in a script this
(cd "$amdir/archive" && zip -rm "$amdir/archive/a.zip" "$amdir/archive/*")
causes zip warning: name not matched
when I echo that and copy paste it to a command line it works fine
any idea why that doesn't work in a bash script on linux ?
You are already in $amdir/archive after your cd.
So your zip tries to find another directory $amdir/archive, when already being there.
I can reproduce the error message, when I try to do a zip -rm on a nonexisting directory.
In addition you should consider the remark from Inian: by quoting the * you escape it and therefore have no bash pattern matching - see for example here
So the second part should simply read:
zip -rm a.zip *
Related
Example:
Here's list of files in "/tmp/test_dir"
file1
zip -r Test_Files.zip *
When I unzip Test_Files.zip I'm getting the below output
Current working directory "/tmp/test_dir"
/tmp/test_dir/file1
What I'm expecting when I unzip Test_Files.zip;
/tmp/test_dir/Test_Files/file1
Can anyone help how do i get expected result as mentioned above?
Use unzip. You can use -o to overwrite the existing files and -q to make it quiet. In doubt? Just use terminal and type in unzip (or try /usr/bin/unzip) to see helpful information.
I'm new to Bash scripting. My script intended role is to access a provided path and then apply some software (RTG - Real time Genomics) commands on the data provided in that path. However, when i try to execute the bash from CLI, it gives me following error
ERROR:There were invalid input file paths
The path I have provided in the script is accurate. That is, In the original directory, where the program 'RTG' resides, I have made folders accordingly like /data/reads/NA19240 and placed both *_1.fastq and *_2.fastq files inside NA19240.
Here is the script:
#!/bin/bash
for left_fastq in /data/reads/NA19240/*_1.fastq; do
right_fastq=${left_fastq/_1.fastq/_2.fastq}
lane_id=$(basename ${left_fastq/_1.fastq})
rtg format -f fastq -q sanger -o ${lane_id} -l ${left_fastq} -r ${right_fastq} --sam-rg "#RG\tID:${lane_id}\tSM:NA19240\tPL:ILLUMINA"
done
I have tried many workarounds but still not being able to bypass this error. I will be really grateful if you guys can help me fixing this problem. Thanks
After adding set -aux in bash script for debugging purpose, I'm getting following output now
adnan#adnan-VirtualBox[Linux] ./format.sh
+ for left_fastq in '/data/reads/NA19240/*_1.fastq'
+ right_fastq='/data/reads/NA19240/*_2.fastq'
++ basename '/data/reads/NA19240/*'
+ lane_id='*'
+ ./rtg format -f fastq -q sanger -o '*' -l '/data/reads/NA19240/*_1.fastq' -r '/data/reads/NA19240/*_2.fastq' --sam-rg '#RG\tID:*\tSM:NA19240\tPL:ILLUMINA'
Error: File not found: "/data/reads/NA19240/*_1.fastq"
Error: File not found: "/data/reads/NA19240/*_2.fastq"
Error: There were 2 invalid input file paths
You need to set the nullglob option in the script, like so:
shopt -s nullglob
By default, non-matching globs are expanded to themselves. The output you got by setting set -aux indicates that the file glob /data/reads/NA19240/*_1.fastq is getting interpreted literally. The only way this would happen is if there were no files found, and nullglob was disabled.
In the original directory, where the program 'RTG' resides, I have
made folders accordingly like /data/reads/NA19240 and placed both
*_1.fastq and *_2.fastq files inside NA19240.
So you say, your data folders are in the original directory (whatever that may be), but in the script you wrongly specify them to be in the root directory (by the leading /).
Since you start the script in the original directory, just drop the leading / and use a relative path:
for left_fastq in data/reads/NA19240/*_1.fastq
I'm trying to run a script that installs some files in a directory a user specifies. Once the user specifies the directory, I'd like to transfer the main file to that directory so it can perform so more tasks there before ultimately deleting itself once complete.
#prompt for directory in which to build project
read -p "Drag and drop the directory in which you'd like to build this project: "
echo "reply is $REPLY"
cp ./myScript.sh $REPLY
/bin/bash $REPLY/myScript.sh
I've got the script to execute the file from this question. I tried doing it with source $REPLY/myScript.sh as well as simply sh $REPLY/myScript.sh. I get the error /path/to/file/ is a directory
It must be that it doesn't known I'm trying to run myScript.sh, but I don't understand how I've given it a directory.
A likely cause is that drag-and-drop is putting whitespace after the directory name.
Thus:
/bin/bash $REPLY/myScript.sh
would be running
/bin/bash /path/to/directory /myScript.sh
A simple fix, if that's only a standard space, would be:
/bin/bash "${REPLY% }/myScript.sh"
You are missing the variable in read command so obiously it will fail as whatever you are reading is not getting stored. You can replace the read command as follows.
#prompt for directory in which to build project
read -p "Drag and drop the directory in which you'd like to build this project: " REPLY
I have a directory containing a set of subdirectories and files. I need to recursively copy all the content of this directory to all the subdirectories of another directory, also recursively.
How do I achieve this, preferably without using a script and only with the cp command?
You can write this in a script but you don't have to. Just write it line by line in the terminal:
# $TARGET is the directory containing subdirectories where you want to STORE the copies
# $SOURCE is the directory containing the subdirectories you want to COPY
for dir in $(ls $TARGET); do
cp -r $SOURCE/* $TARGET/$dir
done
Only uses cp and runs on both bash and zsh.
You can't. cp can copy multiple sources but will only copy to a single destination. You need to arrange to invoke cp multiple times - once per destination - for what you want to do; using, as you say, a loop or some other tool.
The first part of the command before the pipe instruct tar to create an archive of everything in the current directory and write it to standard output (the – in place of a file-name frequently indicates stdout).
tar cf - * | ( cd /target; tar xfp -)
The commands within parentheses cause the shell to change directory to the target directory and untar data from standard input. Since the cd and tar commands are contained within parentheses, their actions are performed together.
The -p option in the tar extraction command directs tar to preserve permission and ownership information, if possible given the user executing the command. If you are running the command as superuser, this option is turned on by default and can be omitted.
Also you can use the following command, but it seems to be quite slower than tar;
cp -a * /target
I'm writing a PHP script that downloads a series of generated files (using wget) into a directory, and then zips then up, using the zip command.
The downloads work perfectly, and the zipping mostly works. I run the command:
zip -r /var/www/oraviewer/rgn_download/download/fcst_20100318_0319.zip /var/www/oraviewer/rgn_download/download/fcst_20100318_0319
which yields a zip file with all the downloaded files, but it contains the full /var/www/oraviewer/rgn_download/download/ directories, before reaching the fcst_20100318_0319/ directory.
I'm probably just missing a flag, or something small, from the zip command, but how do I get it to use fcst_20100318_0319/ as the root directory?
I don't think zip has a flag to do that. I think the only way is something like:
cd /var/www/oraviewer/rgn_download/download/ && \
zip -r fcst_20100318_0319.zip fcst_20100318_0319
(The backslash is just for clarity, you can remove it and put everything on one line.)
Since PHP is executing the command in a subshell, it won't change your current directory.
I have also get it worked by using this command
exec('cd '.$_SERVER['DOCUMENT_ROOT'].' && zip -r com.zip "./"');
cd /home/public_html/site/upload/ && zip -r sub_upload.zip sub_upload/
Use the -j or --junk-paths option in your zip command.
From the zip man page:
-j
--junk-paths
Store just the name of a saved file (junk the path), and do not store
directory names. By default, zip will store the full path (relative
to the current directory).