I have this piece in which the variable TESTS_SUCEEDED disappears or its value is unset. I saw many examples in which variables disappear due to subshell starting in loops, but couldn't find any clue about this behaviour.
${SRCDIR}/3rdParty/bin/alxdatabasemanager
--create-database-with-name=TestAlexandriaDB || exit 1
Src/Tests/Functional/FunctionalTestLibalexandria
TESTS_SUCCEEDED="$?"
#Here variable exists
echo ${TESTS_SUCEEDED}
${SRCDIR}/3rdParty/bin/alxdatabasemanager
--delete-database-with-name=TestAlexandriaDB || exit 1
#FIXME: Variable nonexisten here or value lost??!! Why?
exit ${TESTS_SUCCEDED}
Could anyone tell me what is going on? Thanks in advance.
You are having spelling issues: TESTS_SUCCEEDED and TESTS_SUCEEDED are not the same thing.
Let's line'em up, to clarify:
TESTS_SUCCEEDED
TESTS_SUCEEDED
Related
I have a script like this:
#!/bin/bash
x=${1:-20}
for ((i=1;i<=x;i++))
{
if ((i%3==0))
{
echo 'Fizz'
}
echo $i
}
I get an error color on the last brace in VIM and when I try to run the script I get a "syntax error near unexpected token" for that same brace. Without the nested if statement, this will print 1 through 20 on a new line for each number, which is the expected outcome. If the number is divisible by 3, it should print Fizz instead of that number. I'm not as worried about how to implement the replacement, that should be easy to figure out, but what I don't understand is why I cannot use a brace to close the for loop. If I take out the brace, I get an error that says end of file expected. So what is the proper syntax for ending a for loop with a nested if statement? I've looked around online and here on stack but haven't found a similar format to what I am trying to do. I don't like the
for f in *
format as it is not as easy to read for someone coming from a different coding language and I like to keep my code looking very similar across different languages (I use comments too, but just the same, I try to keep things as similar as possible which is why I used (( )) with the for loop.)
If I comment out the if statement and leave everything else intact, the error disappears and it will print
1
Fizz
2
Fizz
etc.
Any insight into this would be greatly appreciated. Thanks!
So here is what I was able to figure out thanks to #Cyrus:
x=${1:-20}
for ((i=1;i<=x;i++))
do
if ((i%3==0))
then
echo 'Fizz'
else
echo $i
fi
done
In many ways bash is simpler than most other languages but that makes it harder to work with when you are used to "higher" level languages.
So, to help out anyone else that's like me and just starting to code with bash, here is the full program I made, with comments as to why I coded it the way I did. If there are errors in my explanation or my formatting style, please point them out. Thanks! This was kind of fun to write, call me crazy.
# This will literally just print the string inside the single quotes on the screen
echo 'Try running this again but with something like this: fizzbuzz 25 pot kettle black'
# The $0 is the first index, in other words the file name of the executable,
# this will set the default value of x to 20 but will allow the user to input
# something else if they want.
x=${1:-20}
# This is the same but with string variables
f=${2:-FizzBuzz}
g=${3:-Fizz}
b=${4:-Buzz}
# We start the index variable at 1 because it's based on the input,
# otherwise it would echo 0 thru 19
for ((i=1;i<=x;1++))
do
# I recommend using (( )) for if statement arithmetic operations
# since the syntax is similar to other programming languages
if ((i%3==0 && i%5==0)); then echo $f
# you need to use a semicolon to separate if and then if they are
# on the same line, otherwise you can just go to the next line for
# your then statement
else if ((i%3==0)); then echo $g
else if ((i%5==0)); then echo $b
else echo $1
# You need fi in order to finish an if then statement
fi fi fi
done
I've 3 lines and I would like to check the status of each line for a specific variable i.e OFF.
So basically I need to loop this 3 lines and exit the loop when all 3 lines variable status are OFF or else just loop for ever. I've tried the while true do loop but to no avail. Any help or advise will be highly appreciated.
I think this may be done by searching on each line variable OFF and counting set one variable to count=0 and iff line ~/OFF/ count++ , which language do you use.?
So I'm pretty new to bash scripting but so far tldp.org has been a good friend. Anyways I've confused myself and swearing to much so looking for help in clarification: I declare a variable like such
MAXseeds=-1;
sumS=0
I do a bunch of things in my script and get a new value for sumS which is an integer value. I would then like to compare MAXseeds and sumS if sumS is larger make MAXseeds equal to sumS. I do this by:
echo $MAXseeds
echo $sumS
if [ $MAXseeds -lt $sumS ];
then
MAXseeds = $sumS
best_file=$COUNT
fi
echo $MAXseeds
This from what I can tell should work however the terminal output I get when running over this section of script is
-1
492
lookup.sh: line 34: MAXseeds: command not found
-1
Basically I am wondering what I am doing wrong here? why does it respond with command not found? Any explanation to why this is incorrect would be greatly appreciated.
Try this:
if [ $MAXseeds -lt $sumS ];
then
MAXseeds=$sumS
best_file=$COUNT
fi
Without the spaces around "=".
If you put a space after "MAXseeds", then it will be interpreted as a command. And of course, it is not a command, thus you get your error message.
I have a small loop problem as below.
I have 3 different values to be used while running the same script -
va1="some-value1"
va2="some-value2"
va3="some-value3"
Now I want to use these three variable values to be used for running the same command, like -
while (i=0,i<=3,i++)
do
bin/java -s (run something with $var);
done
Now I want $var taking the value of var1, var2 and var3 each time it runs,
so can someone please tell me how do we achieve the above?
I tried doing -
for $1 $2 $3
do
case 1
case 2
case 3
done
OR
while read a b c
do
<code assumed to have loop iteration>
done <<< $(command)
But it isnt working as expected... Would really appreciate your help on this.
Thanks,
Brian
You forgot the 'in' part of the syntax:
for var in $va1 $va2 $va3
do
command $var
done
try
while ((i=0,i<=3,i++))
do
eval bin/java -s \$var$i
done
This is a great example of how to use eval. Note that 1. the value of $i is seen by the shell has it scans the line. Then because $var is escaped like \$var, it is not
'make visible' in the first scan. 2. Eval forces a 2nd scan of the cmd-line, so it sees $var1 or whatever, and that value is substituted into the command-line for execution.
I hope this helps.
P.S. Welcome to StackOverflow and let me remind you of three things we usually do here: 1) As you receive help, try to give it too, answering questions in your area of expertise 2) Read the FAQs, http://tinyurl.com/2vycnvr , 3) When you see good Q&A, vote them up by using the gray triangles, http://i.imgur.com/kygEP.png , as the credibility of the system is based on the reputation that users gain by sharing their knowledge. Also remember to accept the answer that better solves your problem, if any, by pressing the checkmark sign , http://i.imgur.com/uqJeW.png
You can use indirect variable expansion with ${!...}:
va1="some-value1"
va2="some-value2"
va3="some-value3"
for ((i=1;i<=3;i++)); do
varname="va$i"
echo "$varname = ${!varname}"
done
This prints:
va1 = some-value1
va2 = some-value2
va3 = some-value3
I'm trying to write a shell script to automate a job for me. But i'm currently stuck.
Here's the problem :
I have a variable named var1 (a decreasing number from 25 to 0
and another variable named
var${var1} and this equals to some string.
then when i try to call var${var1} in anywhere in script via echo it fails.
I have tried $[var$var1], ${var$var} and many others but everytime it fails and gives the value of var1 or says operand expected error.
Thanks for your help
It's probably better if you use an array, but you can use indirection:
var25="some string"
var1=25
indirect_var="var$var1"
echo ${!indirect_var} # echoes "some string"
There's only one round of variable expansion, so you can't do it directly. You could use eval:
eval echo \${var$var1}
A better solution is to use an array:
i=5
var[$i]='foo'
echo ${var[$i]}
It sounds like you need bash variable indirection. Take a look at the link below.
http://mywiki.wooledge.org/BashFAQ/006