I have a short shell script that I wrote to just create backups.
#!/bin/bash
export MyBackup="MyBackup`date +%m-%d-%H:%M`"
echo $MyBackup
vi /tmp/$MyBackup.txt
rm -rf /tmp/"$MyBackup"
However, the filename that is created is something like MyBackup12-09-08:46?.txt?. The echo command returns the correct string, but the vi command creates a file with ?'s. How do I create the file without these?
Most of the issues ive lookedu p seem to talk about encoding differences, but I would think it would display incorrectly when I echoed if that was the case.
Thanks
Replace vi with touch to just create an empty file.
I ran e ++ff=unix and it seems like each one of the lines has a ^M at the end of them, so I removed them.
Related
I am writing a shell script to sync to a github repo, kick off the build, then take the output file, rename it, and move it to a location where it can be seen by Apache.
It's the renaming of the file that I've got not the faintest how to do within a shell script (I have virtually no experience with shell scripts - my understanding
Compiler will create /var/espbuild/firstpart_1vXX_secondpart.bin
I need to move this file to:
/var/www/html/builds/espbuild/firstpart_1vXX_DATE_secondpart_postfix.bin
1vXX is the version number
DATE is the output of date +%m-%d
postfix is just a string.
I'm not really certain where to start for something like this - I'm sure there's a graceful way, since this is the kind of thing shell scripts are made for, but I know just about nothing about shell scripts.
Thanks in advance
You can get the result of a command into a variable by using $():
DATE=$(date +%m-%d)
Then just use it in the new filename:
INPUT=/var/espbuild/firstpart_1vXX_secondpart.bin
OUTPUT=/var/www/html/builds/espbuild/firstpart_1vXX_${DATE}_secondpart_postfix.bin
mv ${INPUT} ${OUTPUT}
Edit: To get out the version part, here's a quick example:
VERSION=$(grep -o 1v.. <<< ${INPUT})
Then OUTPUT should be set like:
OUTPUT=/var/www/html/builds/espbuild/firstpart_${VERSION}_${DATE}_secondpart_postfix.bin
You can use this in BASH:
f='/var/espbuild/firstpart_1vXX_secondpart.bin'
s="${f##*/}"
s2=${s##*_}
dest="/var/www/html/builds/espbuild/${s%_*}_$(date '+%m-%d')_${s2%.*}_postfix.bin"
echo "$dest"
/var/www/html/builds/espbuild/firstpart_1vXX_07-14_secondpart_postfix.bin
cp "$f" "$dest"
I'm trying to execute the program as followed.
./chExt1.sh cpp test.CPP
This should rename test.CPP to test.cpp but I don't even think this script is executing at all.
I am consistently getting this "command not found error".
The script is below :
#!/bin/sh
newExtension=$1;
oldFile=$2;
firstPart=`echo $oldFile | sed 's/\(.*\)\..*/\1/'`
newName="$firstPart.$newExtension";
#echo $oldFile
#echo $newName
mv "$oldFile" "$newName"
#echo "$oldFile"
#echo "$firstPart"
#echo "$newName"
I finally fixed the issue. Something went horribly wrong when I FTP'd the text file which contained the script and then just transferred it inside of a .sh in linux. I wrote in from scratch in emacs and that cleared everything up.
Based on your comment, do this in vi to remove the extra control characters. I have had this problem before when editing files in gedit or when editing in Windows and then using on a Unix/Linux machine.
To remove the ^M characters at the end of all lines in vi, use:
:%s/^V^M//g
The ^v is a CtrlV character and ^m is a CtrlM. When you type this, it will look like this:
:%s/^M//g
In UNIX, you can escape a control character by preceeding it with a CtrlV. The :%s is a basic search and replace command in vi. It tells vi to replace the regular expression between the first and second slashes (^M) with the text between the second and third slashes (nothing in this case). The g at the end directs vi to search and replace globally (all occurrences).
Source
I used the bash commands to append several lines to multiple configuration files:
> for filename in *.ovpn; do
> printf 'configurationscript-security 2\nup /etc/openvpn/update-resolv-conf\ndown /etc/openvpn/update-resolv-conf' >> $filename;
> done
However the control character "^M" appeared at end of each line in the configuration file:
I opened the files in vim, the files before bash commands looked like as folows:
I am curious why "^M" appears at end of each line? Thanks.
It is Windows' carriage return, use dos2unix to convert file. Vim recognize the file format and displays it correctly.
The ^M can also be removed via a regular expression in vim, if dos2unix isn't available.
:%s/^M//g, which can be entered as: Esc:%s/ctrl+Vctrl+M//g
Not sure why this has occurred for you with just a simple printf command on a linux system, maybe have a look that you're picking up the correct version of printf. I've given this a go on a linux system, and the local printf keeps the correct line-endings, as you would expect.
Every time I run a script using bash scriptname.sh from the command line in Debian, I get Command Not found and then the result of the script.
The script works but there is always a Command Not Found statement printed on screen for each empty line. Each blank line is resulting in a command not found.
I am running the script from the /var folder.
Here is the script:
#!/bin/bash
echo Hello World
I run it by typing the following:
bash testscript.sh
Why would this occur?
Make sure your first line is:
#!/bin/bash
Enter your path to bash if it is not /bin/bash
Try running:
dos2unix script.sh
That wil convert line endings, etc from Windows to unix format. i.e. it strips \r (CR) from line endings to change them from \r\n (CR+LF) to \n (LF).
More details about the dos2unix command (man page)
Another way to tell if your file is in dos/Win format:
cat scriptname.sh | sed 's/\r/<CR>/'
The output will look something like this:
#!/bin/sh<CR>
<CR>
echo Hello World<CR>
<CR>
This will output the entire file text with <CR> displayed for each \r character in the file.
You can use bash -x scriptname.sh to trace it.
I also ran into a similar issue. The issue seems to be permissions. If you do an ls -l, you may be able to identify that your file may NOT have the execute bit turned on. This will NOT allow the script to execute. :)
As #artooro added in comment:
To fix that issue run chmod +x testscript.sh
This might be trivial and not related to the OP's question, but I often made this mistaken at the beginning when I was learning scripting
VAR_NAME = $(hostname)
echo "the hostname is ${VAR_NAME}"
This will produce 'command not found' response. The correct way is to eliminate the spaces
VAR_NAME=$(hostname)
On Bash for Windows I've tried incorrectly to run
run_me.sh
without ./ at the beginning and got the same error.
For people with Windows background the correct form looks redundant:
./run_me.sh
If the script does its job (relatively) well, then it's running okay. Your problem is probably a single line in the file referencing a program that's either not on the path, not installed, misspelled, or something similar.
One way is to place a set -x at the top of your script or run it with bash -x instead of just bash - this will output the lines before executing them and you usually just need to look at the command output immediately before the error to see what's causing the problem
If, as you say, it's the blank lines causing the problems, you might want to check what's actaully in them. Run:
od -xcb testscript.sh
and make sure there's no "invisible" funny characters like the CTRL-M (carriage return) you may get by using a Windows-type editor.
use dos2unix on your script file.
for executing that you must provide full path of that
for example
/home/Manuel/mywrittenscript
Try chmod u+x testscript.sh
I know it from here:
http://www.linuxquestions.org/questions/red-hat-31/running-shell-script-command-not-found-202062/
If you have Notepad++ and you get this .sh Error Message: "command not found"
or this autoconf Error Message "line 615:
../../autoconf/bin/autom4te: No such file or directory".
On your Notepad++, Go to Edit -> EOL Conversion then check Macinthos(CR).
This will edit your files. I also encourage to check all files with this command,
because soon such an error will occur.
Had the same problem. Unfortunately
dos2unix winfile.sh
bash: dos2unix: command not found
so I did this to convert.
awk '{ sub("\r$", ""); print }' winfile.sh > unixfile.sh
and then
bash unixfile.sh
Problems with running scripts may also be connected to bad formatting of multi-line commands, for example if you have a whitespace character after line-breaking "\". E.g. this:
./run_me.sh \
--with-some parameter
(please note the extra space after "\") will cause problems, but when you remove that space, it will run perfectly fine.
I was also having some of the Cannot execute command. Everything looked correct, but in fact I was having a non-breakable space right before my command which was ofcourse impossible to spot with the naked eye:
if [[ "true" ]]; then
highlight --syntax js "var i = 0;"
fi
Which, in Vim, looked like:
if [[ "true" ]]; then
highlight --syntax js "var i = 0;"
fi
Only after running the Bash script checker shellcheck did I find the problem.
I ran into this today, absentmindedly copying the dollar command prompt $ (ahead of a command string) into the script.
Make sure you haven´t override the 'PATH' variable by mistake like this:
#!/bin/bash
PATH="/home/user/Pictures/"; # do NOT do this
This was my mistake.
Add the current directory ( . ) to PATH to be able to execute a script, just by typing in its name, that resides in the current directory:
PATH=.:$PATH
You may want to update you .bashrc and .bash_profile files with aliases to recognize the command you are entering.
.bashrc and .bash_profile files are hidden files probably located on your C: drive where you save your program files.
I've run into a really silly problem with a Linux shell script. I want to delete all files with the extension ".bz2" in a directory. In the script I call
rm "$archivedir/*.bz2"
where $archivedir is a directory path. Should be pretty simple, shouldn't it? Somehow, it manages to fail with this error:
rm: cannot remove `/var/archives/monthly/April/*.bz2': No such file or directory
But there is a file in that directory called test.bz2 and if I change my script to
echo rm "$archivedir/*.bz2"
and copy/paste the output of that line into a terminal window the file is removed successfully. What am I doing wrong?
TL;DR
Quote only the variable, not the whole expected path with the wildcard
rm "$archivedir"/*.bz2
Explanation
In Unix, programs generally do not interpret wildcards themselves. The shell interprets unquoted wildcards, and replaces each wildcard argument with a list of matching file names.
if $archivedir might contain spaces, then rm $archivedir/*.bz2 might not do what you
You can disable this process by quoting the wildcard character, using double or single quotes, or a backslash before it. However, that's not what you want here - you do want the wildcard expanded to the list of files that it matches.
Be careful about writing rm $archivedir/*.bz2 (without quotes). The word splitting (i.e., breaking the command line up into arguments) happens after $archivedir is substituted. So if $archivedir contains spaces, then you'll get extra arguments that you weren't intending. Say archivedir is /var/archives/monthly/April to June. Then you'll get the equivalent of writing rm /var/archives/monthly/April to June/*.bz2, which tries to delete the files "/var/archives/monthly/April", "to", and all files matching "June/*.bz2", which isn't what you want.
The correct solution is to write:
rm "$archivedir"/*.bz2
Your original line
rm "$archivedir/*.bz2"
Can be re-written as
rm "$archivedir"/*.bz2
to achieve the same effect. The wildcard expansion is not taking place properly in your existing setup. By shifting the double-quote to the "front" of the file path (which is legitimate) you avoid this.
Just to expand on this a bit, bash has fairly complicated rules for dealing with metacharacters in quotes. In general
almost nothing is interpreted in single-quotes:
echo '$foo/*.c' => $foo/*.c
echo '\\*' => \\*
shell substitution is done inside double quotes, but file metacharacters aren't expanded:
FOO=hello; echo "$foo/*.c" => hello/*.c
everything inside backquotes is passed to the subshell which interprets them. A shell variable that is not exported doesn't get defined in the subshell. So, the first command echoes blank, but the second and third echo "bye":
BAR=bye echo `echo $BAR`
BAR=bye; echo `echo $BAR`
export BAR=bye; echo `echo $BAR`
(And getting this to print the way you want it in SO takes several tries is apparently impossible...)
The quotes are causing the string to be interpreted as a string literal, try removing them.
I've seen similar errors when calling a shell script like
./shell_script.sh
from another shell script. This can be fixed by invoking it as
sh shell_script.sh
Why not just rm -rf */*.bz2? Works for me on OSX.