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.
Related
I'm stumped when writing a simple script.
Essentially the $u variable is does not take u=$USER. Here's the code:
#!/bin/bash
if [ $# > 0 ] ; then
u=$1
else
u=$USER
fi
echo $u
The fix
You have 2 equally viable options:
Use -gt
if [ $# -gt 0 ]
Use double brackets [[ (Does a lexicographic comparison but will work for this case)
if [[ $# > 0 ]]
Why?
When you did if [ $# > 0 ] the > was treated like an output redirection command similar to echo "foo" > file.txt. You might notice you have created a file named 0 someplace after executing:
if [ $# > 0 ]
When deciding between using [...] or [[...]] many find the answer is to use double brackets
Getting fancy
Now if what you'd really like to do is write a script that gives a default value to the u variable if none is provided by the first argument I would recommend using a neat bash syntax trick for implementing default values
u=${1:-${USER}}
I believe I found the answer using double brackets
#!/bin/bash
if [[ $# > 0 ]] ; then
u=$1
else
u=${USER}
fi
echo $u
Not sure I fully understand why it failed with single brackets.
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.
I'm trying to compare the left values of two IP adresses, but i get an error:
":a : int expected..."
Here is the code:
IN="195.152.15.1"
IFS='.' read -ra ADDR <<< "$IN"
IN2="196.151.14.1"
IFS='.' read -ra ADDR2 <<< "$IN2"
a=${ADDR[0]}
b=${ADDR2[0]}
if [ a -eq b ]
then
echo "OK"
fi
I tried several combinations but nothing works, any suggestion ?
p.s: i don't want to use the "=" sign, only the "-eq"
Your problem has been solved in the comments. You need to refer to the variables in the comparison:
if [ "$a" -eq "$b" ]
then
echo "OK"
fi
Quotes around variables are always recommended when using [. With bash, you've got a couple of alternatives:
if [[ $a -eq $b ]] # extended test is smart enough to deal with unquoted variables
if (( a == b )) # inside arithmetic context, a and b unambiguously refer to variables
I've heard that the performance of (( is slightly worse than the other options but I doubt that it's significant enough to worry about.
It's worth mentioning that ALL_CAPS variable names should be avoided in scripts as they are for use by the shell. Also, there's no need to read into an array if you only want the first part of each IP:
IFS=. read -r ip1 junk <<< "$in1"
The rest of the line is written to the variable junk, which can be ignored.
change if [ a -eq b ] to if [ $a -eq $b ]
I am trying to make a bash script with the output based on the input.
My code looks like this:
#!/bin/bash
echo "Letter:"
read a
if a=3
then
echo "LOL"
fi
if a=4
then
echo "ROFL"
fi
But when I enter 3 or 4, I get both LOL and ROFL.
Is there a way for me to get LOL for 3 and ROFL for 4?
Sorry if I'm using incorrect terms and stuff, I'm new to bash scripting.
In bash, a=3 is an assignment, not a test. Use, e.g.:
if [ "$a" = 3 ]
Inside [...], the equal sign tests for string (character) equality. If you want to test for numeric value instead, then use '-eq` as in:
if [ "$a" -eq 3 ]
The quotes around "$a" above are necessary to avoid an "operator" error when a is empty.
bash also offers a conditional expressions that begin with [[ and have a different format. Many like the [[ format better (it avoids, for example, the quote issue mentioned above) but the cost is loss of compatibility with other shells. In particular, note that dash, which is the default shell (/bin/sh) for scripts under Debian-derived distributions, does not have [[.
Bash thinks you're trying to assign a variable by saying a=3. You can do the following to fix this:
Use the = operator whilst referencing the variable with a $, like so: if [[ $a = 3 ]]
Use the -eq operator, which is special and doesn't require you to reference the variable with a $, but may not be compatible with all sh-derived shells: if [[ a -eq 3 ]]. If you wish to use -eq without Bash reference the variable: if [[ $a -eq 3 ]]
Note:
The double square brackets [[ ... ]] are a preferred format with specifically Bash conditionals. [ ... ] is good with any sh-derived shell (zsh, tcsh, etc).
if a=3 will assign value 3 to variable a
unless a is readonly variable, if a=3 always returns TRUE
same for if a=4
To compare variable a with a value, you can do this if [ $a = 3 ]
so the script should change to
#!/bin/bash
echo "Letter:"
read a
if [ $a = 3 ]
then
echo "LOL"
fi
if [ $a = 4 ]
then
echo "ROFL"
fi
Since a is read from user input, there is possibility user key in:
non numeric value
a string with empty space
nothing, user may just press Enter key
so a safer way to check is:
if [ "x$a" = "x3" ]
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 :)