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
Related
I have a shell script myautoappupgrade.sh where I automate a process of application upgrade. The script has to be run on few different servers. Unfortunately, the application is located in slightly different directory on each server - the number for parent directory varies between 1-20. How I can modify the script, so that the directory can be replaced by some sort of variable? I don't want to edit the script for each server because there are many directory queries in the automation script.
example:
cd /ae1/apps/myapp/upgradefiles/
unzip file.zip
./install.sh
the directory slightly changes on another server:
cd /ae2/apps/myapp/upgradefiles/
unzip file.zip
./install.sh
and another..
cd /ae3/apps/myapp/upgradefiles/
unzip file.zip
./install.sh
Try something like this:
#!/bin/bash
num=$1
cd /ae${num}/apps/myapp/upgradefiles/file.zip
unzip file.zip
./install.sh
Then call the script with the number as first argument:
myautoappupgrade.sh 1
The simple and obvious solution is to not hard-code the directory at all. Modify the script so it accepts the parent directory as an argument, or just cd into the parent directory before running the script.
Perhaps something like this:
while read server dir; do
ssh "$server" "cd '$dir' && unzip apps/myapp/upgradefiles/file.zip/file.zip && ./install.sh"
done <<\:
ernie /ae1
bert /ae2
cookiemonster /home/cmonster/anN
:
It would probably be even better if you unzipped into a temporary directory, but hopefully this should get you moving in the right direction.
Of course, if you can be sure that /ae[0-1] is always there and there is only one match,
cd /ae[0-9]/apps/myapp/upgradefiles/file.zip
would do what you are asking.
(Do you really have a file named file.zip inside a directory also named file.zip? I'm guessing actually take away the file.zip from the end of the cd path.)
By simply using:
cd /ae*/apps/myapp/upgradefiles/
The * will expand any character.
I am not trying to execute a Bash script from any directory by adding the script to my Path variable.
I want to be able to execute the script from any directory using the directory path to that file ... but the file I want to execute sources other files, that is the problem.
If I am in directory file with two scripts myFunctions.sh and sourceFunctions.sh
sourceFunctions.sh
#!/bin/bash
source ./myFunctions.sh
echoFoo
myFunctions.sh
function echoFoo()
{
echo "foo"
}
I can run myFunctions.sh and foo will print to console, but If I go up a directory and run myFunctions.sh I get error
cd ..
file/sourceFunctions.sh
-bash: doFoo.sh: command not found
Unless I changed source file/myFunctions.sh to source file/myFunctions.sh in sourceFunctions.sh.
So how can I source independent of my working directory so I can run sourceFunctions.sh from any working directory I want?
Thanks
You have the right idea. Doesn't need to be that complicated though:
source `dirname $0`/myFunctions.sh
I often compute "HERE" at the top of my script:
HERE=`dirname $0`
and then use it as needed in my script:
source $HERE/myFunctions.sh
One thing to be careful about is that $HERE will often be a relative path. In fact, it will be whatever path you actually used to run the script, or "." if you provided no path. So if you "cd" within your script, $HERE will no longer be valid. If this is a problem, there's a way (can't think of it off hand) to make sure $HERE is always an absolute path.
I ended up just using a variable of the directory path to the script itself for the source directory
so
#!/bin/bash
source ./myFunctions.sh
echoFoo
becomes
#!/bin/bash
SCRIPTPATH="$( cd "$(dirname "$0")" ; pwd -P )"
source ${SCRIPTPATH}/myFunctions.sh
echoFoo
source
I'm trying to create few files under folder with script, but not able to.
following is the folder privilege under which some files are to be created
su pnaid
The partial script content, responsible to create folder and create files in it
MKDIR_CMD="/bin/mkdir -p \"${PATH_TO_WRITE}\" > /dev/null 2>&1"
"${MKDIR_CMD}"
echo "Checking Dir Path exist"
if [ -d "${PATH_TO_WRITE}" ]; then
echo "Calling another script to create files under this folder"
"/createFiles.sh \"${PATH_TO_WRITE}\""
else
echo "WARNING: unable to create folder"
The parent folder to the $(PATH_TO_WRITE) has following privileges
drwxr-x---. 2 pnaid pnaid 4096 Dec 3 12:31 work_directory
Each time the statement "WARNING: unable to create folder" is displayed.
I tried creating a folder with pnaid user having 777 permission and feeding that in script instead of "${MKDIR_CMD}", in that case the statement "Calling another script to create files under this folder" is displayed but the other script is not able to write to this folder.
Also the echo statements from createFiles.sh when called from original script are not displayed, is there any way to view it.
If we perform the same commands on shell prompt instead of script, the commands work and desired output is obtained; i.e. folder is created with all the files in it.
Also the same script works if we run it with user root.
I believe this should work across Linux flavors, in this case I'm using CentOS
Please help me resolve this issue and let me know if I have missed mentioning any details.
Thanks
This line:
"${MKDIR_CMD}"
will not work. It treats the entire value of $MKDIR_COMMAND as the name of the program to run, it doesn't split it into the program and arguments, because you put quotes around it. Also, redirections are not processed when expanding a variable.
You need to use eval to re-parse the string:
eval "$MKDIR_CMD"
You have the same problem with:
"/createFiles.sh \"${PATH_TO_WRITE}\""
This should be:
/createFiles.sh "$PATH_TO_WRITE"
These problems don't depend on permissions, I doubt the script really works when run as root.
Here's a related question that shows how to store command parameters best in variables:
Setting an argument with bash
However, the solution there (using an array instead of a string) won't work if you're also storing shell operators like redirection.
I have dozens of scripts, all in different directories. (exported/expanded Talend jobs)
At this moment each job has 1 or 2 scripts, starting with the same lines, most important one:
CD ***path-to-script***
and several lines to set the Java path and start the job.
I want to create a script, which will be ran from all these scripts.
e.g.:
/scripts/talend.sh
And in all talend scripts, the first line will run /scripts/talend.sh, some examples of where these scripts are ran from:
/talend-job1_0.1/talend-job1_0.1/talend-job1/talend-job1.sh
/talend-task2_0.1/talend-task2_0.1/talend-task2/talend-task2.sh
/talend-job3_0.1/talend-job3_0.1/talend-job3/talend-job3.sh
How can I determine where the /scripts/talend.sh is started from, so I can CD to that path from within /scripts/talend.sh.
The Talend scripts are not run from within the directory itself, but from a cronjob, or a different users home directory.
EDIT:
The question was marked as duplicate, but Getting the source directory of a Bash script from within is not answering my question 100%.
Problem is:
- The basic script is being called from different scripts
- Those different scripts can be run from command line, with, and with or without a symbolic link.
- The $0, the $BASH_SOURCE and the pwd all do some things, but no solution mentioned covers all the difficulties.
Example:
/scripts/talend.sh
In this script I want to configure the $PATH and $HOME_PATH of Java, and CD to the place where the Talend job is placed. (It's a package, so that script MUST be run from that location).
Paths to the jobs are, for example:
/u/talend/talendjob1/sub../../talendjob1.sh
/u/talend/talendjob2/sub../../talendjob2.sh
/u/talend/talendjob3/sub../../talendjob3.sh
Multiple jobs are run from a TMS application. This application cannot run these scripts with the whol name (to long, name can only be 6 long), so in a different location I have symbolic links:
/u/tms/links/p00001 -> /u/talend/talendjob1/sub../../talendjob1.sh
/u/tms/links/p00002 -> /u/talend/talendjob1/sub../../talendjob2.sh
/u/tms/links/p00003 -> /u/talend/talendjob1/sub../../talendjob3.sh
/u/tms/links/p00004 -> /u/talend/talendjob1/sub../../talendjob4.sh
I think you get an overview of the complexity and why I want only one basic talend script, where I can leave all basic stuff. But I only can do that, if I know the source of the Talend script, because there I have to be to start that talend job.
These answers (beyond the first) are specific to Linux, but should be very robust there -- working with directory names containing spaces, literal newlines, wildcard characters, etc.
To change to your own source directory (a FAQ covered elsewhere):
cd "$(basename "$BASH_SOURCE")"
To change to your parent process's current directory:
cd "/proc/$PPID/cwd"
If you want to change to the directory passed as the first command-line argument to your parent process:
{ IFS= read -r -d '' _ && IFS= read -r -d '' argv1; } <"/proc/$PPID/cmdline"
cd "$argv1"
That said, personally, I'd just export the job directory to the environment variable in the parent process, and read that environment variable in the children. Much, much simpler, more portable, more accurate, and compliant with best process.
You can store pwd in a variable and then cd to it when you want to go back
This works for me:
In
/scripts/talend.sh
do
cd ${1%/*}
${1%/*} will strip off everything after the last / effectively providing a dirname for $1, which is the path to the script that calls this one.
and than call the script with the line:
/scripts/talend.sh $0.
Calling the script with $0 passes the name of the current script as an argument to the child which as shown above can be used to cd to the correct directory.
When you source /scripts/talend.sh the current directory is unchanged:
The scripts
# cat /scripts/talend.sh
echo "Talend: $(pwd)"
# cat /talend-job1_0.1/talend-job1_0.1/talend-job1/talend-job1.sh
echo Job1
. /scripts/talend.sh
Executing job1
# cd /talend-job1_0.1/talend-job1_0.1
# talend-job1/talend-job1.sh
Job1
Talend: /talend-job1_0.1/talend-job1_0.1
When you want to see the dir where the calling script is in, see get dir of script.
EDIT:
When you want to have the path of the callling script (talend-job1.sh) without having to cd to that dir first, you should get the dir of the script (see link above) and source talend.sh:
# cat /scripts/talend.sh
cd "$( dirname "${BASH_SOURCE[0]}" )"
echo "Talend: $(pwd)"
In talend.sh get the name of the calling script and then the directory:
parent_cmd=$(ps -o args= $PPID)
set -- $parent_cmd
parent_cmd=$(dirname $2)
Update: as pointed by Charles Duffy in the comments below this will cause havoc when used with paths containing white-space or glob patterns.
If procfs is available you could read the content of /proc/$PPID/cmdline or if portability is a concern do a better parsing of the args.
In /scripts/talend.sh:
cd "$(dirname "$0")"
Or:
cd "$(dirname "$BASH_SOURCE")"
Another one is:
cd "$(dirname "$_")"
#This must be the first line of your script after the shebang line
#Otherwise don't use it
Note: The most reliable of the above is $BASH_SOURCE
I haven't written any Shell scripts before, but i have to write a simple shell script to do the following;
I will keep all the required files in a single folder and bundle it with this shell script as a tar file; so when the user runs the shell script, it needs to copy the respective files to the respective destinations.
The execution of copy as follows:
copy the plugin.so file to /usrlib/mozilla/plugins/
copy the .so library files to /usr/local/lib/
copy some header files directories(folders) to /usr/local/include/
and finally, need to do ldconfig.
Basically, you can add in a script any command you are able to type inside the terminal itself. Then, you have two options for executing it:
Execute it from the terminal with sh your_script.sh. You don't even need to give execute permission to it with this solution.
Give it the execute permission and run it with ./your_script.sh.
For the second solution, you have to start the file with what is called a shebang. So your script will look like:
#!/bin/sh
cp path/to/source path/to/destination
cp path/to/source path/to/destination
cp path/to/source path/to/destination
ldconfig
echo "Done!"
Nothing else. Just write the commands one after the other.
The first line is the so-called shebang and tells the shell which interpreter to use for the script.
Note: the extension for shell scripts is usually .sh, but you can actually name your file however you prefer. The extension has no meaning at all.
Good scripting!