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Why should there be spaces around '[' and ']' in Bash?
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Closed 3 years ago.
I'm trying to do test automation with a bash script using if then else statements but I'm running into a few errors. For one, when I try to execute it I'm doing something wrong with the variable assignment with j and k, because it tells me that the command j and the command k aren't found when I try to execute. How do you correctly create variables?
The most confusing thing though is when I try to execute the script I get an error telling me I have an unexpected token near fi, and then it just says 'fi'. What am I doing wrong here?
#!/bin/bash
j = 0
k = 0
echo Test1:
echo -ne "0\nIn\nUG\n" | /u/cgi_web/Tuition/cost
echo Test2:
echo -ne "0\nOut\nUG\n" | /u/cgi_web/Tuition/cost
echo Test3:
echo -ne "0\nIn\nGR\n" | /u/cgi_web/Tuition/cost
echo Test4:
echo -ne "0\nOut\nGR\n" | /u/cgi_web/Tuition/cost
for i in {1..17}
do
echo Test$((i+4)):
if[ "$j" -eq 0 ] && [ "$k" -eq 0 ] then
$j = 1
echo -ne "$i\nIn\nUG\n" | /u/cgi_web/Tuition/cost
elif[ "$j" -eq 1 ] && [ "$k" -eq 0 ] then
$k = 1
echo -ne "$i\nIn\nGR\n" | /u/cgi_web/Tuition/cost
elif[ "$j" -eq 1 ] && [ "$k" -eq 1 ] then
$j = 0
echo -ne "$i\nOut\nUG\n" | /u/cgi_web/Tuition/cost
elif[ "$j" -eq 0 ] && [ "$k" -eq 1 ] then
$k = 0
echo -ne "$i\nOut\nGR\n" | /u/cgi_web/Tuition/cost
fi
done
EDIT: I figure out the variable issue with j and k, I had to remove the spaces in the statement.
Bash if statements require a semi-colon before the then:
if [ condition ] || [ condition ]; then
# code
elif [ condition ] && [ condition ]; then
# code
fi
For example.
To help anyone who might look at this for help in the future, I figured I'd answer my own question with all the syntax errors I found from my own testing and with the helpful responses of others.
To start the variable assignment:
j = 0
you can't have spaces in between, so it would be:
j=0
Also if statements need a space between if and the bracket and need a semicolon after the last bracket before then. Therefore my incorrect if statement
if[ "$j" -eq 0 ] && [ "$k" -eq 0 ] then
becomes
if [ "$j" -eq 0 ] && [ "$k" -eq 0 ]; then
or instead of a semicolon you can have a new line between the bracket, so it would become
if [ "$j" -eq 0 ] && [ "$k" -eq 0 ]
then
Related
Im trying to get an array from grades.txt, and determine what letter grade it should be assigned.
I either get
hw4part2.sh: line 26: [: : integer expression expected
If i use -ge or
hw4part2.sh: line 26: [: : unary operator expected
If i use >=
Below is the code im trying to get working
mapfile -t scores < grades.txt
numOScores=0
numOA=0
numOB=0
numOC=0
numOD=0
numOF=0
DoneWScores=0
A=90
B=80
C=70
D=60
F=59
while [ $DoneWScores -eq 0 ]
do
numOScores=$((numOScores + 1))
if [ "${scores[$numOScores]}" -ge "$A" ]
then
echo "A"
elif [ "${scores[$numOScores]}" -ge "$B" ]
then
echo "B"
elif [ "${scores[$numOScores]}" -ge "$C" ]
then
echo "C"
elif [ "${scores[$numOScores]}" -ge "$D" ]
then
echo "D"
elif [ "${scores[$numOScores]}" -le "$F" ]
then
echo "F"
else
echo "Done/error"
DoneWScores=1
fi
done
If anyone knows what my problem is, that'd be greatly appreciated
Consider this:
#!/usr/bin/env bash
if (( ${BASH_VERSINFO[0]} < 4 )); then
echo "Bash version 4+ is required. This is $BASH_VERSION" >&2
exit 1
fi
letterGrade() {
if (( $1 >= 90 )); then echo A
elif (( $1 >= 80 )); then echo B
elif (( $1 >= 70 )); then echo C
elif (( $1 >= 60 )); then echo D
else echo F
fi
}
declare -A num
while read -r score; do
if [[ $score == +([[:digit:]]) ]]; then
grade=$(letterGrade "$score")
(( num[$grade]++ ))
echo "$grade"
else
printf "invalid score: %q\n" "$score"
fi
done < grades.txt
for grade in "${!num[#]}"; do
echo "$grade: ${num[$grade]}"
done | sort
I'm nearly done writing a script for an assignment, but am having some trouble thinking of how to do this final part.
My problem is within a while loop; it prints out the number based on the IF statements, the number entered will always be an even number.
The IFs aren't connected by else/elif because the number should be able to printed out if it applies to more than 1 of the statements.
I want to print $starting on every loop if it doesn't meet any of the IF conditions, but if it does I don't want to print it. Can anyone see how to do that?
while [[ $starting -lt $ending ]]; do
if [ $((starting %7)) -eq 0 ]
then
echo "$starting red"
fi
if [ $((starting % 11)) -eq 0 ]
then
echo "$starting green"
fi
if [ $((starting % 13)) -eq 0 ]
then
echo "$starting blue"
fi
starting=$((starting + 2))
done
Keep track of whether you've done what you want to do in a variable:
while [[ $starting -lt $ending ]]; do
handled=0
if [ $((starting %7)) -eq 0 ]
then
echo "$starting red"
handled=1
fi
if [ $((starting % 11)) -eq 0 ]
then
echo "$starting green"
handled=1
fi
if [ $((starting % 13)) -eq 0 ]
then
echo "$starting blue"
handled=1
fi
if ! (( handled ))
then
echo "$starting didn't match anything"
fi
starting=$((starting + 2))
done
Add another if at the end that checks if none of the previous if statements are true. if !(starting%7==0 or starting%11==0 or starting%13==0) => echo starting
I'm trying to write a script that will check two error flags, and in case one flag (or both) are changed it'll echo-- error happened. My script:
my_error_flag=0
my_error_flag_o=0
do something.....
if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then
echo "$my_error_flag"
else
echo "no flag"
fi
Basically, it should be, something along:
if ((a=1 or b=2) or (a=1 and b=2))
then
display error
else
no error
fi
The error I get is:
line 26: conditional binary operator expected
line 26: syntax error near `]'
line 26: `if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then'
Are my brackets messed up?
Use -a (for and) and -o (for or) operations.
tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html
Update
Actually you could still use && and || with the -eq operation. So your script would be like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ] || ([ $my_error_flag -eq 1 ] && [ $my_error_flag_o -eq 2 ]); then
echo "$my_error_flag"
else
echo "no flag"
fi
Although in your case you can discard the last two expressions and just stick with one or operation like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ]; then
echo "$my_error_flag"
else
echo "no flag"
fi
You can use either [[ or (( keyword. When you use [[ keyword, you have to use string operators such as -eq, -lt. I think, (( is most preferred for arithmetic, because you can directly use operators such as ==, < and >.
Using [[ operator
a=$1
b=$2
if [[ a -eq 1 || b -eq 2 ]] || [[ a -eq 3 && b -eq 4 ]]
then
echo "Error"
else
echo "No Error"
fi
Using (( operator
a=$1
b=$2
if (( a == 1 || b == 2 )) || (( a == 3 && b == 4 ))
then
echo "Error"
else
echo "No Error"
fi
Do not use -a or -o operators Since it is not Portable.
Please try following
if ([ $dateR -ge 234 ] && [ $dateR -lt 238 ]) || ([ $dateR -ge 834 ] && [ $dateR -lt 838 ]) || ([ $dateR -ge 1434 ] && [ $dateR -lt 1438 ]) || ([ $dateR -ge 2034 ] && [ $dateR -lt 2038 ]) ;
then
echo "WORKING"
else
echo "Out of range!"
You can get some inspiration by reading an entrypoint.sh script written by the contributors from MySQL that checks whether the specified variables were set.
As the script shows, you can pipe them with -a, e.g.:
if [ -z "$MYSQL_ROOT_PASSWORD" -a -z "$MYSQL_ALLOW_EMPTY_PASSWORD" -a -z "$MYSQL_RANDOM_ROOT_PASSWORD" ]; then
...
fi
I was modifying an script didn't know how to write more than one condition in an if statement. I want to connect the two condition with an AND.
if [ envoi1 -eq 2 ];then
if [ envoi2 -eq 0 ];then
echo 'Ahora mismo.'
envoi = 1
fi
else
if [ envoi2 -eq 1 ];then
if [ envoi1 -eq 1 ];then
echo 'Situacion Normal.'
envoi = 1
fi
else
echo 'Raruno'
envoi=`expr $envoi1 + envoi2`
fi
fi
Now i use nested if to do the same but the code it's not so clear for me.
try this:
if [ $envoi1 -eq 2 ] && [ $envoi2 -eq 0 ] ; then
envoi = 1
fi
In bash, you can use [[ as follows:
if [[ $envoi2 -eq 1 && $envoi1 -eq 1 ]]; then
echo "Situacion Normal."
envoi=1
fi
However, [[ is not POSIX and will not work if you are using the /bin/sh shell. So if portability is desired use:
if [ $envoi2 -eq 1 -a $envoi1 -eq 1 ]; then
echo "Situacion Normal."
envoi=1
fi
Also note that when assigning variables you should not have any spaces on either side of the =.
My current script does the following;
It takes integer as a command line argument and starts from 1 to N , it checks whether the numbers are divisible by 3, 5 or both of them. It simply prints out Uc for 3, Bes for 5 and UcBes for 3,5. If the command line argument is empty, it does the same operation but the loop goes to 1 to 20.
I am having this error "Too many arguments at line 11,15 and 19".
Here is the code:
#!/bin/bash
if [ ! -z $1 ]; then
for i in `seq 1 $1`
do
if [ [$i % 3] -eq 0 ]; then
echo "Uc"
elif [ i % 5 -eq 0 ]; then
echo "Bes"
elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0 ]
then
echo "UcBes"
else
echo "$i"
fi
done
elif [ -z $1 ]
then
for i in {1..20}
do
if [ i % 3 -eq 0 ]
then
echo "Uc"
elif [ i % 5 -eq 0 ]
then
echo "Bes"
elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0 ]
then
echo "UcBes"
else
echo "$i"
fi
done
else
echo "heheheh"
fi
Note that [ is actually synonym for the test builtin in shell (try which [ in your terminal), and not a conditional syntax like other languages, so you cannot do:
if [ [$i % 3] -eq 0 ]; then
Moreover, always make sure that there is at least one space between [, ], and the variables that comprise the logical condition check in between them.
The syntax for evaluating an expression such as modulo is enclosure by $((...)), and the variable names inside need not be prefixed by $:
remainder=$((i % 3))
if [ $remainder -eq 0 ]; then
You should probably use something like :
if [ $(($i % 3)) -eq 0 ]
instead of
if [ $i % 3 -eq 0 ]
if [ [$i % 3] -eq 0 ]
Your script could be greatly simplified. For example:
#!/bin/sh
n=0
while test $(( ++n )) -le ${1:-20}; do
t=$n
expr $n % 3 > /dev/null || { printf Uc; t=; }
expr $n % 5 > /dev/null || { printf Bes; t=; }
echo $t
done
gives slightly different error messages if the argument is not an integer, but otherwise behaves the same.