I am not exactly sure if i am using the exits correctly. But when i execute the code with something that prints the usage statement it should stop there.
It should do one or the other. In my case it is doing both.
cmd="$1" ## the command to find
if [[ $# -ne 1 ]]
then
echo "usage: ./findcmd command"
fi
exit=1
path=$(echo $PATH | tr ":" " ")
for dir in $path
do
if [[ -x "$dir/$cmd" && -r "$dir/$cmd" ]]
then
echo "$dir/$cmd"
exit 0
fi
done
echo "$cmd not on $PATH"
exit=0
OUTPUT:
[112] ./findcmd
usage: ./findcmd command
/usr/local/bin/ **this should not be here
[113] ./findcmd ping
/usr/bin/ping
You use exit correctly to stop the script after your dir/cmd gets printed; try using it that way elsewhere.
Keep in mind that your first exit, once correct, will stop the script before the loop whether usage was printed or not.
It should be exit [n].
Within a script, an exit nnn command may be used to deliver an nnn
exit status to the shell (nnn must be an integer in the 0 - 255
range).
And it should be inside the IF/FOR block.
Like this:
cmd="$1" ## the command to find
if [[ $# -ne 1 ]]
then
echo "usage: ./findcmd command"
exit 1
fi
## rest of code to execute if args are correct
Related
I need to compare a string partially to check for a given condition.
Like my $1 will be checked if it has a part of a string BLR
while my file input has $1 entries as BLR21 BLR64 IND23
I only need a true condition when $1 is equal to BLR**
where these stars can be anything.
I used a simple if condition
if($1=="BLR21")
{print $2}
Now this only works when whole BLR21 is there in row.
I need to ckeck not for BLR21 but only BLR.
Please Help
Your question is not great, I hope I understood.
Quick and easy solution
grep BLR input.txt
This will output all the lines in which "BLR" is found, in file input.txt. It will match "BLR" with any prefix and suffix, whatever they might be (spaces, alphanumerical, tabs, ...).
"Complicated" solution
A bit more complicated. It does the same thing, but makes sure input.txt exists, and is in the form of a script.
Input file, input.txt:
BLR21 BLR64 IND23
Your script could be:
#!/bin/bash
#
# Arguments
inputfile="input.txt"
if [[ $# -ne 1 ]]
then
echo "Usage: myscript.bash <STRING>"
exit 1
else
string="$1"
fi
# Validation, and processing...
if [[ ! -f "$inputfile" ]]
then
echo "ERROR: file >>$inputfile<< does not exist."
exit 2
else
grep "$string" "$inputfile"
fi
And to call the script, you do:
./myscript.bash BLR
But really, a simple grep does the job here.
Taking it even further...
#!/bin/bash
#
# Arguments
inputfile="input.txt"
if [[ $# -ne 1 ]]
then
echo "Usage: check.bash <STRING>"
exit 1
else
string="$1"
fi
# Validation, and processing...
if [[ ! -f "$inputfile" ]]
then
echo "ERROR: file >>$inputfile<< does not exist."
exit 2
else
while read -r line
do
if [[ "$line" =~ $string ]]
then
echo "$line"
fi
done <"$inputfile"
fi
Now this one is like going to the moon via mars...
It reads each line of the file, one by one. Then it checks if that line contains the string, using the =~ operator inside the if.
But this is crazy, when a simple grep would do.
I am writing a script where I am grepping for a word and if it is true it should exit script else should continue the script. But my script is not going to else part even when grep output is false.
ps -fu rdb1 | grep humor >> humor_chk
grep humor.pl humor_chk | tail -1
if [ $? -eq 0 ]
then
echo "`date +%D' '%T` Humor is still running. Hence exiting" >> $LOG1
exit
else
ls -lrt /a/b/c/ >> files
fi
I am getting the output as:
01/19/15 05:54:03 Humor is still running. Hence exiting
#!/bin/bash
LOCATION=$1
FILECOUNT=0
DIRCOUNT=0
if [ "$#" -lt "1" ]
then
echo "Usage: ./test2.sh <directory>"
exit 0
fi
I don't actually get what the If statement is saying can anyone help me to explain this?Thank you
$1 refers to the first argument of the bash file. In this case, you can pass your directory path by issuing the following command:
# ./test2.sh /path/of/your/directory
#!/bin/bash
LOCATION=$1 #first argument of the script
FILECOUNT=0
DIRCOUNT=0
if [ "$#" -lt "1" ] #if the number of argument(s) ($#) is less than 1
then
echo "Usage: ./test2.sh <directory>"
exit 0
fi
You can read this article for more information about parameter passing.
Hope it helps.
$1 is the first argument that is passed to the bash script. If you start the script like ./test2.sh argument1 argument2 the $1 will refer argument1.
The if-statement checks, if the count of arguments (that's the $#) is smaller than 1, then it will output the usage statement (as it seems you can't run the script without any argument).
How to write a shell script named "backup.sh" which accepts one parameter, which would be a filename/directory.
Create a backup copy of that with the .bak appended to its name.Show message on success.
If the file/directory does not exist, show a proper message.
i did up to this point.please help me to figure this out
#!/bin/sh
#create_backup.sh
And add a .bak
bak="${backup.sh}.bak"
if [ "$#" -eq 0 ]
then
exit 1;
echo "File Succesfully backuped"
fi
cp ${1} "${1}.back"
echo "File is not found "
exit 0
#!/bin/bash -e
directory=$1
cp -r $directory $directory.bak
echo "Success"
obvious caveats with pathing/error codes/etc
I want to be able to tell if a command exists on any POSIX system from a shell script.
On Linux, I can do the following:
if which <command>; then
...snip...
fi
However, Solaris and MacOS which do not give an exit failure code when the command does not exist, they just print an error message to STDOUT.
Also, I recently discovered that the which command itself is not POSIX (see http://pubs.opengroup.org/onlinepubs/9699919799/idx/utilities.html)
Any ideas?
command -v is a POSIX specified command that does what which does.
It is defined to to return >0 when the command is not found or an error occurs.
You could read the stdout/stderr of "which" into a variable or an array (using backticks) rather than checking for an exit code.
If the system does not have a "which" or "where" command, you could also grab the contents of the $PATH variable, then loop over all the directories and search for the given executable. That's essentially what which does (although it might use some caching/optimization of $PATH results).
One which utility is available as shell script in the Git repository of debianutils package of Debian Linux. The script seems to be POSIX compatible and you could use it, if you take into account copyright and license. Note that there have been some controversy whether or not and how the which utility should be deprecated; (at time of writing) current version in Git shows deprecation message whereas an earlier version added later removed -s option to enable silent operation.
command -v as such is problematic as it may output a shell function name, an alias definition, a keyword, a builtin or a non-executable file path. On the other hand some path(s) output by which would not be executed by shell if you run the respective argument as such or as an argument for command. As an alternative for using the which script, a POSIX shell function using command -v could be something like
#!/bin/sh
# Argument $1 should be the basename of the command to be searched for.
# Outputs the absolute path of the command with that name found first in
# a directory listed in PATH environment variable, if the name is not
# shadowed by a special built-in utility, a regular built-in utility not
# associated with a PATH search, or a shell reserved word; otherwise
# outputs nothing and returns 1. If this function prints something for
# an argument, it is the path of the same executable as what 'command'
# would execute for the same argument.
executable() {
if cmd=$(unset -f -- "$1"; command -v -- "$1") \
&& [ -z "${cmd##/*}" ] && [ -x "$cmd" ]; then
printf '%s\n' "$cmd"
else
return 1
fi
}
Disclaimer: Note that the script using command -v above does not find an executable whose name equals a name of a special built-in utility, a regular built-in utility not associated with a PATH search, or a shell reserved word. It might not find an executable either in case if there is non-executable file and executable file available in PATH search.
A function_command_exists for checking if a command exists:
#!/bin/sh
set -eu
function_command_exists() {
local command="$1"
local IFS=":" # paths are delimited with a colon in $PATH
# iterate over dir paths having executables
for search_dir in $PATH
do
# seek only in dir (excluding subdirs) for a file with an exact (case sensitive) name
found_path="$(find "$search_dir" -maxdepth 1 -name "$command" -type f 2>/dev/null)"
# (positive) if a path to a command was found and it was executable
test -n "$found_path" && \
test -x "$found_path" && \
return 0
done
# (negative) if a path to an executable of a command was not found
return 1
}
# example usage
echo "example 1";
command="ls"
if function_command_exists "$command"; then
echo "Command: "\'$command\'" exists"
else
echo "Command: "\'$command\'" does not exist"
fi
command="notpresent"
if function_command_exists "$command"; then
echo "Command: "\'$command\'" exists"
else
echo "Command: "\'$command\'" does not exist"
fi
echo "example 2";
command="ls"
function_command_exists "$command" && echo "Command: "\'$command\'" exists"
command="notpresent"
function_command_exists "$command" && echo "Command: "\'$command\'" does not exist"
echo "End of the script"
output:
example 1
Command: 'ls' exists
Command: 'notpresent' does not exist
example 2
Command: 'ls' exists
End of the script
Note that even the set -eu that turns -e option for the script was used the script was executed to the last line "End of the script"
There is no Command: 'notpresent' does not exist in the example 2 because of the && operator so the execution of echo "Command: "\'$command\'" does not exist" is skipped but the execution of the script continues till the end.
Note that the function_command_exists does not check if you have a right to execute the command. This needs to be done separately.
Solution with command -v <command-to-check>
#!/bin/sh
set -eu;
# check if a command exists (Yes)
command -v echo > /dev/null && status="$?" || status="$?"
if [ "${status}" = 127 ]; then
echo "<handle not found 1>"
fi
# check if a command exists (No)
command -v command-that-does-not-exists > /dev/null && status="$?" || status="$?"
if [ "${status}" = 127 ]; then
echo "<handle not found 2>"
fi
produces:
<handle not found 2>
because echo was found at the first example.
Solution with running a command and handling errors including command not found.
#!/bin/sh
set -eu;
# check if a command exists (No)
command -v command-that-does-not-exist > /dev/null && status="$?" || status="$?"
if [ "${status}" = 127 ]; then
echo "<handle not found 2>"
fi
# run command and handle errors (no problem expected, echo exist)
echo "three" && status="$?" || status="$?"
if [ "${status}" = 127 ]; then
echo "<handle not found 3>"
elif [ "${status}" -ne 0 ]; then
echo "<handle other error 3>"
fi
# run command and handle errors (<handle not found 4> expected)
command-that-does-not-exist && status="$?" || status="$?"
if [ "${status}" = 127 ]; then
echo "<handle not found 4>"
elif [ "${status}" -ne 0 ]; then
echo "<handle other error 4>"
fi
# run command and handle errors (command exists but <handle other error 5> expected)
ls non-existing-path && status="$?" || status="$?"
if [ "${status}" = 127 ]; then
echo "<handle not found 5>"
elif [ "${status}" -ne 0 ]; then
echo "<handle other error 5>"
fi
produces:
<handle not found 2>
three
./function_command_exists.sh: 34: ./function_command_exists.sh: command-that-does-not-exist: not found
<handle not found 4>
ls: cannot access 'non-existing-path': No such file or directory
<handle other error 5>
The following works in both bash and zsh and avoids both functions and aliases.
It returns 1 if the binary is not found.
bin_path () {
if [[ -n ${ZSH_VERSION:-} ]]; then
builtin whence -cp "$1" 2> /dev/null
else
builtin type -P "$1"
fi
}