What is the gt for here? "if [ $VARIABLE -gt 0 ]; then" - linux

What does the -gt mean here:
if [ $CATEGORIZE -gt 0 ]; then
This is part of a bash script I'm working with.
Also, where can I find a list of "flags" that go in there so I can have for reference in the future?

-gt is an arithmetic test that denotes greater than.
Your condition checks if the variable CATEGORIZE is greater than zero.
Quoting from help test (the [ is a command known as test; help is a shell builtin that provides help on shell builtins):
arg1 OP arg2 Arithmetic tests. OP is one of -eq, -ne,
-lt, -le, -gt, or -ge.
-eq: Equal
-ne: Not equal
-lt: Less than
-le: Less than or equal to
-gt: Greater than
-ge: Greater than or equal to
You could also express the condition in an arithmetic context1 by saying:
if ((CATEGORIZE > 0)); then
instead of
if [ $CATEGORIZE -gt 0 ]; then
1 Quoting from help '((':
(( ... )): (( expression ))
Evaluate arithmetic expression.
The EXPRESSION is evaluated according to the rules for arithmetic
evaluation. Equivalent to "let EXPRESSION".
Exit Status:
Returns 1 if EXPRESSION evaluates to 0; returns 0 otherwise.

-gt means "greater than", compared arithmetically
[ is (peculiarly) an alias of test (with a mandatory last argument of ], to make it look like a pair of brackets).
bash has its own "builtin" version of [/test, so any bash reference (e.g man bash, info bash, or http://www.gnu.org/software/bash/manual/) will document that, or man [/man test should give you the documentation for the standard standalone version.
Specifically, this page gives an overview of the command, as implemented by bash, and this page lists the available operators.
As well as arithmetic and string tests, you may come across the -e test, for "file exists", as in [ -e /hard/coded/path/$variable_filename ]
bash also includes a slightly extended version, in the form of [[ ... ]].

Related

Why do two empty strings compare as not equal?

my#comp:~/wtfdir$ cat wtf.sh
str1=$(echo "")
str2=$(echo "")
if [ $str1 != $str2 ]; then
echo "WTF?!?!"
fi
my#comp:~/wtfdir$ ./wtf.sh
WTF?!?!
my#comp:~/wtfdir$
WTF is going on here?!
How I wrote the above code: Googling "bash compare strings" brought me to this website which says:
You can check the equality and inequality of two strings in bash by using if statement. “==” is used to check equality and “!=” is used to check inequality of the strings.
Yet I'm getting the above?
What am I not understanding? What am I doing wrong?
You aren't running a comparison at all, because you aren't using quotes where they're mandatory. See the warning from http://shellcheck.net/ about unquoted expansions at SC2086.
If both string are empty, then:
[ $str1 != $str2 ]
...evaluates to...
[ != ]
...which is a test for whether the string != is nonempty, which is true. Change your code to:
[ "$str1" != "$str2" ]
...and the exact values of those strings will actually be passed through to the [ command.
Another alternative is using [[; as described in BashFAQ #31 and the conditional expression page on the bash-hackers' wiki, this is extended shell syntax (in ksh, bash, and other common shells extending the POSIX sh standard) which suppresses the string-splitting behavior that's tripping you up:
[[ $str1 != "$str2" ]]
...requires quotes only on the right-hand side, and even those aren't needed for the empty-string case, but to prevent that right-hand side from being treated as a glob (causing the comparison to always reflect a match if str2='*').

Shell Scripting Ternary operator to get string result

In shell scripting, I am using ternary operator like this:
(( numVar == numVal ? (resVar=1) : (resVar=0) ))
I watched shell scripting tutorial by Derek Banas and got the above syntax at 41:00 of the video
https://www.youtube.com/watch?v=hwrnmQumtPw&t=73s
The above code works when we assign numbers to resVar, but if I try to assign a string to resVar, it always returns 0.
(( numVar == numVal ? (resVar="Yop") : (resVar="Nop") ))
and also tried
resVar=$(( numVar == numVal ? (echo "Yop") : (echo "Nop") ))
So which is the right way to do this?
You didn't tell us what shell you use but it's possible you use
bash or something similar. Ternary operator in Bash works only with numbers as
explained in man bash under ARITHMETIC EVALUATION section:
The shell allows arithmetic expressions to be evaluated, under
certain circumstances (see the let and declare builtin commands and
Arithmetic Expansion). Evaluation is done in fixed-width integers
with no check for over- flow, though division by 0 is trapped and
flagged as an error. The operators and their precedence,
associativity, and values are the same as in the C language. The
following list of operators is grouped into levels of
equal-precedence operators. The levels are listed in order of
decreasing precedence.
(...)
expr?expr:expr
conditional operator
And the reason that resVar is assigned 0 when you use "Yop" or
"Nop" is because such string is not a valid number in bash and
therefore it's evaluated to 0. It's also explained in man bash in
the same paragraph:
A null value evaluates to 0.
It's also explained in this Wikipedia
article if you find it
easier to read:
A true ternary operator only exists for arithmetic expressions:
((result = condition ? value_if_true : value_if_false))
For strings there only exist workarounds, like e.g.:
result=$([ "$a" == "$b" ] && echo "value_if_true" || echo
"value_if_false")
(where "$a" == "$b" can be any condition test, respective [, can
evaluate.)
Arkadiusz already pointed out that ternary operators are an arithmetic feature in bash, not usable in strings. If you want this kind of functionality in strings, you can always use arrays:
$ arr=(Nop Yop)
$ declare -p arr
declare -a arr='([0]="Nop" [1]="Yop")'
$ numVar=5; numVal=5; resvar="${arr[$((numVar == numVal ? 1 : 0))]}"; echo "$resvar"
Yop
$ numVar=2; numVal=5; resvar="${arr[$((numVar == numVal ? 1 : 0))]}"; echo "$resvar"
Nop
Of course, if you're just dealing with two values that can be in position 0 and 1 in your array, you don't need the ternary; the following achieves the same thing:
$ resvar="${arr[$((numVar==numVal))]}"
you can use this simple expression :
resVar=$([ numVar == numVal ] && echo "Yop" || echo "Nop")
Running with Gohti's idea to make the script more readable:
#!/bin/bash
declare -a resp='([0]="not safe" [1]="safe")'
temp=70; ok=$(( $temp > 60 ? 1 : 0 ))
printf "The temperature is $temp Fahrenheit, it is ${resp[$ok]} to go outside\n";
temp=20; ok=$(( $temp > 60 ? 1 : 0 ))
printf "The temperature is $temp Fahrenheit, it is ${resp[$ok]} to go outside\n";

Having trouble with simple Bash if/elif/else statement

I'm writing bash scripts that need to work both on Linux and on Mac.
I'm writing a function that will return a directory path depending on which environment I'm in.
Here is the pseudo code:
If I'm on a Mac OS X machine, I need my function to return the path:
/usr/local/share/
Else if I'm on a Linux machine, I need my function to return the path:
/home/share/
Else, you are neither on a Linux or a Mac...sorry.
I'm very new to Bash, so I apologize in advance for the really simple question.
Below is the function I have written. Whether I'm on a Mac or Linux, it always returns
/usr/local/share/
Please take a look and enlighten me with the subtleties of Bash.
function get_path(){
os_type=`uname`
if [ $os_type=="Darwin" ]; then
path="/usr/local/share/"
elif [ $os_type=="Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
echo $path
}
You need spaces around the operator in a test command: [ $os_type == "Darwin" ] instead of [ $os_type=="Darwin" ]. Actually, you should also use = instead of == (the double-equal is a bashism, and will not work in all shells). Also, the function keyword is also nonstandard, you should leave it off. Also, you should double-quote variable references (like "$os_type") just in case they contain spaces or any other funny characters. Finally, echoing an error message ("...not supported") to standard output may confuse whatever's calling the function, because it'll appear where it expected to find a path; redirect it to standard error (>&2) instead. Here's what I get with these cleaned up:
get_path(){
os_type=`uname`
if [ "$os_type" = "Darwin" ]; then
path="/usr/local/share/"
elif [ "$os_type" = "Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported" >&2
exit 1
fi
echo "$path"
}
EDIT: My explanation of the difference between assignments and comparisons got too long for a comment, so I'm adding it here. In many languages, there's a standard expression syntax that'll be the same when it's used independently vs. in test. For example, in C a = b does the same thing whether it's alone on a line, or in a context like if ( a = b ). The shell isn't like that -- its syntax and semantics vary wildly depending on the exact context, and it's the context (not the number of equal signs) that determines the meaning. Here are some examples:
a=b by itself is an assignment
a = b by itself will run a as a command, and pass it the arguments "=" and "b".
[ a = b ] runs the [ command (which is a synonym for the test command) with the arguments "a", "=", "b", and "]" -- it ignores the "]", and parses the others as a comparison expression.
[ a=b ] also runs the [ (test) command, but this time after removing the "]" it only sees a single argument, "a=b" -- and when test is given a single argument it returns true if the argument isn't blank, which this one isn't.
bash's builtin version of [ (test) accepts == as a synonym for =, but not all other versions do.
BTW, just to make things more complicated bash also has [[ ]] expressions (like test, but cleaner and more powerful) and (( )) expressions (which are totally different from everything else), and even ( ) (which runs its contents as a command, but in a subshell).
You need to understand what [ means. Originally, this was a synonym for the /bin/test command. These are identical:
if test -z "$foo"
then
echo "String '$foo' is null."
fi
if [ -z "$foo" ]
then
echo "String '$foo' is null."
fi
Now, you can see why spaces are needed for all of the parameters. These are parameters and not merely boolean expressions. In fact, the test manpage is a great place to learn about the various tests. (Note: The test and [ are built in commands to the BASH shell.)
if [ $os_type=="Darwin" ]
then
This should be three parameters:
"$os_type"
= and not ==
"Darwin"
if [ "$os_type" = "Darwin" ] # Three parameters to the [ command
then
If you use single square brackets, you should be in the habit to surround your parameters with quotation marks. Otherwise, you will run into trouble:
foo="The value of FOO"
bar="The value of BAR"
if [ $foo != $bar ] #This won't work
then
...
In the above, the shell will interpolate $foo and $bar with their values before evaluating the expressions. You'll get:
if [ The value of FOO != The value of BAR ]
The [ will look at this and realize that neither The or value are correct parameters, and will complain. Using quotes will prevent this:
if [ "$foo" != "$bar" ] #This will work
then
This becomes:
if [ "The value of FOO" != "The value of BAR" ]
This is why it's highly recommended that you use double square brackets for your tests: [[ ... ]]. The test looks at the parameters before the shell interpolates them:
if [[ $foo = $bar ]] #This will work even without quotation marks
Also, the [[ ... ]] allows for pattern matching:
if [[ $os_type = D* ]] # Single equals is supported
then
path="/usr/local/share/"
elif [[ $os_type == L* ]] # Double equals is also supported
then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
This way, if the string is Darwin32 or Darwin64, the if statement still functions. Again, notice that there has to be white spaces around everything because these are parameters to a command (actually, not anymore, but that's the way the shell parses them).
Adding spaces between the arguments for the conditionals fixed the problem.
This works
function get_path(){
os_type=`uname`
if [ $os_type == "Darwin" ]; then
path="/usr/local/share/"
elif [ $os_type == "Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
echo $path
}

How to compare two floating-point values in shell script

I had to do a division in shell script and the best way was:
result1=`echo "scale=3; ($var1 / $total) * 100"| bc -l`
result2=`echo "scale=3; ($var2 / $total) * 100"| bc -l`
but I want to compare the values of $result1 and $result2
Using if test $result1 -lt $result2 or if [ $result1 -gt $result2 ] didn't work :(
Any idea how to do that?
You can compare floating-point numbers using expr(1):
: nr#yorkie 3724 ; expr 3.1 '<' 3.3
1
: nr#yorkie 3725 ; expr 3.1 '<' 3.09
0
You can also have bc do the comparisons as well as the calculations:
if [ "$(echo $result1 '<' $result2 | bc -l)" -eq 1 ];then ... fi
Finally, ksh93 can do arithmetic evaluation $(($result1 < $result2)) with floating-point numbers, although bash cannot.
note that you've gotta be a bit careful when dealing with floating point numbers and if you are testing for equality you really want to decide on some precision and then compare using that. Something like:
if (abs(x1-x2) < 0.0001) then equal # pseudo-code
the reason being that with computers we're dealing with limited-precision binary fractions not true mathematical reals. Limiting the precision in bc with the scale=3 will have this effect.
I'd also advise against trying to do this stuff in shell script. It's not that you can't do it but you'll have to fork off lots of little sub commands to do the tricky bits and that's slow to execute and generally a pain to write - you spend most of your time trying to get the shell to do what you want rather than writing the code you really want. Drop into a more sophisticated scripting language instead; my language of choice is perl but there are others. like this...
echo $var1 $var2 $total | perl -ne 'my ($var1, $var2, $tot) = split /\s+/; if ($var1/$tot == $var2/$tot) { print "equal\n"; }'
also note that you're dividing by the same value ($total in your question) so the whole comparison can be done against the numerators (var1 and var2) provided $total is positive
Posting a new answer since I cannot yet comment...
#Norman Ramsey's answer is not quite accurate:
expr will perform an integer or string comparison, not a floating-point comparison.
Here's what the man page says:
expr1 {=, >, >=, <, <=, !=} expr2
Return the results of integer comparison if both arguments are integers; otherwise, returns the results of string comparison using the locale-specific collation sequence.
(just try expr 8.9 '<' 10 and get 0 where it should be 1).
bcworks great, but isn't always installed.
So another alternative is using perl -e:
perl -e 'print expression' will print 1 if expression is true and nothing (empty string) otherwise.
e.g. perl -e 'print 8.9 < 10' - prints "1", while perl -e 'print 2>4' prints nothing.
And when used in if statement:
if [ $(perl -e "print $result1 < $result2") ];then ... fi

Compare integer in bash, unary operator expected

The following code gives
[: -ge: unary operator expected
when
i=0
if [ $i -ge 2 ]
then
#some code
fi
why?
Your problem arises from the fact that $i has a blank value when your statement fails. Always quote your variables when performing comparisons if there is the slightest chance that one of them may be empty, e.g.:
if [ "$i" -ge 2 ] ; then
...
fi
This is because of how the shell treats variables. Assume the original example,
if [ $i -ge 2 ] ; then ...
The first thing that the shell does when executing that particular line of code is substitute the value of $i, just like your favorite editor's search & replace function would. So assume that $i is empty or, even more illustrative, assume that $i is a bunch of spaces! The shell will replace $i as follows:
if [ -ge 2 ] ; then ...
Now that variable substitutions are done, the shell proceeds with the comparison and.... fails because it cannot see anything intelligible to the left of -gt. However, quoting $i:
if [ "$i" -ge 2 ] ; then ...
becomes:
if [ " " -ge 2 ] ; then ...
The shell now sees the double-quotes, and knows that you are actually comparing four blanks to 2 and will skip the if.
You also have the option of specifying a default value for $i if $i is blank, as follows:
if [ "${i:-0}" -ge 2 ] ; then ...
This will substitute the value 0 instead of $i is $i is undefined. I still maintain the quotes because, again, if $i is a bunch of blanks then it does not count as undefined, it will not be replaced with 0, and you will run into the problem once again.
Please read this when you have the time. The shell is treated like a black box by many, but it operates with very few and very simple rules - once you are aware of what those rules are (one of them being how variables work in the shell, as explained above) the shell will have no more secrets for you.
Judging from the error message the value of i was the empty string when you executed it, not 0.
I need to add my 5 cents. I see everybody use [ or [[, but it worth to mention that they are not part of if syntax.
For arithmetic comparisons, use ((...)) instead.
((...)) is an arithmetic command, which returns an exit status of 0 if
the expression is nonzero, or 1 if the expression is zero. Also used
as a synonym for "let", if side effects (assignments) are needed.
See: ArithmeticExpression
Your piece of script works just great. Are you sure you are not assigning anything else before the if to "i"?
A common mistake is also not to leave a space after and before the square brackets.

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