This question already has answers here:
Compound 'if' statements with multiple expressions in Bash
(4 answers)
An if statement with AND operator [duplicate]
(2 answers)
Closed 2 years ago.
My shell script looks like below.
#!/bin/bash
STR1="STRING1"
STR2="STRING2"
if [[ "sample string contains some value" == *"contains"* ]]; then
if [ "$STR1" != "$STR2" ]; then
echo "do something"
fi
fi
How do I can merge two if line into single if?
My script searches for a substring in a string and compares the other two strings to do something.
Related
This question already has answers here:
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Difference between single and double square brackets in Bash
(7 answers)
Why is "[[ 10 < 2 ]]" true when comparing numbers in bash? [duplicate]
(1 answer)
Closed 6 months ago.
I am trying some simple input validation in bash. Basically this part of my script outputs "press [index] to select [array element]." However I cannot seem to get it to stop and exit gracefully when an invalid input is entered. From my research so far this SHOULD work:
declare -a scenarios=()
scenarios+=("Scenario_123")
scenarios+=("Scenario_456")
scenarios+=("Scenario_789")
for i in ${!scenarios[#]}; do
echo -e "select $i for ${scenarios[$i]}"
done
read ScenInd
echo ${#scenarios[#]} #[${ScenInd}=~^[0-9]+$, =~ ^[[:digit:]]+
if ! [${ScenInd}=~ ^[[:digit:]]+$ ] || [${ScenInd} < 0] || [${ScenInd} >= ${#scenarios[#]}];
then
echo "INVALID SELECTION"
exit
fi
but when I run it and enter 8, I get '[8=~: command not found'
What have I done wrong and how do I fix this? I have tried this both with [${ScenInd}=~^[0-9]+$ and =~ ^[[:digit:]]+ yet the results are the same.
Thanks in advance
This question already has answers here:
String length of bash
(3 answers)
Closed 1 year ago.
I have two string variables that I need to compare by length inside of if. So I need to do something like this:
if [ length($string_one) != length($string_two) ]:
fi
if [[ ${#string} != ${#string_two} ]]; then
run command
fi
This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
Closed 3 years ago.
I have written a script for word guessing in bash.
But is not working properly.
w = "###"
read -p "input word: " var
while [ "$w" != "$var" ]
do
echo "Wrong."
read -p "input word: " var
done
echo "Right answer"
You shouldn't have spaces when declaring varaible:
code="banana"
This question already has answers here:
How do I compare two string variables in an 'if' statement in Bash? [duplicate]
(12 answers)
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 6 years ago.
#!/bin/bash
a=0
while ["$a" -lt 10]
do
a=`expr "$a" + 1`
echo $a
done
This is the error:
a.sh: 3: a.sh: [0: not found
you need to keep spaces around your brackets:
while [ "$a" -lt 10 ]
since bash interpretes strings; it uses spaces to separate words. When you write ["$a"; then bash reads: [0 (after $a is replaced with 0)
This question already has answers here:
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 6 years ago.
I am having trouble with executing the following bash script:
#!/bin/bash
response=" "
while ["$response" != "q"]; do
echo -n "Please enter a response"; read response
done
ALSO
!/bin/bash
response="x"
if ["$response" = "x"]
then
echo "the value is x"
fi
What could the possible errors be?
Your while and if statements are spaced wrong.
while [ "$response" != "q" ]; do etc
You need a space between the bracket and the double quote.