Below is a snapshot of a file called ".bashrc":
I'm beginner in bash and What i'm trying to do in bash is to check if the last two lines inside the file exist and correctly written like for example :
if [ export PATH=/opt/ads2/arm-linux64/bin:$PATH ]
then
echo "found system variable lines"
else
echo "systemvariables do not exists, please insert it in .bashrc"
fi
However, this doesn't seem to be trivial since the tow lines to be shared are not pure string lines.
Thanks in advance
Use grep to find stuff in file contents.
# if file .bashrc contains the line exactly export PATH=....
if grep -Fxq 'export PATH=/opt/ads2/arm-linux64/bin:$PATH' .bashrc ; then
echo "found system variable lines"
else
echo "systemvariables do not exists, please insert it in .bashrc"
fi
Read man grep and decide if you want or not the -F and -x options in grep. For sure research and learn regex - I recommend regex crosswords available on the net. Research also difference between single quoting and double quoting in shell. Remember to check scripts with http://shellcheck.net
Related
I would like to know how to access the contents of a variety of txt files by passing arguments into shell scripts. I'll have different files and I'm expecting to execute with this command:
./script.sh FileA.txt
What should I put into my shell script so that I can access and manipulate the contents of the files?
I tried this but it outputs 0:
echo "$#"
I also tried these, but both output nothing:
for i in $1
do
echo "$i"
done
echo "$1"
To sum up the contents see this link to understand bash arguments more https://tecadmin.net/tutorial/bash-scripting/bash-command-arguments/ . Also as #Barmar said, to iterate through a list of arguments of unknown quantity use for i in "$#" .
edit
and as #Barmar said, $1 is simply the name of the argument. So echoing $1 will just echo the name.
I don't understand your question fully. Lets assume you have list of file names in a text file "FileA.txt".
And you wanted to run some commands for each file in the "FileA.txt" file.
Can you try below:
for i in `cat $1`
do
echo $i
done
I am new to Ubuntu and learning bash script by googling around. I want to know how to load image files from a folder and save it in an array in bash script.
Probably am not doing a really smart search, but if anyone knows how to do it already, can you please help?
I am planning to get the path from the command line argument, so $1 will have the path, as far as I have read.
Thus, I have this code
#!/bin/bash
for f in "$1"
do
echo "$f"
done
But the output just prints 1 file instead of all 36 files. Can you please help me here?
Note : the input am giving is of this format
/path/*.png
That glob (/path/*.png) has already been expanded by the shell when your script is called.
You have all the filenames in $# (the array of all the positional parameters to the script/function).
Try
echo "$#"
to see them or
for file in "$#"; do
echo "$file"
done
The default list for in is $# so you can use for file; do in place of for file in "$#"; do if you want.
I'm new to shell programming. I intend to get directory name after zip file was extracted. The print statement of it is
$test.sh helloworld.zip
helloworld
Let's take a look at test.sh:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
However I got the Bad substitution error from the compiler.
And when I mention about 'shell'.I just talking about shell for I don't know the difference between bash or the others.I just using Ubuntu 10.04 and using the terminal. (I am using bash.)
If your shell is a sufficiently recent version of bash, that parameter expansion notation should work.
In many other shells, it will not work, and a bad substitution error is the way the shell says 'You asked for a parameter substitution but it does not make sense to me'.
Also, given the script:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
The second line exports variable length with value echo for the command that is generated by running expr index "$1" .zip. It does not assign to length. That should be just:
length=$(expr index "${1:?}" .zip)
where the ${1:?} notation generates an error if $1 is not set (if the script is invoked with no arguments).
The last line should be:
echo ${a:0:$length}
Note that if $1 holds filename.zip, the output of expr index $1 .zip is 2, because the letter i appears at index 2 in filename.zip. If the intention is to get the base name of the file without the .zip extension, then the classic way to do it is:
base=$(basename $1 .zip)
and the more modern way is:
base=${1%.zip}
There is a difference; if the name is /path/to/filename.zip, the classic output is filename and the modern one is /path/to/filename. You can get the classic output with:
base=${1%.zip}
base=${base##*/}
Or, in the classic version, you can get the path with:
base=$(dirname $1)/$(basename $1 .zip)`.)
If the file names can contain spaces, you need to think about using double quotes, especially in the invocations of basename and dirname.
Try running it with bash.
bash test.sh helloworld.zip
-likewise-
"try changing the first line to #!/bin/bash" as comment-answered by – #shellter
Try that in bash :
echo $1
len=$(wc -c <<< "$1")
a="${1}.zip"
echo ${a:0:$len}
Adapt it to fit your needs.
I would like to copy the contents of a variable (here called var) into a file.
The name of the file is stored in another variable destfile.
I'm having problems doing this. Here's what I've tried:
cp $var $destfile
I've also tried the same thing with the dd command... Obviously the shell thought that $var was referring to a directory and so told me that the directory could not be found.
How do I get around this?
Use the echo command:
var="text to append";
destdir=/some/directory/path/filename
if [ -f "$destdir" ]
then
echo "$var" > "$destdir"
fi
The if tests that $destdir represents a file.
The > appends the text after truncating the file. If you only want to append the text in $var to the file existing contents, then use >> instead:
echo "$var" >> "$destdir"
The cp command is used for copying files (to files), not for writing text to a file.
echo has the problem that if var contains something like -e, it will be interpreted as a flag. Another option is printf, but printf "$var" > "$destdir" will expand any escaped characters in the variable, so if the variable contains backslashes the file contents won't match. However, because printf only interprets backslashes as escapes in the format string, you can use the %s format specifier to store the exact variable contents to the destination file:
printf "%s" "$var" > "$destdir"
None of the answers above work if your variable:
starts with -e
starts with -n
starts with -E
contains a \ followed by an n
should not have an extra newline appended after it
and so they cannot be relied upon for arbitrary string contents.
In bash, you can use "here strings" as:
cat <<< "$var" > "$destdir"
As noted in the comment by Ash below, #Trebawa's answer (formulated in the same room as mine!) using printf is a better approach than cat.
All of the above work, but also have to work around a problem (escapes and special characters) that doesn't need to occur in the first place: Special characters when the variable is expanded by the shell. Just don't do that (variable expansion) in the first place. Use the variable directly, without expansion.
Also, if your variable contains a secret and you want to copy that secret into a file, you might want to not have expansion in the command line as tracing/command echo of the shell commands might reveal the secret. Means, all answers which use $var in the command line may have a potential security risk by exposing the variable contents to tracing and logging of the shell.
For variables that are already exported, use this:
printenv var >file
That means, in case of the OP question:
printenv var >"$destfile"
Note: variable names are case sensitive.
Warning: It is not a good idea to export a variable just for the sake of printing it with printenv. If you have a non-exported script variable that contains a secret, exporting it will expose it to all future child processes (unless unexported, for example using export -n).
If I understood you right, you want to copy $var in a file (if it's a string).
echo $var > $destdir
When you say "copy the contents of a variable", does that variable contain a file name, or does it contain a name of a file?
I'm assuming by your question that $var contains the contents you want to copy into the file:
$ echo "$var" > "$destdir"
This will echo the value of $var into a file called $destdir. Note the quotes. Very important to have "$var" enclosed in quotes. Also for "$destdir" if there's a space in the name. To append it:
$ echo "$var" >> "$destdir"
you may need to edit a conf file in a build process:
echo "db-url-host=$POSTGRESQL_HOST" >> my-service.conf
You can test this solution with running before export POSTGRESQL_HOST="localhost"
I just started learning PHP. I'm following phpacademy's tutorials which I would recommend to anyone. Anyways, I'm using XAMPP to test out my scripts. I'm trying to write a bash script that will start XAMPP and then open firefox to the localhost page if it finds a specific string, "XAMPP for Linux started.", that has been redirected from the terminal to the file xampp.log. I'm having a problem searching the file. I keep getting a:
grep: for: No such file or directory
I know the file exists, I think my syntax is wrong. This is what I've got so far:
loaded=$false
string="XAMPP for Linux started."
echo "Starting Xampp..."
sudo /opt/lampp/lampp start 2>&1 > ~/Documents/xampp.log
sleep 15
if grep -q $string ~/Documents/xampp.log; then
$loaded=$true
echo -e "\nXampp successfully started!"
fi
if [$loaded -eq $true]; then
echo -e "Opening localhost..."
firefox "http://localhost/"
else
echo -e "\nXampp failed to start."
echo -e "\nHere's what went wrong:\n"
cat ~/Documents/xampp.log
fi
In shell scripts you shouldn't write $variable, since that will do word expansion on the variable's value. In your case, it results in four words.
Always use quotes around the variables, like this:
grep -e "$string" file...
The -e is necessary when the string might start with a dash, and the quotes around the string keep it as one word.
By the way: when you write shell programs, the first line should be set -eu. This enables *e*rror checking and checks for *u*ndefined variables, which will be useful in your case. For more details, read the Bash manual.
You are searching for a string you should put wihtin quotes.
Try "$string" instead of $string
There are a couple of problems:
quote variables if you want to pass them as a simple argument "$string"
there is no $true and $false
bash does variable expansion, it substitutes variable names with their values before executing the command. $loaded=$true should be loaded=true.
you need spaces and usually quotes in the if: if [$loaded -eq $true] if [ "$loaded" -eq true ]. in this case the variable is set so it won't cause problems but in general don't rely on that.