I'm basically setting up two sshfs mounts and I have it set up so I run one command but type my password twice.Is there an easy to way to input a password using bash and pass that variable to another process asking for a password?
When I use the scp command, here is a command prompt "Input password:",Now I want to use a one line command to perform the scp command.I tested the "echo "password" | scp xxx yu@10.xxx.xxx.xxx:/", but failed.
I'm trying to call a specific variable based on a user selection. For example:
Code: Select a file:
[1] foo.tar [2] bar.tar
Enter a selection: I have already coded each possible selection to have its own variable. If the user selects 2 I need to select $SELECTED_TAR2, or if they select 1 I need to select $SELECTED_TAR1 and then do something like this behind the scenes:
On one of my servers I see this when I log in. What does this mean and how can I get it to go away? Everything seems to work fine, but none of my other machines give this error.
I'm creating a bash script to do some tasks for me. I would like the script to be run at a set time of everyday. My first question is if it is possible that if one of the commands in the script requires sudo, is there a way to get around it with out making sudo not require a password. Such as, is there a way to include the password in the script? If that is the case, I can always just set the file as read only by sudo. I've been looking for a way to do this, with no success. if I have a command that wants input, how do I give it to the program. For example, if I want to make a zip file that is encrypted, the command would go as:
Code:
zip -r example * -e
now how would I get the script to insert my wanted password.
I have a script in which it searches the password file for a username based on the user's input. Is there a way I can take the bad input and keep it in a variable. (one that keeps adding all the non-existent users)so how would I get non-existent users into another variable?
I'm trying to read content of file to variable and use this variable in for loop. The problem is, when I have c++ comment style in file - /*. Spaces in line are also interpreted as separated lines.
For example:
Code:
Changing $files to "$files" eliminate these problems but causes that whole content of variable is treated as one string (one execution of loop).
I need to create a single line of output from multiple and variable lines of input in a Linux bash shell script.
My input file looks like this:
Where there may be any number of umsecondaryphonenumber lines; if there is not a umsecondaryphonenumber line for a telephonenumber, I don't want to write any output.
So, the output file should look like:
The script I have so far is:
My question is - how do print each of the elements of an array in one record - i.e. what do I put in place of howdoiprintarray?
I want to create a variable that when passed as a parameter to another bash script will keep its string quotes (so it stays as one parameter). What ways can I achieve this cleanly?
In my script, and I would like to concatenate 2 variables names, to give me the true variable.I've 3 variables X1, X2 and X3, and I invoked them inside a for loop.
When I run this command from shell, it runs ok export REVS=`svn info svn+ssh://svn.myone.ca/var/svn/story/trunk/lib |grep 'Last Changed Rev:'| awk -F: '{print $2}'` However when I save it into a file called test.sh (of course, I chmod it with +x), I got error "export: 2: bad variable name"
Here is the file: #!/bin/bash export REVS=`svn info svn+ssh://svn.myone.ca/var/svn/story/trunk/lib |grep 'Last Changed Rev:'| awk -F: '{print $2}'` I am using ubuntu.
I have a file (.tmpfile) and inside it is a string which i only know part of, the rest being a random group of characters... I would like to know how to pull the whole string out of the file and into a variable.
Now, I have one script called "defcon" defcon gets the current DEFCON level and outputs it using echo.
Code:
#!/bin/bash DEFCON=`curl -s http://members.tripod.com/~Swat_25/defcon.html | sed -n '/^$/!{s/<[^>]*>//g;p;}' | sed '/^$/d' | grep '[12345]$'` echo "The current DEFCON level is $DEFCON"
The second script ("tweet") updates my twitter account.
What I want to do is be able to update my twitter account with the current defcon status (this is really more of a learning thing than something I actually want to be doing). The original script for tweet replaced $@ with $1, but if I use:
tweet `defcon`
it only uses the first word in the string, similarly if I used $2 or $3.So I changed it to $@. The normal function still works, but typing:
tweet `defcon`
updates twitter with nothing.
EDIT I should mention the /dev/null is there to catch the output of curl, otherwise it won't run silently. It still updates twitter normally with the send to /dev/null
Now in my bash script, I want to get the output /home/user instead of $HOME once read. So far, I have managed to get the $HOME variable but I can't get it to echo the variable. All I get is the output $HOME.
If I have an interactive program, e.g top, which, after launch, can be fed with key strokes and changes its behavior, for example, after launching top and then pressing "1", you can see the CPU utilization per core. I was wondering if there is a programmatical way to do so, so I don't have to press "1" every time. I would think it is an easy task, but so far my attempts have failed.
run_repeatedly "programX -o "./messy/path/output-$NUM.txt"" The echo inside the loop prints "...-$NUM.txt"; obviously I'm aiming to have bash substitute the iteration number so that I end up with many output files not 1.
I'm writing a script for asterisk to monitor trunk failure, i do a loop for every trunk it got nad would like to name variable like server1=, server2= naming the server upgoing as the trunk is. here is the scripts:
[Code]....
what i would like to do is name the variable server, username and status with the count variable, like this server$COUNT to have server1 when on trunk one, bu as soon as i add the $COUNT after the server, it seems to try to make it a command, it says that:
Code: ./test.sh: line 45: server1=74.63.41.218: command not found
When Linux ask for my password in command line,my keyboard becomes unresponsive just for that moment or let me put it this way,until i need to write my password my keyboard is unresponsive,but for all other things is fine.or for example,i can input my password when i have a graphic interface but i can't with command line or i can't login to my computer from command lineexcuse me if my question is doll,i moved from windows to ubuntu just lastweek,( & i am flying high in sky with ubuntu & digging a very deep hole for windows too!!.)
Ive been using linux for a while but I am just getting into shell scripting, im currently trying to get a simple script for finding and copying files powered by the command:
Code:
This works fine from the command line but when put in a script such as:
Code:
Code:
with the keyboard inputs for $fc1 and $fc2 being *.doc and ~/test respectivly. The only problem i can see is the xargs -ivar "var" part possibly needing $var to be defined?
right now i am writing bash script for simple everyday todo tasks.script consisit of two files, fisrt is just script (which can delete/append /clear) and second one is todolist.txt which stores my notes.I am litlebit confused about sed!!for example, If my todolist.txt have these lines:
- Writing my Homework - take my girlfriend to launch - Take a break
How can sed take my input $@ and delete all line with name "Homework"..i trayed with many sed combination like:
Code: sed s/$@//g sed 'd/$@/' and many more combination with ""/'' or bracket but nothing helped