Test bash command line arguments - bash

I don't know bash well but this seems pretty basic, yet I'm stuck on it. I'm using the bash installed on Mac OS X. I'm simply trying to test 1 command line argument and this is what I have and it doesn't work.
if [$1 -eq 'clean']
then
echo "Your argument is 'clean'!"
fi
Every time I've tried it, bash gives me a command not found error.
I'm obviously doing something wrong, what is it?

Couple of issues here:
Spaces around [ and ] are required in shell
-eq is used for comparing integers not for strings
Try this instead:
if [[ "$1" == "clean" ]]; then
echo "Your argument is 'clean'!"
fi
If you are using bash then [[ and ]] are more efficient than [ and ]

Related

Verifying bash script inputs

I've just started using Linux as part of my computer science degree.
I'm writing some very simple Bash scripts and I've become a tad bit stuck.
I would like the script I'm attempting to write to be able to differentiate between "non valid inputs ie letters" from "valid inputs ie numbers from a specific range"
Currently the script "works" although I'm having troubles with another echo that I would like only to "echo" when the below line is "not true", is there a simple way to write this? I'm not specifically looking for efficient code, just code that I can learn from and understand at my amateur level.
So, long story short, is it possible to obtain information from the command line below, so that I can have a simple "not true" variable that I can use in another "else" or "elif" command?
For reference line 1 is to detect alphabetical inputs, and line 2 being the line of code I would like to write as "not true" for use in another part of my script.
let xx=$a+1-1 2>/dev/null; ret=$?
if [ $a -ge 7 ] && [ $a -le 70 ] && [ $xx -eq $xx ] && [ $ret -eq 0 ]
I'm not sure I'm explaining it very well, so any help would be appreciated. :)
Welcome to Stack Overflow. :)
Start by reading the docs. I don't mean that in any way to be mean - it's just the best way to go about this.
c.f. this manual
Then read through the BashFAQs
Also, this site is really your friend. Start by familiarizing yourself with how to ask a question well.
For your question, if I read it right:
typeset -i xx # accepts only digits now.
If the input is foo, the value defaults to 0, so now just check the range.
if (( xx >= 7 && xx <= 70 )); then : value is ok
else echo "Value must be a number from 7 to 70"
exit 1
fi
Good luck. :)
One problem with the "variable with integer attribute" is that it still doesn't protect you from invalid input:
$ declare -i aNumber
$ aNumber=1234X
bash: 1234X: value too great for base (error token is "1234X")
See 6.5 Shell Arithmetic for how bash interprets values to be numbers (scroll down to the paragraph starting with "Integer constants follow the C language definition")
In my experience, the best way to check for valid numeric input is with string-oriented pattern matching.
if [[ $1 =~ ^[+-]?[0-9]+$ ]]; then
echo "input $1 is an integer"
fi
In addition to extended regular expressions, bash's advanced pattern matching can also be used within [[...]]
if [[ $1 == ?([+-])+([0-9]) ]]; then
echo "input $1 is an integer"
fi
((...)) is preferred over let. See the let builtin
command for details.
Also the shellcheck wiki entry.

Confusion about bash file exists test operator

The following returns nothing:
which asdf
So why does the if statement get triggered here?
x=$(which asdf)
if [ -f $x ]; then echo "exists"; fi
You didn't quote $x, so your test becomes [ -f ], which is true because -f is a non-empty string.
if [ -f "$x" ]; then
Though Chepner has given good solution, in case you want to look for an alternate approach then try following once.
which asdf 2>&1 >/dev/null && echo "exists"
Looks like you are trying to check if a command exists. It is better to use the command builtin instead of which in that context, like this:
if command -v asdf; then
echo "exists"
fi
To learn more about command, try help command.
Related:
How to check if a program exists from a Bash script?
Is double square brackets [[ ]] preferable over single square brackets [ ] in Bash?

Command online argument in shell script

Hi I have written small shell script, I am not able to understand the behavior of that script. can any one help me to understand that script.
Script:
#!/bin/bash
if [ -z $1 ]
then
echo "fail"
else
echo "success"
fi
While executing the script .
./test.sh one
It exuting the else statement instead of main statement , even though its passing the argument.
can any one explain me this behavior to understand
The -z test in bash is checking if a string is an empty (zero length) value.
Since you're passing an argument to the script $1 is not empty and therefore -z $1 evaluates to false, executing the else portion of your script.
Side note: Since you're working with strings I recommend you to quote variables as follows:
if [ -z "$1" ]; then
echo "String is empty / No argument given"
else
echo "String is not empty / Argument given"
fi
Edit:
As pointed out by user1934428 it's probably better to use [[ instead of [. This, among others, eliminates the need for quoting. See more differences here.
if [[ -z $1 ]]; then
...
However, be aware that this is a bash extension and won't work in sh scripts.

How do I test if a string starts with another in bash?

Very similar but not duplicate : https://stackoverflow.com/a/2172367/57883
I'm in Git Bash 3.1 (at least that's what comes up in the prompt when I type bash inside git bash.
and $ test [["DEV-0" == D*]] || echo 'fail' prints fail.
if [['DEV-0-1' == DEV* ]]; then echo "yes"; says [[DEV-0-1: command not found
I'm trying to test if git branch returns something that starts with DEV. but I can't seem to apply the answer. is it because all my attempts are using a string literal on the left instead of a variable value?
I've also tried it on ideone http://ideone.com/3IyEND
and no luck.
It's been ~14 years since I was good with a linux prompt.
What am I missing for a string starts with test in bash?
You missed a space there:
if [[ 'DEV-0-1' == DEV* ]]; then echo "yes"; fi
^^
I'd probably rather do the check like this:
s1=DEV-0-1
s2=DEV
if [ "${s1:0:${#s2}}" == "$s2" ]; then
echo "yes"
fi

could someone help me figure out why this doesn't work?

i=0
if [$i -eq 0]
then
echo "i is equal to 0"
else
echo "NOT EQUAL <><><><><><><><><><><><><><><><><><><>"
fi
it is part of a bash script and it always takes the else branch. I'm completely new to bash so its probably something silly
you need [ $i instead of [$i.
This is because the [ is a builtin command and $i should be it's first parameter. If you miss the space between command and parameter, then the shell will look for [$i command and after evaluation will tell you that there is no [0 command to be executed.
You need spaces after '[' and before ']'. '[' is a command.

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