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=$COUNT
fi
Without the spaces around "=".
If you put a space after "MAXseeds", then it will be interpreted as a command. And of course, it is not a command, thus you get your error message.
Related
I have a script like this:
#!/bin/bash
x=${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 $1
# You need fi in order to finish an if then statement
fi fi fi
done
this is my first stackoverflow question, regarding bash scripting. I am a beginner in this language, so be kind with me.
I am trying to write a comparison script. I tried to store all the outputs into variables, but only the last one is stored.
Example code:
me:1234567
you:2345678
us:3456789
My code:
#!bin/bash
while read -r forName forNumber
do
aName="$forName"
echo "$aName"
aNumber="$forNumber"
echo "$aNumber"
done < "exampleCodeFile.txt"
echo "$aNumber"
For the first time, everything will be printed out fine. However, the second echo will only print out "3456789", but not all the numbers again. Same with $aName. This is a problem because i have another file, which i stored a bunch of numbers to compare $aNumber with, using the same method listed above, called $aMatcher, consisting:
aMatcher:
1234567
2345678
3456789
So if i tried to run a comparison:
if [ "$aNumber" == "$aMatcher" ]; then
echo "match found!"
fi
Expected output (with bash -x "scriptname"):
'['1234567 == 1234567']'
echo "match found!"
Actual output (with bash -x "scriptname"):
'['3456789 == 3456789']'
echo "match found!"
Of course my end product would wish to list out all the matches, but i wish to solve my current issue before attempting anything else. Thanks!
When you run your following code
aNumber="$forNumber"
You are over-writing the variable $aNumber for every line of the file exampleCodeFile.txt rather than appending.
If you really want the values to be appended, change the above line to
aNumber="$aNumber $forNumber"
And while matching with $aMatcher, you again have to use a for/while loop to iterate through every value in $aNumber and $aMatcher.
I found a nifty little shell script that I wanted to use from this website here. I have followed everything step-by-step, but receive the following error when running this on my CentOS box.
./deploy: line 3: =: command not found
Line 3 only contains...
$ERRORSTRING = "Error. Please make sure you've indicated correct parameters"
I've tried toying around with a bit, but don't understand why it won't accept the "=" character. Is there something wrong with the script, or is it merely something different in the way that my server processes the script?
Thanks!
Gah, that script is full of bad scripting practices (in addition to the outright error you're running into). Here's the outright error:
$ERRORSTRING = "Error. Please make sure you've indicated correct parameters"
As devnull pointed out, this should be:
ERRORSTRING="Error. Please make sure you've indicated correct parameters"
A couple of lines down (and again near the end), we have:
echo $ERRORSTRING;
...which works, but contains two bad ideas: a variable reference without double-quotes around it (which will sometimes be parsed in unexpected ways), and a semicolon at the end of the line (which is a sign that someone is trying to write C or Java or something in a shell script). Use this instead:
echo "$ERRORSTRING"
The next line is:
elif [ $1 == "live" ]
...which might work, depending on whether the value of $1 has spaces, or is defined-but-blank, or anything like that (again, use double-quotes to prevent misparsing!). Also, the == comparison operator is nonstandard -- it'll work, because bash supports it in its [ ... ] builtin syntax, but if you're counting on having bash extensions available, why not use the much cleaner [[ ... ]] replacement? Any of these would be a better replacement for that line:
elif [ "$1" = "live" ]
elif [[ $1 == "live" ]]
elif [[ "$1" == "live" ]]
Personally, I prefer the last. The double-quotes aren't needed in this particular case, but IMO it's safest to just double-quote all variable references unless there's a specific reason not to. A bit further down, there's a elif [ $2 == "go" ] that the same comments apply to.
BTW, there's a good sanity-checking tool for shell scripts at www.shellcheck.net. It's not quite as picky as I am (e.g. it doesn't flag semicolons at the ends of lines), but it pointed out all of the actual errors in this script...
"Devnulls" answer was correct -- I had to remove the spaces around the "=" and remove the "$" from that line as well. The end result was...
ERRORSTRING="Error. Please make sure you've indicated correct parameters"
I've upvoted Devnull and gniourf_gniourf's comments.
Thank you to all whom have assisted!
Although this error is quite common and explained almost everywhere on the web I've decided to ask a new question since I can't get a clue for this specific case.
I'm trying to get some data out of Zimbra Collaboration Suite and the only way I can do it is via bash.
Being my first time with bash I find it a bit hard to deal with.This is the code:
#!/bin/bash
all_account=`zmprov -l gaa`;
declare -i szquota
szquota=524288000
for account in ${all_account}
do
mb_size=`zmmailbox -z -m ${account} gms`;
set -i size;
declare -i quota
declare -i quota2
for size in ${mb_size}
do
if [ $((10#$size)) -gt $((10#$szquota)) ] ; then
quota=`zmprov ga ${account} zimbraMailQuota`;
quota2="10#`zmprov ga ${account} zimbraMailQuota`";
echo "${account},${mb_size},$quota2\n";
fi
done
done
and this is the response:
line 12: 137,08: value too great for base (error token is "08")
I need to print all the accounts that have a quota of more than 500MB, and the output should be like this: account/quota/used quota.
Since mb_size is an array of values I can't figure out how I could convert its content to a decimal base as I did with the other values?
It probably is much simpler than my mind makes it look but I really can't get out of this trouble.
Kind regards
EDIT:
Thanks #Alfe!
I've modified the code like this:
#!/bin/bash
all_account=`zmprov -l gaa`;
szquota=524288000
for account in ${all_account}
do
mb_size=`zmmailbox -z -m ${account} gms`;
declare -i quota
declare -i quota2
for sizeStr in ${mb_size}
do
if [ $size -gt $((10#$szquota)) ] ; then # <--- line 13
quota=`zmprov ga ${account} zimbraMailQuota`;
quota2="10#`zmprov ga ${account} zimbraMailQuota`";
echo "${account},${mb_size},$quota2\n";
fi
done
done
but it returns another error:
line 13: [: -gt: unary operator expected
I've also tried to enclose the values inside the if clause between quotation marks but if I'm not wrong bash interprets the content of quotation marks as a string and gives back this:
line 13: [: : integer expression expected
I'm sure I'm getting closer to a solution but I'm still stuck at the moment.
You declared the variable size as integer with set -i size. Later you try to assign a value which then is checked for being a valid integer, it isn't, then you get the error. In your case, one of the values in ${mb_size} is the string 08 which is then interpreted as a bad octal value.
I suggest you let the loop run over a different variable which is not declared as int (so the for statement won't posed a problem), then as first statement you assign the string value of the loop variable properly so that it does not get interpreted as octal:
for sizeStr in ${mv_size}
do
size=$((10#$sizeStr))
if [ $size -gt $((10#$szquota)) ]
then
…
You could of course also just remove the declaration of the variable as integer. This probably would also solve the issue.
This is part of a bigger project but I can't get this part to work and I'm having a brain fart.
#!/bin/bash
echo -n "Do you wish to download/checkout the source code? > "
read text
if ["$text" = "Yes"]
then
do something
else
do something else
fi
It should simply be reading in what the user types and then go through a simple conditional. but I get this error
./check.sh: line 6: [Yes: command not found
I thought I had formatted the shell script correctly but I guess not.
Add spaces after brackets:
if [[ "$text" = "Yes" ]]
When performing operations on strings it's always a good idea to use double square brackets. It will make your code work properly with spaces and new lines.