I have the following file (above) which seems to be an Unix pipe
alt text http://dl.getdropbox.com/u/175564/problemFile.png
How can you make the pipe a default text file?
Unless there is some other complication that hasn't been mentioned, the easiest is to use the big hammer:
$ rm outside
$ touch outside
If there is a process currently using the file, you will need to kill the process first, then restart it so it uses the new file. Otherwise, the pipe will stay open but invisible until the process finally dies.
How can you make the pipe a default
text file, such that the first letter
changes to d?
The first character of 'd' means that that entry is a directory - not a normal file.
Related
I want to change my PATH variable in zsh.
Problem: I don't understand where in the .zshrc file I have to make modifications.
Normally, I would look for the assignment to the PATH variable and set the values from scratch how I would like them to be (leaving all the systems binaries directories untouched).
The first lines in my .zshrc file are as follows:
# If you come from bash you might have to change your $PATH.
# export PATH=$HOME/bin:/usr/local/bin:$PATH
# Path to your oh-my-zsh installation.
export ZSH="/Users/Sam/oh-my-zsh"
export PATH=$PATH:/Applications/Postgres.app/Contents/Versions/13/bin
etc.
My actual PATH variable is:
/Library/Frameworks/Python.framework/Versions/3.9/bin:/Library/Frameworks/Python.framework/Versions/3.8/bin:/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/Applications/Postgres.app/Contents/Versions/13/bin
I want to delete the directory where python3.8 is in, it's redundant.
My questions:
Do I have to change line 2 or line 7 in my .zshrc file?
Line 2 is commented out...is it executed anyway at the start of the terminal?
I have tried to comment out line 7. But the postgres directory still remained in my PATH variable which I don't understand.
The .zshrc is located in the home dir. The dot at the beginning keeps it hidden. Type ls -a from ~ directory to see it. To edit just run vim, nvim, etc as usual.
nvim ~/.zshrc
This is the command for Neovim. For your editor, sub nvim for the proper command.
Once you get in, you need only add the same export command that you would add from the command line.
export PATH=$PATH:/whatever/you/are/adding
EDIT
To remove a path variable:
First, run the command:
echo $PATH
from the command line.
Next, Copy the output to clipboard.
Finally, at the very end of the .zshrc file, add the following:
export PATH=<paste-what-you-copied-here>
Because you didn't reference $PATH after the =, this will set the path to EXACTLY what you pasted, no more, no less. Adding $PATH: as in the first example will just add whatever to the end of what is already there.
Since this gives you access to every item in the path array, deleting is just a matter of a literal highlight/select and of whatever you want deleted.
Finally, be sure that there is only one place in the file where you are editing PATH. If there are more than one, the result can be confusing to say the least.
That said, I believe the script runs top-to-bottom, so only the last mention should persist. You can take advantage of this in some situations, but for this purpose, one will suffice. XD
Be careful when you decide to fiddle with the PATH in .zshrc: Since the file is processed by every interactive subshell, the PATH would get longer and longer for each subshell, with the same directory occuring in it several times. This can become a nightmare if you later try to hunt down PATH-related errors.
Since you are using zsh, you can take advantage that the scalar variable PATH is mirrored in the array variable path, and that you can ask zsh to keep entries in arrays unique.
Hence, the first thing I would do is put a
typeset -aU path
in your .zshrc; this (due to mirroring) also keeps the entries in PATH unique. You can put this statement anywhere, but I have it for easier maintenance before my first assignment to PATH or path.
It is up to you to decide where exactly you add a new entry to PATH or path. The entries are searched in that order which is listed in the variable. You have to ask yourself two questions:
Are some directories located on a network share, where you can sometimes expect access delays (due to bad network conditions)? Those directories should better show up near the end of the path.
Do you have commands which occur in more than one directoryin your path? In this case, a path search will always find the first occurance only.
Finally, don't forget that your changes will be seen after zsh processes the file. Therefore, you could create a new subshell after having edited the file, or source .zshrc manually.
I want to write a bash script that takes a user input (which will be a filename) and replaces a path to a file inside a css file with that filename. For simplicity, the two files will be in the same folder and in the css code only the filename at the end of the path should be changed.
I thought of using regex to match any line of code that has a specific pattern and then change the end of it. I know about sed, but since the filename always changes I'm not sure how to solve this problem other than regex. I also thought of adding a variable in the css file that holds the filename as a value and then adding that variable at the end of the path, but I'm not sure then how to access that variable from a bash script.
Any recommendations on how to tackle this problem?
Thanks!
Edit Adding more Information:
Here is the line in the css file I want to edit. The part to be changed is the fileName.png at the end. Since it will change I thought of using a regex to "find" the correct spot in the css file.
background: #2c001e url(file:////usr/share/backgrounds/fileName.png/);
A regex matching only this line in this specific file is the following. It could probably be simplified, but I don't see a reason why since it should work too:)
(background)\:\s\#.{6}\s(url)\((file)\:\/{4}(usr)\/(share)\/backgrounds\/.+\.(png)\/\)\;
So, there are some ways to do that. You can check topic in links below. sed command is also good idea. But before executing it, you can build a new variable (or multiple variables) to use them in regex sed -e syntax.
Getting the last argument passed to a shell script
Maybe, if you will add some input and output examples, I could be more specific in this case.
To replace the input in the file at run-time you could use this line in a script
sed "s/stringToReplace/$1/g" templateFile >fileToUse
the $1 is referencing the 2nd bash script argument (the first being $0, the name of the invoking script). stringToReplace would be written in verbatim in the templateFile.
You could also use a script with two runtime arguments ($1, $2), and you would change the original contents of the fileToUse using the -i option. But this requires storage of the last file path to be used as argument $1.
I'm using oh-my-zsh which pipes the output of some functions like git diff and git log into less, whilst this is great for reading the output in the terminal. If I need to refer back to it it isn't possible after quitting with :q
Is there an option to preserve the current view on the file in my terminal after quitting?
Secondly, If there is an option where would I need to edit my oh-my-zsh config to ensure anything piped to less passes this option?
To prevent less from clearing the screen on exit you can start it with the option -X:
less -X FILE
If you want to pass this option automatically to every instance of less, you can set the LESS environment variable accordingly in your ~/.zshrc:
export LESS="-X"
Note:
If your shell has syntax coloring enabled, the -X option will cause your less output to display those color change escape sequences as inline ESC text.
This can be fixed by also passing the raw-control-chars display option, -r. For example:
export LESS="-Xr"
This also includes instances where less is started by another program, for example man. If you want to disable this option for a single command, you can just prepend LESS=. For example
LESS= man less
For Git specifically, this can be handled with the following
git config --global color.ui true
git config --global core.pager 'less -Xr'
When I invoked "tail -f myfile.txt", the new line added using the following command output the new line, but not the line added/saved using vi. Does anyone know why ?
$echo "this is new line" >> myfile.txt
Thanks.
It has something to do w/the fact that while you are editing the file, vi keeps your changes in a second file (.myfile.txt.swp in this case).
When you save the changes, it's likely that vi is replacing the original file w/the second file. This means the file that tail was watching is no longer valid.
To prove this, try your echo command after saving the file with vi. When you do that, the output won't be displayed by tail.
The tail program opens a file, seeks to the end, and (with "-f") waits, then checks again if that open file has anything new to read.
vi does not append to a file. It makes a copy, (not a "swap", which is something else altogether) writes it out, and then moves the new file to have the same name as the old file.
tail is still watching the old file, not looking up a file by that file name every time.
In addition, tail uses the location in the file, so if you delete 10 characters and add 15, the next loop of 'tail' will emit the next 5 it thinks are new because they are after its placeholder.
Run 'tail --follow=name ...' to get tail to look up the file every loop by name, instead of watching the location on disk of a file it opens at start.
I am taking a intro to Unix class and am stuck on the final assignment. I need to write a script to change the file extension of a filename that is input when the script is run. The new file extension is also input when the script is run. The script is call chExt1.sh . Our first trial of the script is run as follows
./chExt1.sh cpp aardvark.CPP
The script is suppose to change the second input file extension to the file extension given in the first input. It is not suppose to matter what file extension is given with the file name or what file extension is given as the new extension, nor is it only for changing uppercase to lowercase. In hope to make this very clear if given the following:
./chExt1.sh istink helpme.plEaSe
The script would change helpme.plEaSe to helpme.istink . I have searched on this forum and in google and have had no look with trying the different examples I found. Below is some of the examples I have tried and what I currently have.
Current
#!/bin/sh
fileExtension="$1"
shift
oldName="$2"
shift
newName=${oldName%%.*}${fileExtension}
echo $newName
The echo is just to see if it works, and if I get it working I'm going to add an mv to save it.
Others that I have tried:
newName=`${oldName%.*}`
newName=`${oldName#.*}`
sed 's/\.*//' $oldName > $newName
I can't seem to find some of the other sed I have used but they involved alot of backslashes and () with .* in there. I did not try the basename command cause I don't know the file extension to be entered and all I the examples I saw required that you specify what you wanted removed and I can't. I did not list all the different quote variations that I used but I have tried alot. My instructions say to use the sed command since we should know how to use that from class but when I try to do it I don't isolate just the ending of the file and I believe (cause it takes so long to finish) that it is going through the whole file and looking for .'s and anything after cause I kept doing .* as the pattern. Thanks for anyhelp you can give.
shift shifts the positional parameters, so after calling shift the second parameter ($2) is now the first ($1). The second shift is not necessary, because you are done accessing the parameters. You need to either remove the shift
#!/bin/sh
fileExtension="$1"
oldName="$2"
newName=${oldName%%.*}${fileExtension}
echo $newName
or change $2 to $1.
#!/bin/sh
fileExtension="$1"
shift
oldName="$1"
newName=${oldName%%.*}${fileExtension}
echo $newName
However, you are still missing a dot from your new file name. That is left as an exercise for the reader.