This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
Closed 6 years ago.
I want to rewrite a configuration file when asked from a bash script. Here is my code.
function quality {
echo $1 > ~/.livestreamerrc
echo ".livestreamer was modified!"
}
best="stream-types=hls
hls-segment-threads=4
default-stream=best
player=vlc --cache 5000"
read -p "Set quality: " INPUT
if [[ "$INPUT" == "!best" ]]; then
quality $best
fi
This code does the following to .livestreamer file though.
$cat ~/.livestreamerrc
stream-types=hls
Why?
Change it to
quality "$best" # double quotes to avoid word splitting
and then
echo "$1" > ~/.livestreamerrc
Note : Worth checking the [ shellcheck ] documentation.Also, fully uppercase variables like INPUT are reserved for the system.
Related
This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
When should I double-quote a parameter expansion? [duplicate]
(1 answer)
Closed 5 months ago.
If I have a file containing newlines, the below script will output the file as is, with newlines:
#!/bin/bash
FOO=$(cat filename.yaml)
echo "$FOO"
but
#!/bin/bash
FOO=$(cat filename.yaml)
FOO=$(echo $FOO)
echo "$FOO"
outputs the file all on one line. How come?
I do not recommend storing the contents of entire files in a single variable. In my experience that can have unpredictable results.
/usr/bin/env bash -x
index=$(wc -l filename.yaml | cut -d' ' -f1)
count=1
next () {
[[ "${count}" -lt "${index}" ]] && main
[[ "${count}" -eq "${index}" ]] && exit 0
}
main () {
line=$(sed -n "${count}p" filename.yaml)
echo "var${count}=${line}" >> varfile
count=$(($count+1))
next
}
next
If you source varfile at the start of another script, it will give you every line from that file, in its' own variable.
This question already has answers here:
How to check if a string contains a substring in Bash
(29 answers)
Closed 6 months ago.
I have a string given below:
string1 = "Hello there, my name is Jack.
Hello there, my name is Jack.
Hello there, my name is Jack."
I'm taking the following input from the string:
read string2
I want to check whether the string2(which is a variable) is present in string1.
I tried running the below command:
output=$(echo $string1 | grep -o "$string2")
echo $output
eg: Let string2="name"
The output is empty when I'm running this command.
Can someone tell me where am I going wrong?
#!/bin/bash
string1="Hello there, my name is Jack"
string2="name"
if [[ $string1 == *"$string2"* ]]; then
echo "$string2 found"
else
echo "$string2 not found"
fi
Alternate method with POSIX-shell grammar:
string1='Hello there, my name is Jack'
string2='name'
case "$string1" in
*"$string2"*) printf '%s found\n' "$string2";;
*) printf '%s not found\n' "$string2";;
esac
This question already has answers here:
How do I compare two string variables in an 'if' statement in Bash? [duplicate]
(12 answers)
Closed 2 years ago.
here is how i tried it
while IFS= read line
do
var=$(cut -d ":" -f 3 $line)
if [ "$var" = "L2" ]
then :here is my action:
fi
done < myfile.txt
What i want to do is read a file line by line, read the third word of each line, and do a special action if the third word = a certaine string, i've tried a lot of syntax but it doesn't work. i've also tried to echo "$var" just to see if my variable get the right value, and it does. i don't know what to do anymore
It is better to use double brackets for if condition & for String comparison double equals (==)
And the line which has "cut" command wouldn't have worked. Please find below the corrected code which is working.
while IFS= read line
do
echo "Line is $line"
var=`echo $line | cut -d ":" -f 3`
echo $var
if [[ "$var" == "L2" ]]
then
echo "Some Action"
fi
done < myfile.txt
This question already has answers here:
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 6 years ago.
I am having trouble with executing the following bash script:
#!/bin/bash
response=" "
while ["$response" != "q"]; do
echo -n "Please enter a response"; read response
done
ALSO
!/bin/bash
response="x"
if ["$response" = "x"]
then
echo "the value is x"
fi
What could the possible errors be?
Your while and if statements are spaced wrong.
while [ "$response" != "q" ]; do etc
You need a space between the bracket and the double quote.
This question already has answers here:
Extract substring in Bash
(26 answers)
Closed 8 years ago.
I have a little bash script and have $file and $file2 variable. This is file and I want to get first 3 letters of this file name. And I want to compare them:
I tried:
curfile=$(basename $file)
curfilefirst3=${curfile:0:3}
curfile2=$(basename $file2)
curfile2first3=${curfile2:0:3}
if ((curfilefirst3 == curfile2first3 )); then
....
But I think have problems, how can I fix?
Thank you.
You're missing $ for the strings in the comparison and you'll need to wrap each string with " and the entire expression with []:
file="this.txt"
file2="that.txt"
curfile=$(basename $file)
curfilefirst3=${curfile:0:3}
curfile2=$(basename $file2)
curfile2first3=${curfile2:0:3}
echo $curfile2first3
echo $curfilefirst3
if [ "$curfile2first3" == "$curfilefirst3" ]
then
echo "same!"
else
echo "different!"
fi
It might be a good idea to read up on bash conditionals
Substring Extraction
${string:position} Extracts substring from $string at $position.
However, if should use [ and not ( as in:
if [ $curfilefirst3 == $curfile2first3 ]; then
The corrected version:
#!/bin/bash
file=abcdef
file2=abc123456
curfile=$(basename $file)
curfilefirst3=${curfile:0:3}
curfile2=$(basename $file2)
curfile2first3=${curfile2:0:3}
echo $curfilefirst3
echo $curfile2first3
if [ $curfilefirst3 = $curfile2first3 ]; then
echo same
else
echo different
fi
It prints same
So, works