How to create dynamic alias (Conditional command) in bash? - bash

I want to create shortcut for 'code file' (VS Code) if there's a second argument, or 'clear' if there's none in a single command line, but i don't know how the bash syntax works.
By looking at the mkcd (mkdir & cd) shortcut that i created:
function mkcd {
if [ ! -n "$1" ]; then
echo "Enter a directory name"
elif [ -d $1 ]; then
echo "\`$1' already exists"
else
mkdir $1 && cd $1
fi
}
I tried to do the same, but the error shows 'syntax error near unexpected token `else''
function c {
if $1 then
code $1
else
clear
fi
}

Your syntax error is that if $1 then is missing a semi-colon (e.g. if $1; then) but read on... There are no "shortcut"s in UNIX. There are scripts, functions, and aliases. It looks like you want to create a function named c to call some command named code when an argument is given, some other command named clear otherwise. That'd be:
c() {
if (( $# > 0 )); then
code "$*"
else
clear
fi
}

Related

bash script syntax error at function call and passing array as argument

I'm new to bash so assume that I don't understand everything in this simple script as I've been putting this together as of today with no prior experience with bash.
I get this error when I run test.sh:
command substitution: line 29: syntax error near unexpected token `$1,'
./f.sh: command substitution: line 29: `index_of($1, $urls))'
FILE: f.sh
#!/bin/bash
urls=( "example.com" "example2.com")
error_exit()
{
echo "$1" 1>&2
exit 1
}
index_of(){
needle=$1
haystack=$2
for i in "${!haystack[#]}"; do
if [[ "${haystack[$i]}" = "${needle}" ]]; then
echo "${i}"
fi
done
echo -1
}
validate_url_param(){
index=-2 #-2 as flag
if [ $# -eq 0 ]; then
error_exit "No url provided. Exiting"
else
index=$(index_of($1, $urls)) #ERROR POINTS TO THIS LINE
if [ $index -eq -1 ]; then
error_exit "Provided url not found in list. Exiting"
fi
fi
echo $index
}
FILE: test.sh
#!/bin/bash
. ./f.sh
index=$(validate_url_param "example.com")
echo $index
echo "${urls[0]}"
I've lost track of all of the tweaks I tried but google is failing me and I'm sure this is some basic stuff so... thanks in advance.
The immediate error, just like the error message tells you, is that shell functions (just like shell scripts) do not require or accept commas between their arguments or parentheses around the argument list. But there are several changes you could make to improve this code.
Here's a refactored version, with inlined comments.
#!/bin/bash
urls=("example.com" "example2.com")
error_exit()
{
# Include script name in error message; echo all parameters
echo "$0: $#" 1>&2
exit 1
}
# A function can't really accept an array. But it's easy to fix:
# make the first argument the needle, and the rest, the haystack.
# Also, mark variables as local
index_of(){
local needle=$1
shift
local i
for ((i=1; i<=$#; ++i)); do
if [[ "${!i}" = "${needle}" ]]; then
echo "${i}"
# Return when you found it
return 0
fi
done
# Don't echo anything on failure; just return false
return 1
}
validate_url_param(){
# global ${urls[#]} is still a bit of a wart
if [ $# -eq 0 ]; then
error_exit "No url provided. Exiting"
else
if ! index_of "$1" "${urls[#]}"; then
error_exit "Provided url not found in list. Exiting"
fi
fi
}
# Just run the function from within the script itself
validate_url_param "example.com"
echo "${urls[0]}"
Notice how the validate_url_param function doesn't capture the output from the function it is calling. index_of simply prints the result to standard output and that's fine, just let that happen and don't intervene. The exit code tells us whether it succeeded or not.
However, reading the URLs into memory is often not useful or necessary. Perhaps you are simply looking for
grep -Fx example.com urls.txt

Shell script syntax error (if, then, else)

I have been trying to make a shell script in bash that will display the following:
You are the super user (When I run the script as root).
You are the user: "user" (When I run the script as a user).
#!/bin/bash/
if { whoami | grep "root" }; then
echo $USER1
else
echo $USER2
fi
I keep recieving these syntax error messages:
script.sh: line 2: syntax error near unexpected token `then'
script.sh: line 2: `if { whoami | grep "root" }; then'
Could someone help me out?
If braces are used to chain commands then the last command must have a command separator after it.
{ foo ; bar ; }
userType="$(whoami)"
if [ "$userType" = "root" ]; then
echo "$USER1"
else
echo "$USER2"
fi
pay attention with your first line, the correct syntax for she-bang is:
#!/bin/bash
everything you put there, is the interpreter of your script, you can also put something like #!/usr/bin/python for python scripts, but your question is about the if statement, so you can do this in two ways in shell script using
if [ test ] ; then doSomething(); fi
or
if (( test )) ; then doSomething(); fi
so to answer your question basically you need to do this
#!/bin/bash
if [ `id -u` -eq 0 ] ; then
echo "you are root sir";
else
echo "you are a normal user"
fi
if (( "$USER" = "root" )); then
echo "you are root sir";
else
echo "you are a normal user"
fi
note that you could use a command using `cmd` or $(cmd) and compare using -eq (equal) or = (same), hope this help you :-)

How to run my bash functions in terminal using a parent name?

* the wording of the question is terrible, sorry!
I have some bash functions I create
test() {echo "hello wold"}
test2() {echo "hello wold"}
Then in my .bashrc I source the file that has the above function . ~/my_bash_scripts/testFile
In the terminal I can run test and get hello world.
is there a way for me to add parent variable that holds all my functions together. For example personal test, personal test2.
Similar to every other gem out there, I downloaded a tweeter one. All it's methods are followed by the letter t, as in t status to write a status, instead of just status
You are asking about writing a command-line program. Just a simple one here:
#!/usr/bin/env bash
if [[ $# -eq 0 ]]; then
echo "no command specified"
exit
elif [[ $# -gt 1 ]]; then
echo "only one argument expected"
exit
fi
case "$1" in
test)
echo "hello, this is test1"
;;
test2)
echo "hello, this is test2"
;;
*)
echo "unknown command: $1"
;;
esac
Then save it and make it an executable by run chmod +x script.sh, and in your .bashrc file, add alias personal="/fullpath/to/the/script.sh".
This is just very basic and simple example using bash and of course you can use any language you like, e.g. Python, Ruby, Node e.t.c.
Use arguments to determine final outputs.
You can use "$#" for number of arguments.
For example,
if [ $# -ne 2 ]; then
# TODO: print usage
exit 1
fi
Above code exits if arguments not euqal to 2.
So below bash program
echo $#
with
thatscript foo bar baz quux
will output 4.
Finally you can combine words to determine what to put stdout.
If you want to flag some functions as your personal functions; no, there is no explicit way to do that, and essentially, all shell functions belong to yourself (although some may be defined by your distro maintainer or system administrator as system-wide defaults).
What you could do is collect the output from declare -F at the very top of your personal shell startup file; any function not in that list is your personal function.
SYSFNS=$(declare -F | awk '{ a[++i] = $3 }
END { for (n=1; n<=i; n++) printf "%s%s", (n>1? ":" : ""), a[n] }')
This generates a variable SYSFNS which contains a colon-separated list of system-declared functions.
With that defined, you can check out which functions are yours:
myfns () {
local fun
declare -F |
while read -r _ _ fun; do
case :$SYSFNS: in *:"$fun":*) continue;; esac
echo "$fun"
done
}

Bash function arguments not passed as expected

I have an unexpected behaviour in my script:
#!/bin/bash
checkFolders "first folder"
checkFolders() {
checkEmptyVar $1 "folder to be copied"
}
checkEmptyVar() {
echo "0: $0 1: $1 2: $2"
[[ $(isNotEmpty $1) = false ]] && (echo "Specify a path for $2") && exit
echo "post exit"
}
The function checkEmptyVar echoes the following:
0: ./lcp.sh 1: folder to be copied 2:
I expected to have passed "folder-to-be-copied" as $1of checkEmptyVar, what is happening?
You have numerous problems:
$0 is not the first argument to a function; $1 is. $0 is the name of the current script.
You must quote parameter expansions to prevent them from being split into multiple words on embedded whitespace.
Functions must be defined before they are used.
The correct script is
#!/bin/bash
checkFolders() {
checkEmptyVar "$1" "folder to be copied"
}
checkEmptyVar() {
[[ $(isNotEmpty "$1") = false ]] && echo "Specify a path for $2" && exit
echo "post exit"
}
checkFolders "first folder"
Further, it would be better to have isNotEmpty return a non-zero value instead of outputting the string false, so that you could write
checkEmptyVar () {
if isNotEmpty "$1"; then
echo "Specify a path for $2" >&2 # Use standard error, not standard output
exit 1
fi
echo "post exit"
}
(I suspect you could replace isNotEmpty with [[ -n $1 ]] anyway.)
This script isn't doing what you think it is. Your function definitions are happening too late.
When you call checkFolders on the first line you are calling a version of that function from the pre-existing environment and not the one defined later in that script.
If you run command -V checkFolders from the shell you are running this script from I expect you'll get output somewhat like:
checkFolders is a function
checkFolder ()
{
checkEmptyVar "folder to be copied"
}
though anything is possible there.
You should also always quote variables when you use them to prevent the shell from word-splitting their contents.
Without doing that when you call checkFolders "first folder" that ends up calling checkFolders first folder "folder to be copied" which isn't what you want at all.
Invert the order of your functions and calls and fix your variable quoting and you should see what you expect.

Equivalent of __FILE__ and __LINE__ in Bash

Is there any variable in bash that contains the name of the .sh file executed? The line number would be great too.
I want to use it in error messages such as:
echo "ERROR: [$FILE:L$LINE] $somefile not found"
#!/bin/bash
echo $LINENO
echo `basename $0`
$LINENO for the current line number
$0 for the current file. I used basename to ensure you only get the file name and not the path.
UPDATE:
#!/bin/bash
MY_NAME=`basename $0`
function ouch {
echo "Fail # [${MY_NAME}:${1}]"
exit 1
}
ouch $LINENO
You have to pass the line as a parameter if you use the function approach else you will get the line of the function definition.
I find the "BASH_SOURCE" and "BASH_LINENO" built-in arrays very useful:
$ cat xx
#!/bin/bash
_ERR_HDR_FMT="%.23s %s[%s]: "
_ERR_MSG_FMT="${_ERR_HDR_FMT}%s\n"
error_msg() {
printf "$_ERR_MSG_FMT" $(date +%F.%T.%N) ${BASH_SOURCE[1]##*/} ${BASH_LINENO[0]} "${#}"
}
error_msg "here"
error_msg "and here"
Invoking xx yields
2010-06-16.15:33:13.069 xx[11]: here
2010-06-16.15:33:13.073 xx[14]: and here
You just need to
echo $LINENO
echo $(basename $0)
Here's how to do it in a reusable function. if the following is in a file named script:
#!/bin/bash
debug() {
echo "${BASH_SOURCE[1]##*/}:${FUNCNAME[1]}[${BASH_LINENO[0]}]" > /dev/tty
}
debug
This produces the output:
script:main[5]
Which indicates the line on which debug was called.
The following will print out the filename, function, line and an optional message.
Also works in zsh for extra goodness.
# Say the file, line number and optional message for debugging
# Inspired by bash's `caller` builtin
# Thanks to https://unix.stackexchange.com/a/453153/143394
function yelp () {
# shellcheck disable=SC2154 # undeclared zsh variables in bash
if [[ $BASH_VERSION ]]; then
local file=${BASH_SOURCE[1]##*/} func=${FUNCNAME[1]} line=${BASH_LINENO[0]}
else # zsh
emulate -L zsh # because we may be sourced by zsh `emulate bash -c`
# $funcfiletrace has format: file:line
local file=${funcfiletrace[1]%:*} line=${funcfiletrace[1]##*:}
local func=${funcstack[2]}
[[ $func =~ / ]] && func=source # $func may be filename. Use bash behaviour
fi
echo "${file##*/}:$func:$line $*" > /dev/tty
}

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