This question already has answers here:
Case insensitive comparison of strings in shell script
(14 answers)
Closed 1 year ago.
I am trying to add users from a csv file to a group in a bash script running on CentOS 8. The group names are "Faculty" and "Students", which I am forcing them to be lowercase. The following did not work. It defaults to the "else" clause, even when $groupName is "Faculty" (I would "echo" before the if statement).
if [ "$groupName" = "Faculty" ]
then
goodGroup="faculty"
else
goodGroup="student"
fi
However, it worked when I gave it a substring of only the capital letter:
if [ "${groupName:0:1}" = "F" ]
then
goodGroup="faculty"
else
goodGroup="student"
fi
Using the second method gives me the outcome I need, I am just curious why the first bit of code did NOT work. All the answers I've seen on StackOverflow say that's the syntax for comparing strings, so I can't see what I'm doing wrong.
Ways to force a variables value to lowercase in bash without having to check for specific values:
#!/usr/bin/env bash
# Using declare to give a variable the lowercase attribute
declare -l groupName1
groupName1=Faculty
printf "%s\n" "$groupName1"
# Using parameter expansion
groupName2=FACULTY
printf "%s\n" "${groupName2,,}" # or "${groupName2#L}"
Related
This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
Closed 2 years ago.
Execute the following command in bash shell:
export sz1='"authorities" : ["uaa.resource"]'
Now, try echo $sz1
I expect to see the following output:
"authorities" : ["uaa.resource"]
But instead I get this:
"authorities" : c
The interesting thing is that I have dozens of servers where I can execute this type of variable assignment and it works except on this server. This server has exactly the same OS version, profile, bash version etc. What could be the reason for this behavior?
Always quote your variables. Use
echo "$sz1"
When you don't quote the variable, word splitting and wildcard expansion is done on the variable expansion. On ["uaa.resource"] is a wildcard that will match any of the following filenames:
"
u
a
.
r
e
s
o
u
c
On that one machine you have a file named c, so the wildcard matches and gets replaced with that filename.
This question already has answers here:
Assignment of variables with space after the (=) sign?
(4 answers)
Assing a variable and use it inside of if statement shell scripting
(2 answers)
Difference between sh and Bash
(11 answers)
Closed 3 years ago.
So there is a kernel adiutor for android, that let's you add custom controls with shell scripting. I'm trying to add a switch, but I have trouble setting the switch up correctly, when it's active, and when it's not. So there is a (text)file I'm trying to read (you will see it in the code), whether it's 0 or 1 inside, and that determines the switch on-off state.
I've tried with cat, read, everything, but honestly I think the problem is that I'm not familiar with sh scripting, and there is a problem with my syntax. Sometimes the script won't return anything when using cat. Also, su is available so that's not a problem, also the file has the correct permissions.
#!/system/bin/sh
var= $(<sys/class/lcd/panel/mdnie/hdr)
if ( "$var" = 0) then
echo 0
else echo 1
fi
The problem with my code is that right now it returns 1 (on), even when the file has a 0.
When assigning a variable in shell, there must be no space after the assignment sign. Also, make sure you use the correct syntax for conditions (and be aware of whitespace sensitivity):
var=$(cat sys/class/lcd/panel/mdnie/hdr)
if [ "$var" = "0" ]; then
# if [ "$var" -eq 0 ], if you want numeric comparison (won't really matter here)
echo 0
else
echo 1
fi
This question already has answers here:
Why does a space in a variable assignment give an error in Bash? [duplicate]
(3 answers)
Closed 4 years ago.
I'm newer on bash scripting ,I have a global variable that I want to change his value insead a loop in my script but still get an error that commande not found
this my script :
SCRIPT_BASE = "/home/scripts/test-Scripts"
CURRENT_SCRIPT_PATH = ""
declare -a arr=("A" "B" "C" "D")
for i in "${arr[#]}"
do
if [ $i == "A" ]; then
CURRENT_SCRIPT_PATH = $SCRIPT_BASE
echo -e "Current Path : $CURRENT_SCRIPT_PATH"
fi
done
when I run this script I get that CURRENT_SCRIPT_PATH commande not found
Thanks in advance for any help
In bash you should be really cautious about spaces in if conditions but also when you assign a value to a variable.
Replace in your code the following tree lines:
SCRIPT_BASE="/home/scripts/test-Scripts"
CURRENT_SCRIPT_PATH=""
CURRENT_SCRIPT_PATH=$SCRIPT_BASE
If you keep a space after the variable name bash will interpret it as a command and as you do not have commands SCRIPT_BASE, CURRENT_SCRIPT_PATH, CURRENT_SCRIPT_PATH in your current $PATH you have the error command not found that is produced.
This question already has answers here:
Check number of arguments passed to a Bash script
(10 answers)
Add a bash script to path
(5 answers)
Closed 5 years ago.
So far
I have the following code:
#!/bin/bash
echo "Adding new path...."
if [[$# -eq1] || [$# -eq2]]
then
if [$# -eq2]
then
export PATH=$PATH:/$1:/$2
fi
if [$# -eq1]
then
export PATH=$PATH:/$1
fi
else echo "Incorrect number of parameters. No more than two directories can be added at once."
fi
echo $PATH
exit 0
When I run this script passing it one parameter i get an error:
"./addDir: line 3: [[1: command not found
./addDir: line 3: [1: command not found "
when I run it with 2 parameters instead of "1" it says "2"
What's going on?
You're missing some spaces. Basically, if you're trying to use the [...] construction, you need to have spaces before and after each bracket - think of [ as being the name of a command, in the same way as echo, and ] as being an argument to that command. (In fact, there might actually be a /bin/[ program on your system.) Just as you can't type echofoo and expect it to run the echo program, similarly you can't type [[$# if you expect it to run [.
In your case, you'd need to do things like
if [ $# -eq 2 ]; ...
And for the compound test you're doing in line 3, I don't think you can use [ and ] within the test. In other words, don't use those brackets for grouping; it has to be [ something ] where the something doesn't contain any brackets. Read the relevant section of the bash man page for the full details of what you can put there.
There is also a shell construct [[ ... ]] which does basically the same thing but has different syntax. You could use that instead, but be aware that it's very different from [ ... ].
I am using a bash script to execute a program. The program must take the following argument. (The program is gnuplot.)
gnuplot -e "filename='output_0.csv'" 'plot.p'
I need to be able to assemble the following string: "filename='output_0.csv'"
My plan is to assemble the string STRING=filename='output_0.csv' and then do the following: gnuplot -r "$STRING" 'plot.p'. Note I left the words STRING without stackoverflow syntax style highlighting to emphasise the string I want to produce.
I'm not particularly proficient at bash, and so I have no idea how to do this.
I think that strings can be concatenated by using STRING="$STRING"stuff to append to string? I think that may be required?
As an extra layer of complication the value 0 is actually an integer which should increment by 1 each time the program is run. (Done by a for loop.) If I have n=1 in my program, how can I replace the 0 in the string by the "string value" or text version of the integer n?
A safest way to append something to an existing string would be to include squiggly brackets and quotes:
STRING="something"
STRING="${STRING}else"
You can create the "dynamic" portion of your command line with something like this:
somevalue=0
STRING="filename='output_${somevalue}.csv'"
There are other tools like printf which can handle more complex formatting.
somevalue=1
fmt="filename='output_%s.csv'"
STRING="$(printf "$fmt" "$somevalue")"
Regarding your "extra layer of complication", I gather that this increment has to happen in such a way as to store the value somewhere outside the program, or you'd be able to use a for loop to handle things. You can use temporary files for this:
#!/usr/bin/env bash
# Specify our counter file
counter=/tmp/my_counter
# If it doesn't exist, "prime" it with zero
if [ ! -f "$counter" ]; then
echo "0" > $counter
fi
# And if it STILL doesn't exist, fail.
if [ ! -f "$counter" ]; then
echo "ERROR: can't create counter." >&2
fi
# Read the last value...
read value < "$counter"
# and set up our string, per your question.
STRING="$(printf "filename='output_%d.csv'" "${value}")"
# Last, run your command, and if it succeeds, update the stored counter.
gnuplot -e "$STRING" 'plot.p' && echo "$((value + 1))" > $counter
As always, there's more than one way to solve this problem. With luck, this will give you a head start on your reading of the bash man page and other StackOverflow questions which will help you learn what you need!
An answer was posted, which I thought I had accepted already, but for some reason it has been deleted, possibly because it didn't quite answer the question.
I posted another similar question, and the answer to that helped me also answer this question. You can find said question and answer here: bash: Execute a string as a command