I am writing a bash script , I used 1 variable in my bash file like below
list=`/home/ea/students'
I wrote below link in my bash but I got error
cat $list /admin.txt
Do you know how can I connect variable and string together?
Firstly you need to use single quotes (''') around strings, not backticks ('`')
list='/home/ea/students'
To append a string to a variable, do the following:
list=${list}/admin.txt
Demo:
echo $list
/home/ea/students/admin.txt
Try this:
list='/home/ea/students'
...
cat "${list}/admin.txt"
You can go with either:
cat "$list/admin.txt"
In this case the braces '{}' are not mandatory
as the / is not a valid identifier name character.
... or, if you need a variable, recent bash versions provide more concise way for appending:
bash-4.1$ list=/home/ea/students
bash-4.1$ list+=/admin.txt
bash-4.1$ printf '%s\n' "$list"
/home/ea/students/admin.txt
Related
I'm trying to figure out how to use a variable containing an ANSI-C quoting string as an argument for a subsequent bash command.
The string in the variable itself is a list of files (can virtually be a list of anything).
For example, I have a file containing a list of other files, for example test.lst containing :
>$ cat test.lst
a.txt
b.txt
c.txt
I need to pass the file content as a single string so I'm doing :
test_str=$(cat test.lst)
then converts to ANSI-C quoting string:
test_str=${test_str#Q}
So at the end I have :
>$ test_str=$(cat test.lst)
>$ test_str=${test_str#Q}
>$ echo $test_str
$'a.txt\nb.txt\nc.txt'
which is what I'm looking for.
Then problem arises when I try to reuse this variable as a string list in another bash command.
For example direct use into a for loop :
>$ for str in $test_str; do echo $str; done
$'a.txt\nb.txt\nc.txt'
What I expect at this step is that it prints the same thing as the content of the original test.lst
I also tried expanding it back but it leaves leading $' and trailing '
>$ str=${test_str#E}
>$ echo $str
$'a.txt b.txt c.txt'
I also tried printf and some other stuffs to no avail. What is the correct way to use such ANSI-C quoting variable into a bash command ?
How about:
eval echo "${test_str}"
I believe that an ANSI-C quoted string is meant to be evaluated by bash command line parser.
In the first place, why do you need quoting? Just keep the data untouched stored as elements of an array:
mapfile -t filelist < test.lst
# iterate through the list
for file in "${filelist[#]}"; do printf '%s\n' "$file"; done
lets say I have a string variable variable
var="This line is with OldText"
I want to find and replace the text inside this variable
The method that I tried is,
echo $var | sed -i "s/OldText/NewText/g" >> result.log
in this case this gives an error saying "no input files".
The expected output is ,
"This line is with NewText"
what is the correct method to do this using sed, awk or any other method.
Using sed
$ var=$(sed "s/OldText/NewText/" <<< $var)
$ echo $var
This line is with NewText
You don't use -i, as that's for changing a file in place. If you want to replace the value in the variable, you need to reassign to it.
If you are using Bash shell, you could:
$ echo ${var/OldText/NewText}
This line is with NewText
so
$ var=${var/OldText/NewText}
See this for more.
I am extracting a part from an existing file and storing it as a string in a variable.The string looks something like this.
var="*a<br>*b<br>*c"
Now as * is a special character in unix it doesnot work in further operations(like sed,grep) until I put an escape character infront of every *
Thats why,I am doing something like this -
echo $var | sed 's/\*/\\*/g'
On running this command in bash we get
echo $var | sed 's/\*/\\*/g'
\*a<br>\*b<br>\*c
which is the desired output,but when I try to store this in a variable, I am getting back my original variable like so
var=`echo $var | sed 's/\*/\\*/g'`
echo $var
*a<br>*b<br>*c
I am assuming this happens because the variable ignores the backslashes interpreting them as escape characters. How can I retain the backslashes and store them as in a variable?
The problem is caused by backticks. Use $( ) instead, and it goes away:
var="*a<br>*b<br>*c"
var=$(printf '%s\n' "$var" | sed 's/\*/\\*/g')
printf '%s\n' "$var"
(Why is this problem caused by backticks? Because the only way to nest them is to escape the inner ones with backslashes, so they necessarily change how backslashes behave; whereas $( ), because it uses different starting and ending sigils, can be nested natively).
That said, if your shell is one (like bash) with ksh-inspired extensions, you don't need sed at all here, as the shell can perform simple string replacements natively via parameter expansion:
var="*a<br>*b<br>*c"
printf '%s\n' "${var//'*'/'\*'}"
For background on why this answer uses printf instead of echo, see Why is printf better than echo? at [unix.se], or the APPLICATION USAGE section of the POSIX specification for echo.
I am trying to use a variable to store the parameters, here is the simple test:
#!/bin/bash
sed_args="-e \"s/aaaa/bbbb/g\""
echo $sed_args`
I expected the output to be
-e "s/aaaa/bbbb/g"
but it gives:
"s/aaaa/bbbb/g"
without the "-e"
I am new to bash, any comment is welcome. Thanks, maybe this is already answered somewhere.
You need an array to construct arguments dynamically:
#!/usr/bin/env bash
sed_args=('-e' 's/aaaa/bbbb/g')
echo "${sed_args[#]}"
When you use the variable without double quotes, it gets word split by the shell even before echo sees the value(s). Then, the bash's builtin echo interprets -e as a parameter for itself (which is normally used to turn on interpretation of backslash escapes).
When you double quote the variable, it won't be split and will be interpreted as a single argument to echo:
echo "$sed_args"
For strings you don't control, it's safer to use printf as it doesn't take any arguments after the format string:
printf %s "$string"
So I am trying to make a script that contains egrep and accepts a numeric variable
#!/bin/bash
var=$1
list="egrep "^.{$var}$ /usr/share/dict/words"
cat list
For example, if var is 5, I would like this script to print out every line with 5 characters. For some reason the script does not do that. Help would be greatly appreciated!
Your script doesn't work because there are several problems with these lines:
list="egrep "^.{$var}$ /usr/share/dict/words"
cat list
The first line isn't complete, it's missing a closing quote,
Even if you fixed it, you're assigning a literal string to list, not the output of a command,
RE and filename should be separated
cat doesn't print a variable's content, echo does that.
So:
#!/bin/bash
var="$1"
list="$(egrep '^.{'"$var"'}$' /usr/share/dict/words)"
echo "$list"
should work.
Or even better, you can use just an awk command:
awk 'length==5' /usr/share/dict/words
with $1 or any other variable:
awk -v n="$1" 'length==n' /usr/share/dict/words