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  • Parsing result of Diff in Shell Script

    - by Saobi
    I want to compare two files and see if they are the same or not in my shell script, my way is: diff_output=`diff ${dest_file} ${source_file}` if [ some_other_condition -o ${diff_output} -o some_other_condition2 ] then .... fi Basically, if they are the same ${diff_output} should contain nothing and the above test would evaluate to true. But when I run my script, it says [: too many arguments On the if [....] line. Any ideas?

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  • Beginning with shell scripting

    - by Kevin Wyman
    I am fresh into Ubuntu and one of my goals is shell scripting for personal (and maybe public) use. I'm a novice, though I do understand some of the basics (e.g. what a variable, string, loop, etc... is) but to get the most of scripting I need to learn in-depth. I figure the best way to do that is to jump right into scripting and ask questions only pertinent to the stage I am at in my attempted script. Scenario: I have edited my sudoers file to allow my non-root user to run sudo commands without being prompted for a password. Question: In vim, what would be the best code to use for a function that checks whether this condition is [true], If not, prompt the user if they want the script to edit and save the sudoers file to make this condition [true]? Layout - If condition is true, carry-on with rest of script. If condition is not true, the script silently edits/adds the line: %sudo ALL=(ALL:ALL) NOPASSWD: ALL in the sudoers file, saves and then continues on with the next part of the script. Any help with this would be greatly appreciated and assist me in my journey to writing shell scripts.

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  • PHP system() help

    - by sea_1987
    Hello, I have this shell script #!/bin/sh ############################################################# # Example startup script for the SecureTrading Xpay4 client # # Install Xpay4 into /usr/local/xpay4 # # To run this script automatically at startup, place the # # following line at the end of the bootup script. # # eg. for RedHat linux: /etc/rc.d/rc.local # # # # /usr/local/xpay4/xpay4.sh # ############################################################# # Configuration options # Path to java executable JAVAPATH=/System/Library/Frameworks/JavaVM.framework/Versions/1.6.0/Home ########## Do not alter anything below this line ########## echo "Starting Xpay4. Please ensure the Xpay4 client is not already running" $JAVAPATH/java -jar /usr/local/xpay4/Xpay4.jar /usr/local/xpay4/xpay4.ini & And I am trying to run it using, system("/x/sh/shell.sh"); I am doing this when a user navigates to a certain page on my site, however I am getting just a white blank screen is there a way to error check with system(), I am currently using error_reporting(E_ALL | E_STRCIT) and that is applied site wide

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  • How do I get long command lines to wrap to the next line?

    - by BrianH
    Edit It was my .bashrc file. I've copied the same profile from machine to machine, and I used special characters in my $PS1 that are somehow throwing it off. I'm now sticking with the standard bash variables for my $PS1. Thanks to @ændrük for the tip on the .bashrc! ...End Edit... Something I have noticed in Ubuntu for a long time that has been frustrating to me is when I am typing a command at the command line that gets longer (wider) than the terminal width, instead of wrapping to a new line, it goes back to column 1 on the same line and starts over-writing the beginning of my command line. (It doesn't actually overwrite the actual command, but visually, it is overwriting the text that was displayed). It's hard to explain without seeing it, but let's say my terminal was 20 characters wide (Mine is more like 120 characters - but for the sake of an example), and I want to echo the English alphabet. What I type is this: echo abcdefghijklmnopqrstuvwxyz But what my terminal looks like before I hit the key is: pqrstuvwxyzghijklmno When I hit enter, it echos abcdefghijklmnopqrstuvwxyz so I know the command was received properly. It just wrapped my typing after the "o" and started over on the same line. What I would expect to happen, if I typed this command in on a terminal that was only 20 characters wide would be this: echo abcdefghijklmno pqrstuvwxyz Background: I am using bash as my shell, and I have this line in my ~/.bashrc: set -o vi to be able to navigate the command line with VI commands. I am currently using Ubuntu 10.10 server, and connecting to the server with Putty. In any other environment I have worked in, if I type a long command line, it will add a new line underneath the line I am working on when my command gets longer than the terminal width and when I keep typing I can see my command on 2 different lines. But for as long as I can remember using Ubuntu, my long commands only occupy 1 line. This also happens when I am going back to previous commands in the history (I hit Esc, then 'K' to go back to previous commands) - when I get to a previous command that was longer than the terminal width, the command line gets mangled and I cannot tell where I am at in the command. The only work-around I have found to see the entire long command is to hit "Esc-V", which opens up the current command in a VI editor. I don't think I have anything odd in my .bashrc file. I commented out the "set -o vi" line, and I still had the problem. I downloaded a fresh copy of Putty and didn't make any changes to the configuration - I just typed in my host name to connect, and I still have the problem, so I don't think it's anything with Putty (unless I need to make some config changes) Has anyone else had this problem, and can anyone think of how to fix it? Thanks in advance! Brian

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  • How do I get long command lines to wrap to the next line?

    - by BrianH
    Edit It was my .bashrc file. I've copied the same profile from machine to machine, and I used special characters in my $PS1 that are somehow throwing it off. I'm now sticking with the standard bash variables for my $PS1. Thanks to @ændrük for the tip on the .bashrc! ...End Edit... Something I have noticed in Ubuntu for a long time that has been frustrating to me is when I am typing a command at the command line that gets longer (wider) than the terminal width, instead of wrapping to a new line, it goes back to column 1 on the same line and starts over-writing the beginning of my command line. (It doesn't actually overwrite the actual command, but visually, it is overwriting the text that was displayed). It's hard to explain without seeing it, but let's say my terminal was 20 characters wide (Mine is more like 120 characters - but for the sake of an example), and I want to echo the English alphabet. What I type is this: echo abcdefghijklmnopqrstuvwxyz But what my terminal looks like before I hit the key is: pqrstuvwxyzghijklmno When I hit enter, it echos abcdefghijklmnopqrstuvwxyz so I know the command was received properly. It just wrapped my typing after the "o" and started over on the same line. What I would expect to happen, if I typed this command in on a terminal that was only 20 characters wide would be this: echo abcdefghijklmno pqrstuvwxyz Background: I am using bash as my shell, and I have this line in my ~/.bashrc: set -o vi to be able to navigate the command line with VI commands. I am currently using Ubuntu 10.10 server, and connecting to the server with Putty. In any other environment I have worked in, if I type a long command line, it will add a new line underneath the line I am working on when my command gets longer than the terminal width and when I keep typing I can see my command on 2 different lines. But for as long as I can remember using Ubuntu, my long commands only occupy 1 line. This also happens when I am going back to previous commands in the history (I hit Esc, then 'K' to go back to previous commands) - when I get to a previous command that was longer than the terminal width, the command line gets mangled and I cannot tell where I am at in the command. The only work-around I have found to see the entire long command is to hit "Esc-V", which opens up the current command in a VI editor. I don't think I have anything odd in my .bashrc file. I commented out the "set -o vi" line, and I still had the problem. I downloaded a fresh copy of Putty and didn't make any changes to the configuration - I just typed in my host name to connect, and I still have the problem, so I don't think it's anything with Putty (unless I need to make some config changes) Has anyone else had this problem, and can anyone think of how to fix it? Thanks in advance! Brian

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  • idioms for returning multiple values in shell scripting

    - by Wang
    Are there any idioms for returning multiple values from a bash function within a script? http://tldp.org/LDP/abs/html/assortedtips.html describes how to echo multiple values and process the results (e.g., example 35-17), but that gets tricky if some of the returned values are strings with spaces in. A more structured way to return would be to assign to global variables, like foo () { FOO_RV1="bob" FOO_RV2="bill" } foo echo "foo returned ${FOO_RV1} and ${FOO_RV2}" I realize that if I need re-entrancy in a shell script I'm probably doing it wrong, but I still feel very uncomfortable throwing global variables around just to hold return values. Is there a better way? I would prefer portability, but it's probably not a real limitation if I have to specify #!/bin/bash.

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  • Connect to a network via Command Linke

    - by justasking
    Hi guys, I want to be able to connect to a network via command line in Windows. My goal is to script out remoting into my work computer. I hate having to always manually connect to the VPN connection and then rdping into my work machine. I want to just have a script which will do both of it for me. I know how to rdp via command line, I just need to know how to connect to my VPN via command line. Thanks!

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  • Command to zip a directory using a specific directory as the root

    - by Slokun
    I'm writing a PHP script that downloads a series of generated files (using wget) into a directory, and then zips then up, using the zip command. The downloads work perfectly, and the zipping mostly works. I run the command: zip -r /var/www/oraviewer/rgn_download/download/fcst_20100318_0319.zip /var/www/oraviewer/rgn_download/download/fcst_20100318_0319 which yields a zip file with all the downloaded files, but it contains the full /var/www/oraviewer/rgn_download/download/ directories, before reaching the fcst_20100318_0319/ directory. I'm probably just missing a flag, or something small, from the zip command, but how do I get it to use fcst_20100318_0319/ as the root directory?

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  • Executing shell commands from Java

    - by Lauren?iu Dascalu
    Hello, I'm trying to execute a shell command from a java application, on the GNU/Linux platform. The problem is that the script, that calls another java application, never ends, although it runs successfully from bash. I tried to debug it: (gdb) bt #0 0xb773d422 in __kernel_vsyscall () #1 0xb7709b5d in pthread_join (threadid=3063909232, thread_return=0xbf9cb678) at pthread_join.c:89 #2 0x0804dd78 in ContinueInNewThread () #3 0x080497f6 in main () I tried with: ProcessBuilder(); and Runtime.getRuntime().exec(cmd); Looks like it waits for something to finish. Any ideas? Thanks, Lauren?iu

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  • run program multiple times using one line shell command

    - by teehoo
    I have the following gifs on my linux system: $ find . -name *.gif ./gifs/02.gif17.gif ./gifs/fit_logo_en.gif ./gifs/halloween_eyes_63.gif ./gifs/importing-pcs.gif ./gifs/portal.gif ./gifs/Sunflower_as_gif_small.gif ./gifs/weird.gif ./gifs2/00p5dr69.gif ./gifs2/iss013e48788.gif ...and so on What I have written is a program that converts GIF files to BMP with the following interface: ./gif2bmp -i inputfile -o outputfile My question is, is it possible to write a one line command using xargs, awk, find etc. to run my program once for each one of these files? Or do I have to write a shell script with a loop?

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  • Create database in Shell Script - convert from PHP

    - by snaken
    I have the following PHP code that i use to create a databaase and grant permissions to a user: $con = mysql_connect("IP.ADDRESS","user","pass"); mysql_query("CREATE DATABASE ".$dbuser."",$con)or die(mysql_error()); mysql_query("grant all on ".$dbuser.".* to ".$dbname." identified by '".$dbpass."'",$con) or die(mysql_error()); I want to perform these same actions but from within a shell script. Is it just something like this: MyUSER="user" MyPASS="pass" MYSQL -u $MyUSER -h -p$MyPASS -Bse "CREATE DATABASE $dbuser;' MYSQL -u $MyUSER -h -p$MyPASS -Bse "GRANT ALL ON ${DBUSER}.* to $DBNAME identified by $DBPASS;"

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  • Reading into directory, unix shell script

    - by dubya
    I am trying to analyze the files/directories inside of a directory using a shell script, for example if the file is readable, if it is a file, if it is a directory, etc. My script is set up to take a directory as input. so I would type 'file.sh directoryname'. However, when I create a for loop to analyze the files, it analyzes the files in my current working directory rather than the specified directory name. This is my broken code: file=$1 set mypath = $file for file in $mypath * do if [ -d $file ] dirCount=`expr $dirCount + 1` fi done Why does this read the working directory instead of the specified directory? Any help is appreciated. Thanks

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  • Shell script to count files, then remove oldest files

    - by Nic Hubbard
    I am new to shell scripting, so I need some help here. I have a directory that fills up with backups. If I have more than 10 backup files, I would like to remove the oldest files, so that the 10 newest backup files are the only ones that are left. So far, I know how to count the files, which seems easy enough, but how do I then remove the oldest files, if the count is over 10? if [ls /backups | wc -l > 10] then echo "More than 10" fi

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  • How do i compare 2 strings in shell?

    - by Thomas
    I want the user to input something at the command line either -l or -e. so e.g. $./report.sh -e I want an if statement to split up whatever decision they make so i have tried... if [$1=="-e"]; echo "-e"; else; echo "-l"; fi obviously doesn't work though Thanks

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  • Cannot install gnome shell extensions

    - by gnome
    I upgrade from 10.04 to 12.04, install gnome 3 and remove unity. My gnome version is 3.4.1. The GNOME Shell extensions is installed and enabled. When I use firefox to visit each extension page in https://extensions.gnome.org/, for example, https://extensions.gnome.org/extension/5/alternative-status-menu/, there is no where to install the extensions. I am told that one can install extensions by visite these extension pages. This is the guide I followed.

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  • Weird graphical issues with Gnome Shell in 12.10

    - by Chintan Parikh
    I finally got Gnome Shell working on Ubuntu 12.10, but there are some weird graphical issues. It seems as if the text is corrupted or something. Here's a screenshot of what I'm talking about: http://i.imgur.com/OqGyl.png It happens on a bunch of different applications, but not when I'm using Ubuntu. Does anyone know why or how to fix it? My graphics card is an AMD Radeon HD 6540G2, and I'm using the open source drivers (fglrx doesn't work with 12.10 yet).

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  • read file in shell script

    - by moata_u
    how can i read file in shell script , then assign each line to an variable that i can use later ,,,(am thinking in way to load an default setting from file) i already try : process (){ } FILE='' read -p "Please enter name of default file : " FILE if [ ! -f $FILE ]; then echo "$FILE : does not exists " exit 1 elif [ ! -r $FILE ]; then echo "$FILE : can not read " fi exec 0<"$FILE" n=0 while read -r line do (assign each line to an variable) done

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  • Newlines not being interpreted when passed to php via the command line

    - by CarbonX
    I have a PHP script that I'm invoking from another shell script that sends an automated email with a message generated from the shell script. Problem is, when I send the message all the newline characters are printed into the message. How do I get them to be interpreted? sendmail.sh: /path/to/phpscript/sendmail.php "Some Message With Newlines\nHello World.\n" sendmail.php: $message = $argv[1] . "\nNewline"; $smtp->send($to, $from, $message); The odd thing is the \n after the $argv variable is interpreted and actually prints Newline on a new line, but the newlines in the $argv variable don't, I have tried wrapping the variable in double quotes among other things but so far to no avail.

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  • Problem of * in Command line argument

    - by Nithya
    Hi, This is simple question only... i did a program in java that accepts input via command line arguments. I get input of two numbers and operator in command line. multiplying two numbers, i have to give input as 5 3 *. But if i give like this, i cant get answer. Y its not accept * in cmd.... waiting for reply guys...

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  • Korn Shell - Test with variable that may be not set

    - by C. Ross
    I have the following code in KornShell FAILURE=1 SUCCESS=0 isNumeric(){ if [ -n "$1" ]; then case $1 in *[!0-9]* | "") return $FAILURE; * ) return $SUCCESS; esac; else return $FAILURE; fi; } #... FILE_EXT=${FILE#*.} if [ isNumeric ${FILE_EXT} ]; then echo "Numbered file." fi #... In some cases the file name not have an extension, and this causes the FILE_EXT variable to be empty, which causes the following error: ./script[37]: test: 0403-004 Specify a parameter with this command. How should I be calling this function so that I do not get this error?

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  • Open Terminal with multiple tabs and execute application

    - by user172001
    I am new to linux shell scripting. I want to write a shell script which will open terminal with multiple tabs; it should run rtsp client app in each tab. For this, I have gone through question here in this forum and tried to code like bellow, tab="--tab-with-profile=Default -e " cmd="java RunRTSPClient" for i in 1 2 3 4 5 do # foo="$foo $tab $cmd" done gnome-terminal $foo exit 0 This is running and opens the terminal window with tabs but suddenly it will close. I am not getting any errors.

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  • using grep with pipe and ampersand to filter errors from find

    - by HKK
    I am using cygwin to find a file on the cygdrive. However I need to suppress the permission denied messages (otherwise the results get hidden in the error messages). The following command works: find -name 'myfile.*' |& grep -v "Permission denied" I don't understand why the ampersand needs to be put into this command, would have expected this to work but it doesn't. find -name 'myfile.*' | grep -v "Permission denied" Please explain the meaning of the ampersand.

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  • find grep pipe and ampersand

    - by HKK
    I am using cygwin to find a file on the cygdrive. However I need to suppress the permission denied messages (otherwise the results get hidden in the error messages). The following command works: find -name 'myfile.*' |& grep -v "Permission denied" I don't understand why the ampersand needs to be put into this command, would have expected this to work but it doesn't. find -name 'myfile.*' | grep -v "Permission denied" Please explain the meaning of the ampersand.

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