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Why does [$value -lt 10] in shell result in command not found? [duplicate]
(3 answers)
Closed 4 years ago.
I hope you all are doing great. I am new to Linux shell scripting. I am trying to learn basic shell scripting and i started with with if else condition and I am stuck in it since a week. I have read as many articles as I could but I am not understanding what is the problem here. I know this must a very basic for you guys but please help me :)
Here is the script that I wrote
#!/bin/sh
count=102
if [$count -gt 100]
then
echo "$count is greater than 100"
else
echo "$count is less than 100"
fi
When I execute this script, I am getting below error.
./count.sh: line 5: [102: command not found
102 is less than 100
*I also tried taking user input not just with integer but with string also, still getting the same error. please help :)
Thank you in advance
you need to provide space after [ and before ]. For e.g
[ $count -gt 100 ]
Sample code :
count=102
if [ $count -gt 100 ]
then
echo "$count is greater than 100"
else
echo "$count is less than 100"
fi
Related
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I am new to bash scripting and trying to write a loop to run a curl command but however getting errors. Please suggest the write syntax.
#!/usr/bin/env sh
start = $1
end = $2
start=$(date -d $start +%Y-%m-%d)
end=$(date -d $end +%Y-%m-%d)
while [[ $start -le $end ]]
do
echo $start
curl -H "command, which can't be shared"
start=$(date -d"$start + 1 day" +"%Y-%m-%d")
done
The command line to run the script used sh filename.sh "2022-02-01" "2022-02-25".
However, the first error I received is windows cannot find '2022-02-01'. Make sure you typed the name correctly, and then try again.
Please suggest the errors.
This question already has answers here:
Validate the number of arguments passed in bash from read
(2 answers)
Closed 2 years ago.
I'm creating a bash script. However, I need the script to validate that the user has to enter two arguments. Not 1 and nothing more than 3.
echo -n "Enter two values"
read val1 val2
try this:
#!/bin/bash
if [ $# -eq 2 ]; then
echo "var1: $1 var2: $2"
else
echo "error: need to set two argument"
fi
run:
./scriptName.sh first second
This question already has answers here:
if, elif, else statement issues in Bash
(5 answers)
Closed 3 years ago.
i have probably some nooby question
how can I call a defined function in a if statement in bash?
Have a look on my example:
#!/bin/bash -x
Variable_A=xyz
function_x() {
echo "hello everyone"
}
#here I want to define a if statement after the function
if Variable_A == working;then
function_x()
#here I have a while for the parameters in bash example:
while [[ $# -gt 0 ]]
do
key="$1"
case $key in
-s|--status)
Variable_A="$2"
shift
shift
;;
esac
done
so the idea is when I run the script let's call it test.sh it runs the script exp in bash:
./test.sh working
I am not sure how to create something like this, maybe some of you can help me out on this.
Thank you in advance
Your if should look like:
#here I want to define a if statement after the function
if [ "$Variable_A" = "working" ]
then
function_x
fi
You need to end every if statement with fi and functions are called without ().
https://www.thegeekdiary.com/bash-if-loop-examples-if-then-fi-if-then-elif-fi-if-then-else-fi/
Another important note on if statements. You need a space before and after [ or ].
Your if logic should also be after the script parms are read and Variable_A is set.
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When to wrap quotes around a shell variable?
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Closed 4 years ago.
when I run this script in my Ubuntu:
for item in *
do
if [ -d $item ]
then
echo $item
fi
done
I got this output:
output
I don't understand why I get "[: if-then: unexpected operator"
Escape your variables. Read about it here.
for item in *
do
if [ -d "$item" ]
then
echo "$item"
fi
done
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Closed 4 years ago.
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if [ -e $1/file.txt ];then
echo HAHA
fi
if [[ -e "$1/file.txt" ]]; then
echo "It exists"
fi