I have a shell script that runs a test repeatedly :
#!/bin/tcsh
set x = 1
while ($x <= 10000)
echo $x
./test
# x += 1
end
I am trying to adapt it to break the loop and stop running if the test failed, i.e. the test executable returned with a non-zero status. I thought the following change would work.
#!/bin/tcsh
set x = 1
set y = 0
while ($x <= 10000 && $y == 0)
echo $x
# y = ./test
# x += 1
end
But, I get error #: Expression syntax
Can you please tell me what did I do wrong, and how to capture the return value of ./test in a variable so I can break the loop, or some other way to break the loop upon encountering the test failure
I'm not a fan of scripting in csh, and I highly advise against it. However, in this case, csh seems to do the right thing, and you can simply do:
#!/bin/tcsh
set x = 1
while ($x <= 10000)
echo $x
./test || break
# x += 1
end
Related
I am working on updating a script to append the loop number iteration to a filename. However I can't find any documentation that will help me update the variable value. Instead, it appends the operation as a string.
(base) me#axoneme:~$ tcsh
axoneme:~> set j=100
axoneme:~> set j=$j+1
axoneme:~> echo $j
100+1
How can I do arithmetic in tcsh such that the 102nd loop will just be set j= $j + 1 where $j =101
# sign seems to be used in tcsh
axoneme:~> set j = 100
axoneme:~> # j = 100 + 1
axoneme:~> echo $j
101
I'm new to Bash and I've been having issues with creating a script. What this script does is take numbers and add them to a total. However, I can't get total to work.It constantly claims that total is a non-variable despite it being assigned earlier in the program.
error message (8 is an example number being entered)
./adder: line 16: 0 = 0 + 8: attempted assignment to non-variable (error token is "= 0 + 8")
#!/bin/bash
clear
total=0
count=0
while [[ $choice != 0 ]]; do
echo Please enter a number or 0 to quit
read choice
if [[ $choice != 0 ]];
then
$(($total = $total + $choice))
$(($count = $count + 1))
echo Total is $total
echo
echo Total is derived from $count numbers
fi
done
exit 0
Get rid of some of the dollar signs in front of the variable names. They're optional inside of an arithmetic context, which is what ((...)) is. On the left-hand side of an assignment they're not just optional, they're forbidden, because = needs the variable name on the left rather than its value.
Also $((...)) should be plain ((...)) without the leading dollar sign. The dollar sign will capture the result of the expression and try to run it as a command. It'll try to run a command named 0 or 5 or whatever the computed value is.
You can write:
((total = $total + $choice))
((count = $count + 1))
or:
((total = total + choice))
((count = count + 1))
or even:
((total += choice))
((count += 1))
When I am executing the below script, I am getting the following error :-
The script executes infintely and below line is printed everytime.
"line 9: 1=1+2: command not found". Why?
#!/bin/bash
echo "Script 1 - Linux Scripting Book"
x=1
while [ $x -le 45 ]
do
echo x : $x
$x=$x+2
done
echo "End Of Script 1"
exit 0
Also if I change the $x=$x+2 to x+$x+2 then also I am getting the below error.
line 6: [: 1+2: integer expression expected
Same script when executed like this runs fine.
#!/bin/bash
echo "Script 1 - Linux Scripting Book"
x=1
while [ $x -le 45 ]
do
echo x : $x
let x=x+2
done
echo "End Of Script 1"
exit 0
You get line 9: 1=1+2: command not found because 1=1+2 is what $x=$x+2 is expanded into.
Use expr or let or ((...)) for integer calculations and bc for floating point:
let x=x+2
((x=x+2)) #same as above
((x+=2)) #same
((x++)) #if adding just one
((++x)) #if adding just one
x=$((x+2))
x=`expr $x + 2` #space before and after +
x=$(echo $x+2|bc) #using bc
x=$(echo $x+2.1|bc) #bc also works with floating points (numbers with decimals)
Since this part of the question isn't cleared yet, and not fine to post in a comment, I add this partial answer:
x=1; for i in 1 2 3 ; do x=$x+2; echo $x; done
1+2
1+2+2
1+2+2+2
As a side note: Don't use exit 0 at the end of your script without a good reason. When the script is done, it exits by itself without your help. The exit status will be the exit status of the last command performed, in your case a simple echo, which will almost always succeed. In the rare cases it fails, you will probably without intention hide that failure.
If you source the script, the exit will throw you out of your running shell.
But you can rewrite your while loop like this:
x=0
while (($((x)) < 9))
do
echo x : $x
x=$x+2
done
echo $((x))
x : 0
x : 0+2
x : 0+2+2
x : 0+2+2+2
x : 0+2+2+2+2
10
Because that's not the Bourne shell syntax for setting a variable; it looks more like Perl or PHP. The $ is used for parameter expansion and is not part of the variable name. Variable assignment simply uses =, and let evaluates arithmetic expressions (much like $((expression))). Another syntax that should work is x=$((x+2)). Note that these arithmetic evaluations are a bash feature; standard unix shells might require use of external tools such as expr.
I'm trying to write a for loop that goes from 1 to 10, then calculates ( 1 through 10 mod 5) + 2. After that I want to display it like this (1 to 10 mod 5) + 2 = answer. However i'm getting an error at the beginning of the loop which is a syntax error.
for (( i = 0; i <= 10; i++)); do
calculate=(i % 5) + 2
echo ("("i "% 5) + 2" calculate)
done
Try these changes:
calculate=$(( i % 5 + 2 ))
# $(( ... )) is the shell's way to do arithmetic
echo "($i % 5) + 2 = " $calculate
# $x is a way to refer to the value of variable x
# (also inside a double-quoted string)
The for loop header is actually OK.
The script is supposed to print "Enter the nest number to add: " and keep doing this until the user enters a negative number. At which point it is supposed to print the sum of the positive numbers. However as is the loop asks for the next number one time, it is entered, and then does not ask again, the script just stops doing anything and doesn't even reach the next line within the loop.
#!/bin/csh -x
#
# This script adds positive numbers entered by the user, stopping
# when a negative number is added
# Usage: +#, +#, +#... -#.
#
# x=0
# sum = 0
while($x>= 0)
echo -n "Enter the next number to be added: "
# sum = $sum + $<
# x = $<
end
#
exit 0
$< reads a line from stdin. This must be assigned to a variable, else if $< is used a second time the script will expect further input before continuing.
# x=0
# sum = 0
while ($x >= 0)
echo -n "Enter the next number to be added: "
# x = $<
if ($x >= 0) then
# sum = $sum + $x
endif
end
echo $sum
exit 0