" and "%7D".
Here is part of the script
wget -nd -H -p -A jpg,jpeg,png,gif -e robots=off "$url"
The "$url" contains something like "http://webpage.com/di1/dir2/{1..26}.jpg".
I just did a Loop
for i in {1..50}
do
wget "$url$i.jpg"
done
Or: How do I prevent a sudo'ed rsync from infinite firing in a while-loop?
Because that's both what (feels like) is happening and I don't get it.
I am trying to set up a watch for syncing modified files, and it works fine. However, once I introduce the required sudo to the rsync command, a single inotify event causes the rsync command to fire indefinitely.
#!/usr/bin/env bash
inotifywait -m -r --format '%w%f' -e modify -e move -e create -e delete /var/test | while read line; do
sudo rsync -ah --del --progress --stats --update "$line" "/home/test/"
done
When you edit a file, rsync goes in rapid fire mode. But lose the sudo (and use folders to which you have permissions, of course) and the script works as expected.
Why is this?
How do I make this work correctly with the sudo command?
I have the answer, found it by experimenting. But I have no idea why this is. Please someone tell me why sudo in this loop breaks the expected blocking behavior.
Since sudo breaks the script, we can distance ourselves from sudo by using a wrapper:
This is correct:
inotifywait -m -r --format '%w%f' -e modify /var/test | while read line; do
sh -c 'sudo rsync -ah "$line" "/home/test/"'
done
Weird thing is: Pull the sudo out of the wrapper and we have the old faulty behavior again. Very strange.
This is wrong:
inotifywait -m -r --format '%w%f' -e modify /var/test | while read line; do
sudo sh -c 'rsync -ah "$line" "/home/test/"'
done