I'm pretty new with Bash scripting and am having trouble getting my 'while' loop to run. When I echo keywords, a whole list of words prints and then when I echo length, it prints 124. I believe I'm using the while loop and condition correctly, so I can't figure out what I'm doing wrong. Any thoughts?
keywords=$1
length=${#keywords}
echo "$keywords"
echo "$length"
if [ -z "$keywords" ]; then
while [ $length -gt 100 ]; do
echo "$keywords"
echo "$length"
keywords="${keywords%,*}"
length=${#keywords}
done
fi
echo $keywords
The problem is here:
[ -z "$keywords" ]
-z is true if its argument is an empty string. Something of length 124 is definitely far from empty. You probably meant -n.
Next time, please also include the input in the question so we can reproduce the problem.
Related
I am very new here, so I apologize for any my mistakes, and I am sorry for my lack of knowledge (I'm just beginner).
So here it is, i am doing little script in bash with li and I have if statement, here it is
#!/bin/bash
something=$(whiptail --inputbox "Enter some text" 10 30 3>&1 1>&2 2>&3)
if [ $something ?? 'you' ];
then
echo "$something"
else
echo "nope"
fi
To be specific what i want from it - I enter some word/sentence/whatever to whiptail, and if it contains some of of you string then prints it, but every times it goes else ;_;.Please help.
EDIT now it works, thanks but I need to check if string contains word.
if [[ $string =~ .*My.* ]]
doesn't seem to work
I don't get it at all, losing hope and searching the web i've encontered on
#!/bin/bash
OPTION=$(whiptail –title “Menu Dialog” –menu “Choose your option” 15 60 4 \ “1” “Grilled ham” \ “2” “Swiss Cheese” \ “3” “Charcoal cooked Chicken thighs” \ “4” “Baked potatos” 3>&1 1>&2 2>&3)
exitstatus=$?
if [ $exitstatus = 0 ];
then echo “Your chosen option:” $OPTION
else echo “You chose Cancel.”
fi
And I've just pasted this script to check how it works and modify it, it isn't mine script and it supposed to work, but it says “You chose Cancel.”
What you may be looking for are the string comparison operators like == or !=. For example,
if [ "$something" == "you" ]; then
echo "$something"
else
echo "nope"
fi
If $something equals you then echo $something; else echo nope.
Or, as David C.Rankin mentioned in his comment you can check the string variable to prove whether a string is empty or not. For example,
if [ -z "$something"] ;then
String is empty
if [ -n "$something" ]; then
String is non-empty
For more information on this check the TEST manual page.
At the moment I am working on a blackjack game using shell script. I have most of the script working with functions however the method I am using to find out if the player/computer goes bust doesn't seem to work. Could anyone point me in the right direction. (I am new to shell script.) When running it it will throw syntax errors around the lines that begin elif and sometimes if. It also prints all of the 'echo' outputs in bustConfirm instead of only the one that is true.
Also yes, one of my functions is called bustCheck.
bustConfirm(){
bust='bust'
under='under'
if [ $userBust -eq $bust -a $systemBust -eq $bust ]
then
echo "You both went bust! Be more careful!"
endGameRepeat
elif [ $userBust -eq $bust -a $systemBust -eq $under ]
echo $userName "went bust! Congratulations" $systemName"!"
endGameRepeat
elif [ $userBust -eq $under -a $systemBust -eq $bust ]
then
echo $systemName "went bust! Congratulations" $userName"!"
endGameRepeat
else
echo "Nobody went bust! Well played!"
endGameScores
fi
}
bustCheck(){
if [ "$userScore" -gt 21 ]
then
echo $userName "is bust!"
userBust='bust'
else
userBust='under'
fi
if [ "$systemScore" -gt 21 ]
then
echo $systemName "is bust!"
systemBust='bust'
else
systemBust='under'
fi
bustConfirm
}
The idea is that I wanted to use an && in the bustConfirm function and then an || to get the player is bust or system is bust result if only one of them was bust.
Also just a pointer but in the bustCheck I am seeing userBust and systemBust to contain the words bust or under. I created the variables bust and under for the bustConfirm function.
systemScore, userScore, systemName and userName are set before when the script is running.
Hope I've given enough detail and formatted it properly, first proper post so I apologize if not!
Taking a quick look, I see that the first if statement doesn't have a space after the opening square bracket.
I also recommend you put quotes around your variable names in if statements. This is due to the way shell actually works. The bash shell is extremely intelligent, and before your program has a chance to do anything, it grabs the line, does its magic, and then presents the line to the processor.
For example:
foo=""
if [ $foo = "" ]
then
echo "Foo is blank"
fi
Seems simple enough. However, what happens is that your shell will grab the line, substitute the value of $foo for the string "$foo", and then execute the line. Since $foo is blank, your if statement will become:
if [ = "" ] # That's not right!
then
echo "Foo is blank"
fi
By using quotes, this:
foo=""
if [ "$foo" = "" ]
then
echo "Foo is blank"
fi
becomes:
foo=""
if [ "" = "" ]
then
echo "Foo is blank"
fi
And that is valid. Another thing you can do is use the new test format that uses double square brackets:
foo=""
if [[ $foo = "" ]]
then
echo "Foo is blank"
fi
This will always work even without the extra quotes, and is now recommended unless you have to have your program compatible with the original Bourne shell syntax.
One more thing you can do in debugging your shell script is to use set -xv which turns on verbose debugging. Each statement, before it is executed will be printed, then it will print again after the shell fills in variables, patterns, etc., and then execute. It's a great way to debug your program. Just put set -xv on the line before you want this verbose debugging mode and use set +xv to turn it off. (Yes, the - turns it on and + turns it off.)
Thanks alot David, great answer, could you also tell me what the best way to get the && or equivalent of it within this as I need to find out if they are both bust, or just one etc
As already mentioned in a comment, you can use either one of these two forms:
if [ "$foo" = "bar" ] && [ "$bar" = "foo" ]
or
if [[ $foo = "bar" && $bar = "foo" ]]
I'm trying to delete a large amount of files from my computer, and I'm trying to write a bash script to do so using the rm command. What I want to know is how to do equality in bash, and why my code (posted below) won't compile. Thank you for your help!
#!/bin/bash
# int-or-string.sh
b="0000"
c="linorm"
f=500
e1=2
e2=20
e3=200
e4=2000
for i in {0..10000}
do
a=$(($f*$i))
if ["$i" -eq "$e1"]
then
b="000"
echo $b$
fi
if ["$i" -eq "$e2"]
then
b='00'
fi
if ["$i" -eq "$e3"]
then
b='0'
fi
if ["$i" -eq "$e4"]
then
b =''
fi
if [bash$ expr "$i" % "$e3$ -ne 0]
then
d = $b$c$a
rm d
fi
done
Shell scripts aren't compiled at all.
You need spaces after your [ and before your ].
if [ "$i" -eq "$e1" ]
There's an errant bash$ in there you probably don't want at all. It should probably be a $() operator:
if [ $(expr "$i" % "$e3") -ne 0 ]
You can't have spaces around the = in bash. For example, change b ='' to b='' and d = $b$c$a to d=$b$c$a.
echo $b$ looks like it should be echo $b.
Shell script does not compile it is a scripting language.
Try to fix this line :
if [bash$ expr "$i" % "$e3$ -ne 0]
Make it like below :
if [ $(expr "$i" % "$e3$") -ne 0 ]
You need spaces around the square brackets. The [ is actually a command, and like all commands needs to be delineated by white space.
When you set values for variables in shell, you do not put spaces around the equals signs.
Use quotation marks when doing comparisons and setting values to help delineate your values.
What happens if none of the if conditions are true, and $b isn't set.
What is the logic behind this code. It seems to be a bunch of random stuff. You're incrementing $ from 1 to 10000, but only setting the value of $b on only four of those values. Every 200 steps, you delete a file, but $b may or may not be set even though it's part of the file name.
Did you write this program yourself? Did you try to run it? What errors were you getting? Did you look at the lines referenced by those errors. It looks like you included the bash$ prompt as part of the command.
There were plenty of errors, and I've cleaned most of them up. The cleaned up code is posted below, but it still doesn't mean it will do what you want. All you said is you want to delete "a large amount of files" on your computer, but gave no other criteria. You also said "What I want to know is how to do equality in bash" which is not the question you stated in you header.
Here's the code. Note the changes, and it might lead to whatever answer you were looking for.
#!/bin/bash
# int-or-string.sh
b="0000"
c="linorm"
f=500
e1=2
e2=20
e3=200
e4=2000
for i in {0..10000}
do
a=$(($f*$i))
if [ "$i" -eq "$e1" ]
then
b="000"
elif [ "$i" -eq "$e2" ]
then
b='00'
elif [ "$i" -eq "$e3" ]
then
b='0'
elif [ "$i" -eq "$e4" ]
then
b=''
fi
if ! $(($i % $e3))
then
d="$b$c$a"
rm "$d"
fi
done
ERRORS:
Spaces around the [ and ]
The rm "$d" command was originallyrm dwhich would just remove a file namedd`.
if/then statement converted to if/else if.
Rewrote [ $(expr "$1" % "$e3") -ne 0 ].
No need for expr since BASH has $((..)) syntax.
No need for test command ([) since if automatically evaluates zero to true and non-zero to false.
Added quotes.
I'm working with bash and I'm trying to do something like this:
A=1
while [ $A=1 ]; do
read line
echo $line | grep test >/dev/null
A=$?
echo $A
done
This script never ends even when the grep finishes successfully and A is set to 0. What am I missing here? Below is a sample of the output.
$ ./test.sh
asdf
1
test
0
hm...
1
You need to use the correct comparison operator. Bash has different operators for integer and string comparison.
In addition, you need the correct spacing in the comparison expression.
You need
while [ $A -eq 1 ]; do
See here for more
I find Bash's syntax pretty awful - have you tried something like:
while [ $A -eq 1 ] ... ?
It may be trying to re-assign 1 to $A or something strange like that.
Try:
while [ $A -eq 1 ]; do
Most of the answers have focused on the integer/string and spacing problem, which is fine, but your code looks so unidiomatic that IMO it should be completely re-factored. Let's say the idea is to process lines until one line matches the regex 'test':
while read line; do
if [[ "$line" =~ test ]] && break
# do something with $line
done
Of course this can be simplified further if you take advantage of text processing tools like sed:
sed -e '/test/,$d'
you can do this instead. No need to call external grep.
while true; do
read line
case "$line" in
*test* ) break;;
esac
done
echo $line
Have you not tried this
while [ $A == "1" ]
....
done
Edit: Whoops since Dan mentioned my error I graciously admit my mistake and have edited this accordingly - Thanks Dan for the heads up...
while [ $A -eq 1 ]
....
done
Sorry! :(
Hope this helps,
Best regards,
Tom.
All of your answers are in the Advanced Bash-Scripting guide. It is awesome.
So basically I am trying to check the arguments that are passed into the script. If it has three arguments and the third argument is a 1, then I want it to continue. I also want it to continue if it has four arguments and the third argument is not a 1.
So basically I thought that I could just do...
if ([ $# -ne 3 ] and [ "$3" -ne "2" ])
then
exit 0
fi
However it seems that Bash does not have and's to use for if's,so then I figured that I could just use nested if's, however now it's complaining still. So this is what I have currently...
if [ $# -ne 3 ]
then
if [ "$3" -ne "1" ]
then
echo "Improper number of arguments.
FORMAT make-csv-data <STUDY> <TAG> <MODE> <SELECT>
Select can be left off if you want all data (Mode=1)
"
exit 0
fi
fi
if [ $# -ne 4 ]
then
if [ "$3" -ne "2" ]
then
echo "Improper number of arguments.
FORMAT make-csv-data <STUDY> <TAG> <MODE> <SELECT>
Select can be left off if you want all data (Mode=1)
"
exit 0
fi
fi
So where am I going wrong? Can I not nest if statements in Bash? Is there a super-zen way of doing this that I'm missing altogether?
Thanks for the any help you could give me.
New Problem...
Now, for some reason or another, the code isn't working at all. There are no errors or anything, it just doesn't work. It doesn't check the number of arguments. I've run the script with no arguments at all and it just skips it like it's not even there.
Weird part is that I was sure that the code was working yesterday. Come back today, not so much. Any ideas on what the problem is? (Sorry, but I have to remove the accepted answer on this.)
if [[ $# = 3 && "$3" != "1" ]]
then
echo "Improper number of arguments.
FORMAT make-csv-data <STUDY> <TAG> <MODE> <SELECT>
Select can be omitted if all data is required (Mode=1)
"
exit 0
fi
if [[ $# > 4 ]]
then
echo "Improper number of arguments.
FORMAT make-csv-data <STUDY> <TAG> <MODE> <SELECT>
Select can be omitted if all data is required (Mode=1)
"
exit 0
fi
EDIT II:
There are a few things that the Bash shell isn't liking about this script that I'm trying to do. I'll probably end up rewriting it in another scripting language and do a few more things that I have in mind for the project. Thanks for the help in any case.
if [ $# -ne 3 -a "$3" -ne "1" ]; then
exit 0
fi
For reference
-a = and
-o = or
Or, you could just use use:
if [[ $# != 3 && "$3" != "1" ]]; then
Please see:
http://bash-hackers.org/wiki/doku.php/commands/classictest#and_and_or
and
http://bash-hackers.org/wiki/doku.php/syntax/ccmd/conditional_expression
Since you're just checking exit/return values with "if", you need to provide something, e.g. a command, that provides meaningful ones based on your tests. [ is such a command, another possibility is the [[ keyword.
The actual correct examples already were mentioned by scragar above, I don't want to just repeat them :)