This question already has answers here: Closed 11 years ago. Possible Duplicate: unix shell, getting exit code with piped child
I started to code shell scripting for about 2 days currently I have the following code echo \"Permission to Access to Remote Machine\"
I am using bash on Ubuntu.I would like to have a shell script open a program 开发者_开发技巧and continue on to the next line of the shell script, even though the program has not terminated.Adding an &
I want to loop through all the files in a given directory and return their version number and exe name. I have tried digging into the shell to see if I can pull this off, howeve开发者_开发技巧r I have
my script: #!/bin/bash . /home/was/.bash_profile export PATH=/usr/kerberos/bin:/usr/local/bin:/bin:/usr/bin:/usr/X11R6/bin:/home/was/bin:/wasdata/oracle/10.2.0/client_1/bin:
I am trying to get a bash script that g开发者_C百科enerates JSDoc for given parameters like this
I am trying to print the value of VARI in the same line followed by a comma, so that i can have a csv file of these values, but i m not able to save the value of
I have following set of values in a tab delimited file (only a part of values is shown here... file has 2 columns)
This script was written to make changes to the build.prop file on rooted devices.It will run on ubuntu but throws the following error when it is ran on a device. 6: Syntax error: expecting \"in\"
When I was new to shell scripting, I used a lot of short tests instead of if statements, like false && true.开发者_StackOverflow中文版
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