I'm receiveing an error on a simple script when using if/else statement.
The code:
#!/bin/sh
count=100
if [$count > 3]; then
echo "Test IF"
fi
The error: /bin/ash: line 6: [100: not found
#!/bin/sh
count=100;
if [ "$count" -gt 3 ]; then
echo "Test IF";
fi
Correct your syntax: spaces must be used around [ and ], parameter expansions must be quoted, and -gt is appropriate for numeric comparisons inside of [ ]. > in sh is used as redirection operator; if you want to use it in arithmetical comparison, you must use the bash-only syntax
$(( $count > 3 ))
#!/bin/sh
if [ $var -eq 12 ]; then
echo "This is a numeric comparison if example"
fi
if [ "$var" = "12" ]; then
echo "This is a string if comparison example"
fi
if [[ "$var" = *12* ]]; then
echo "This is a string regular expression if comparison example"
fi
The if statement in shell uses the command [.
Since [ is a command (you could also use 'test'), it requires a space before writing the condition to test.
To see the list of conditions, type:
man test
You'll see in the man page that:
s1 > s2 tests if string s1 is after string s2
n1 gt n2 tests if integer n1 is greater than n2
In your case, using > would work, because string 100 comes after string 3, but it is more logical to write
if [ $count -gt 3 ]; then
echo "test if"
fi
this will also do!
#!/bin/sh
count=100
if [ $count -gt 3 ];
then
echo "Test IF"
fi
if [[ "$x" == "true" ]]; then
echo "value matched"
else
echo "value doesn't matched"
fi
Related
Would someone please care to explain how on earth this is supposed to make sense:
Contents of insanity.sh:
#!/bin/bash
ARG=""
if [ -n $ARG ]; then
echo string is greater than zero
fi
if [ -z $ARG ]; then
echo string is empty
fi
Running the script:
[USERNAME#login001 clusterUtils]$ ./insanity.sh
string is greater than zero
string is empty
Currently using this tutorial at the moment.
It is happening because you're not quoting $ARG inside [ ... ].
Without quoting your code is effectively running as:
if [ -n ]; then
echo string is greater than zero
fi
if [ -z ]; then
echo string is empty
fi
Any non-empty string between [ ... ] will evaluate to true, hence both if conditions are successful.
Fix: It is recommended to use [[ ... ]] as you're using bash:
arg=""
if [[ -n $arg ]]; then
echo 'string is greater than zero'
fi
if [[ -z $arg ]]; then
echo 'string is equal to zero, empty'
fi
[[ ... ]] doesn't require your variables to be quoted like [ ... ] sine [ is an external command and [[ ... ]] is a builtin bash construct.
Also avoid all uppercase variables in your script to avoid conflicts with reserved env variables.
I am trying to solve a hackerrank exercise.
If n is odd, print Weird
If n is even and in the inclusive range of 2 to 5, print Not Weird
If n is even and in the inclusive range of 6 to 20, print Weird
If n is even and greater than 20, print Not Weird
My code is as follows:
read n
if [ $n%2==0 ]; then
if [ $n -ge 6 ] && [ $n -le 20 ]; then
echo "Weird"
else
echo "Not Weird"
fi
else
echo "Weird"
fi
When I give the input as 3, the result I get is Not Weird which is not correct same for 1 I get Not Weird. However, when I try this:
read n
if [ $(($n%2)) -eq 0 ]; then
if [ $n -ge 6 ] && [ $n -le 20 ]; then
echo "Weird"
else
echo "Not Weird"
fi
else
echo "Weird"
fi
I get the right result. What is the difference?
[ ] (or test) builtin:
==, or to be POSIX compliant =, does a string comparison
-eq does a numeric comparison
Note: == and -eq (and other comparisons) are parameters to the [ command, so they must be separated by whitespace, so $n%2==0 is invalid.
[[ ]] keyword:
is as [ except that it does pattern matching. Being a keyword rather than a builtin, expansion with [[ is done earlier in the scan.
(( )) syntax
Carries out arithmetic evaluation as with the let builtin. Whitespace separators are not mandatory. Using a leading $ to expand a variable is not necessary and is not recommended since it changes the expansion order.
For truth evaluation inside if-else, bash provides ((..)) operators with no need of a $ on the front.
n=5
if (( (n % 2) == 0 )); then
echo "Something"
if (( n >= 6 )) && (( n <= 20 )); then
echo "Some other thing"
else
echo "Other else thing"
fi
else
echo "Something else"
fi
Read here for more information.
This is my bash scripting code so I want to know How to Rewrite the below Bash script using a “for” loop instead of the “while” loop.
#!/bin/bash
if [ $# -gt 0 ]; then
a=0;
if [ -f RandNos ]; then
rm RandNos;
fi
while [ $a -lt $1 ]
do
a='expr $a + 1';
myrand=$RANDOM;
if [ "$2" "1"]; then
echo "No. $a ==> $myrand";
fi
echo $myrand>>RandNos
done
else
echo "please use with an argument..."
fi
Thanks.
The short of it: for counter-based loops, use the C-like form of the for loop:
for (( a = 0; a < $1; a++ )); do
# ... use $a
done
(This replaces while [ $a -lt $1 ]; do a='expr $a + 1' ...; done.)
See below for more on the rules that apply inside (( ... )).
As for the rest of your code:
Conditional [ "$2" "1"] is broken: it's missing the mandatory space before ]
With that fixed, it'll only work if $2 expands to a unary test operator such as -n.
Perhaps you meant if [[ -z $myrand ]]; then, to check if $RANDOM resulted in a nonempty string?
a='expr $a + 1' - which you don't need anymore with the for loop - doesn't actually invoke expr, because you're using single quotes - you'd need backticks (`) instead, or, preferably, the modern equivalent: $(expr $a + 1). However, with arithmetic evaluation, this could be simplified to (( ++a )).
[ ... ] conditionals work in bash, but they're provided for POSIX compatibility - use [[ ... ]] as the bash-specific alternative, which is more robust, has more features, and is faster.
bash statements only need terminating with ; if you place multiple on a single line
Note that bash considers do ... and then ... separate statements, hence you often see if ...; then and for ...; do.
In general, I encourage you to syntax-check your shell code at http://shellcheck.net - it's a great tool for detecting syntax problems.
Note how different rules apply inside (( ... )) compared to elsewhere in bash:
spaces around the = in the variable assignment are allowed.
referencing a variable without the $ prefix (a++) is allowed.
< performs numerical comparison (whereas inside [[ ... ]] it's lexical) -i.e., it's the more natural equivalent to -lt inside [ ... ] or [[ ... ]].
several other mathematical and even bit-wise operators are supported
...
All these different rules apply when bash operates in an arithmetic context, which applies to (( ... )), $(( ... )), array subscripts, and other cases.
For all the rules, run man bash and read the ARITHMETIC EVALUATION section.
Simply rewriting it with a for loop results in:
#!/bin/bash
if [ $# -gt 0 ]; then
if [ -f RandNos ]; then
rm RandNos;
fi
lim=$(expr $1 - 1)
as=$(seq 0 $lim)
for a in $as
do
a='expr $a + 1';
myrand=$RANDOM;
if [ "$2" "1"]; then # <- Caveat: conditional is BROKEN
echo "No. $a ==> $myrand";
fi
echo $myrand>>RandNos
done
else
echo "please use with an argument..."
fi
But there are several things wrong with the script anyhow. Like the last if statement.
if [ $# -lt 1 ];then
echo "First argument must be number".
exit 1;
fi
for a in `seq $1`
do
...
done
Several things can be improved:
#!/bin/bash
if (( $# )); then # anything but 0 is true
rm -f RandNos # remove if existing, otherwise fail silently
for ((a=0; a<$1; a++)); do
myrand=$RANDOM
# what is the intention here?
(( $2 > 1 )) && echo "No. $a ==> $myrand"
echo "$myrand" >> RandNos
done
else
echo "please use with an argument..."
fi
not sure what your intention was with the [ "$2" "1" ] expression. it is probably not what I made from it.
for ((a=1; a<=$1; a++)); do
may reflect your intended logic better, as you use $a for output only after incrementing it. as pointed out and corrected by #mklement0
!/bin/bash
if [ $# -gt 0 ]; then
a=0;
if [ -f RandNos ]; then
rm RandNos;
fi
for (( i=$a; i<$1; i++ ))
do
myrand=$RANDOM;
if [ "$2" = "1" ]; then
echo "No. $a ==> $myrand";
fi
echo $myrand >> RandNos
done
else
echo "please use with an argument..."
fi
I try to test if a string starts with a certain prefix. But my script seems not work (I would expect the "if" branch will not get run). Can some Bash expert help to take a look? thanks!
Here is my code and test result:
$ cat testb.bash
#!/bin/bash
my_var="abcdefg";
if [[ "${my_var:0:5}"=="order" ]]; then
echo "value of my_var is ${my_var}.";
fi;
if [[ "${my_var:0:5}" -eq "order" ]]; then
echo "value of my_var is ${my_var}.";
fi;
if [ "${my_var:0:5}"="order" ]; then
echo "value of my_var is ${my_var}.";
fi;
$ bash -x testb.bash
+ my_var=abcdefg
+ [[ -n abcde==order ]]
+ echo 'value of my_var is abcdefg.'
value of my_var is abcdefg.
+ [[ abcde -eq order ]]
+ echo 'value of my_var is abcdefg.'
value of my_var is abcdefg.
+ '[' abcde=order ']'
+ echo 'value of my_var is abcdefg.'
value of my_var is abcdefg.
$
Whitespace is significant in this case. As you can see in the -x output, it understands the first condition as
[[ -n "${my_var:0:5}==order" ]]
Moreover, to test for a prefix, you can use a pattern:
[[ $my_var == order* ]]
To test the existence of substring, you can use either of these:
if [[ "$j" =~ string1 ]]; then
if [[ $j == *string1* ]]; then
In your particular case, you miss a space surounding ==, so instead of
if [[ "${my_var:0:5}"=="order" ]]; then
it should be
if [[ "${my_var:0:5}" == "order" ]]; then
^ ^
Finally, note that your condition was evaluated as true because it was evaluating if [ "string" ], which is true if string is not empty:
$ [ "a" ] && echo "yes"
yes
Test
$ cat a
#!/bin/bash
my_var="abcdefg";
if [[ "${my_var:0:5}" == "order" ]]; then
echo "value of my_var is ${my_var}."
elif [[ "${my_var:0:5}" == "abcde" ]]; then
echo "yeahaa"
else
echo "is not"
fi
$ ./a
yeahaa
Ok, i tested your code, you shoud such as the following code:
prefix="pre_order";
pre="pre_"
len=${#pre}
echo $len
if [[ "${prefix:0:len}" == "blahvlah" ]] ; then
echo "dddd"
fi;
Notes:
use == for string comparation
for ${} you should initilize a string variable before ${}
use len=${#pre} for lenght of string.
A POSIX-compliant way to test for a prefix is to attempt to remove the prefix, and compare the result to the original string. If the two are the same, the prefix is not present, the removal fails, and the expression expands to the original string.
prefix=foo
string=foobar
if [ "${string#$prefix}" = "$string" ]; then
printf "$string does not start with $prefix\n"
else
printf "$string starts with $prefix\n"
fi
The Bash Reference Manual says that
[ string ]
and
[ -n string ]
will both return true if the string 's length is not 0
but the fact is not as so
greet=
if [ $greet ]; then
echo '1'
else
echo '2'
fi
if [ -n $greet ]; then
echo '1'
else
echo '2'
fi
the output is
2
1
the Bash Reference Manual just says
-n string
string
True if the length of string is non-zero.
so, what the real difference between the two form?
As #user1502952 said, you need to use double-quotes; but let me explain why. Suppose you execute:
greet=
[ -n $greet ] && echo "it's nonblank"
When the shell parses the [ -n $greet ] part, it expands $greet to the empty string, and then does word splitting. For instance, if $greet expanded to something with spaces in the middle, it would treat each "word" as a separate argument to the [ command. In this case, however, $greet expands to nothing, which contains no "word"s at all, and hence is treated as zero arguments to [ -- it effectively vanishes from the command. So [ -n $greet ] is equivalent to [ -n ], which checks to see if the string "-n" is nonblank. It is, so it evaluates to true.
Compare this with [ -n "$greet" ]: in this case, the double-quotes allow the expansion of $greet, but prevent word splitting. So the [ command actually gets a zero-length second argument, realizes that -n is supposed to be an operator, and gets the expected answer.
when you are using -n option, it is required to use double quotes.
if [ -n "$greet" ]
as the string is empty the above expression evaluates to false, as the length is zero.
if [ "$greet" ]
this also evaluates to false as the string is empty.
Moreover to check for empty string, -z option can be used.
if [ -z "$greet" ]
this will be true as the string is empty.
Check this link too: http://tldp.org/LDP/abs/html/comparison-ops.html
Bash performs word splitting inside [ but not inside [[, so you don't have to quote parameters if you use [[:
$ x=
$ [ -n $x ]; echo $?; [ -n "$x" ]; echo $?
0
1
$ [[ -n $x ]]; echo $?; [[ -n "$x" ]]; echo $?
1
1
type shows [[ $x ]] as [[ -n $x ]]:
$ f() { [[ $x ]]; }; type f
f is a function
f ()
{
[[ -n $x ]]
}