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.
Related
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
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 trying to write a small bash script using mac OS TextEdit that will catch a user input in terminal and based on the file type stipulated in the command line albeit (jpg or gif) iterate through a directory on my desktop named dir1, pull all files of that filetype and place such in a new directory called dir2
i.e The user types jpg into the terminal and the script kicks into life and pulls all of the jpg files situated in dir1 and places such in dir2
What is the leanest and least convoluted way of achieving this conscious that I am new to shell scripting.
I am about to reach for the meds. What can I do to the below code to get it to work.
#!/bin/bash
echo “Good Morning, Please enter your file type for sorting [ENTER]:”
read $FILE
if [[ $file == *.jpg ]]; then
mv ~/DIR1/*jpg* ~/Users/christopherdorman/desktop/dir2/
echo “your files have been successfully processed”
fi
There are a couple confusions here regarding bash variables and syntax. You need to use fi to close your if statement instead of done. Also, you need to capitalize the variable in your if statement, since bash is case sensitive. I believe this is what you are looking for (assuming your input is "jpg" or "gif"):
#!/bin/bash
echo “Good Morning, Please enter your file type for sorting [ENTER]:”
read FILE
if [[ $FILE == "jpg" ]]; then
mv ~/DIR1/*jpg* ~/Users/christopherdorman/desktop/dir2/
echo “your files have been successfully processed”
fi
I run the following code for concatenating files in a directory given as the argument for the script file in bash
for i in $*
do
cat $* > /home/christy/Documents/filetest/catted.txt
done
This produce the error
cat: /home/christy/Documents/filetest/catted.txt: input file is output file
I think there are at least 4 things wrong with your script....
Firstly, your loop will set the value of i to the name of each file in succession, so you would want to actually use i inside your loop, like this:
for i in $*
cat "$i" ....somewhere
done
Secondly, if you use the > redirection, each file will land exactly on top of the previous one, so you should really use the >> redirection will append the current file to the end of the previous one like this
for i in $*
do
cat "$i" >> ...somewhere
done
Thirdly, I think you should use double-quoted "$#" to get all your command-line arguments, rather than plain $*
for i in "$#"
...
Fourthly, you can achieve the exact effect I think you want with this simpler command:
cat "$#" > /home/christy/Documents/filetest/catted.txt
You can't cat a file back onto itself. That's what "input file is output file" means. Because catted.txt shows up in your list of arguments to cat, it is going to try to cat to itself. So, move catted.txt to somewhere other than the source directory.
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.