This question already has answers here:
How to escape single quotes within single quoted strings
(25 answers)
Closed 3 years ago.
I need your help.
I am creating bash script and I have problem in part where I need to add text in line 4 of test.txt file. My text is including variable. I know how to add text with variable, but in this case this variable must be in double quotation.
So, I am using this command:
sed -i "4s/$/Username = "$var1"/g" $dir/output/test.txt
but I get next results in test.txt file:
$Username = example
I tried many options but I can not achive to get variable in double quotation:
$Username = "example"
sed -i "4s/$/Username = \"$var1\"/" $dir/output/test.txt
Global switch is meaningless and can be removed since you're appending to the end of the line
Related
This question already has answers here:
Unix command to prepend text to a file
(21 answers)
Closed 2 years ago.
Real nit picky Linux question.
I have a text file, call it userec. I also have a string variable 'var_a'.
I want to concatenate the string value, let just say it's 'howdy' to the top of the text file.
So something like
echo $var_a | cat usrec > file_out
where it pipes the output from the echo $var_a as a file and adds it to the top of file_out and then adds the rest of the usrec file.
So if the userec file contains just the line 'This is the second line' then the contents of file_out should be:
howdy
This is the second line.
problem is that's not what the command is doing and I do not want to create a variable to store var_a in. This is running from a script and I don't want to create any extra flack to have to clean up afterwards.
I've tried other variations and I'm comming up empty.
Can anyone help me?
If you give cat any file names then it does not automatically read its standard input. In that case, you must use the special argument - in the file list to tell it to read the standard input, and where to include it in the concatenated output. Since apparently you want it to go at the beginning, that would be:
echo $var_a | cat - usrec > file_out
I would simply do :
echo $var_a > file_out
cat usrec >> file_out
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:
How to echo "$x_$y" in Bash script?
(4 answers)
Closed 2 years ago.
Please see the below shell script:
#!/bin/bash
echo "Enter the date in (YYYY-MM-DD) format: "
read dt
i=00
echo "/opt/log-$dt_$i"
Expected Output :
Enter the date in (YYYY-MM-DD) format:
2020-06-18
/opt/log-2020-06-18_00
But Getting the below output:
Enter the date in (YYYY-MM-DD) format:
2020-06-18
/opt/log-00
Please suggest?
You need to change your echo line to this:
echo "/opt/log-${dt}_${i}"
Explanation:
An _ (underscore) is a valid character in a variable name, so bash is looking for $dt_ and $i. Since $dt_ is not defined, it doesn't print it. Bash provides the alternate variable syntax using ${} to explicitly isolate the variables when performing string interpolation like this.
Like this:
echo "/opt/log-${dt}_$i"
# ^ ^
# curly brackets mandatory here
This is because you have to separate the variables with _ that could be part of any variables.
Which editor do you use ? For me in vim, it's clear:
Versus:
(the underscore become white)
This question already has answers here:
Replace a string in shell script using a variable
(12 answers)
sed substitution with Bash variables
(6 answers)
Closed 4 years ago.
I need to replace current value in configuration file with new value which is assigned to variable ,
like
file_name=abc.txt
needs to be replaced like
file_name=xyz.txt
where $file=xyz.txt
I tried
sed -i 's/file_name=.*/file_name=$file/g' conf_file.conf
however the variable is not getting expanded,
it comes like file_name=$file.
any pointers?
This should work,assuming that variable file has value:xyz.txt assigned to it:
sed "s/file_name=.*/file_name=${file}/g" file_name
Output:
file_name=xyz.txt
This question already has answers here:
How to concatenate string variables in Bash
(30 answers)
Closed 9 years ago.
I am trying to download files from a database using wget and url. E.g.
wget "http://www.rcsb.org/pdb/files/1BXS.pdb"
So format of the url is as such: http://www.rcsb.org/pdb/files/($idnumber).pdb"
But I have many files to download; so I wrote a bash script that reads id_numbers from a text file, forms url string and downloads by wget.
!/bin/bash
while read line
do
url="http://www.rcsb.org/pdb/files/$line.pdb"
echo -e $url
wget $url
done < id_numbers.txt
However, url string is formed as
.pdb://www.rcsb.org/pdb/files/4H80
So, .pdb is repleced with http. I cannot figure out why. Does anyone have an idea?
How can I format it so url is
"http://www.rcsb.org/pdb/files/($idnumber).pdb"
?
Thanks a lot.
Note. This question was marked as duplicate of 'How to concatenate strings in bash?' but I was actually asking for something else. I read that question before asking this one and it turns out my problem was with preparing the txt file in Windows not really string concetanation. I edited question title. I hope it is more clear now.
It sounds like your id_numbers.txt file has DOS/Windows-style line endings (carriage return followed by linefeed characters) instead of plain unix line endings (just linefeed). The result is that read thinks the line ends with a carriage return, $line actually has a carriage return at the end, and that gets embedded in the url, causing various confusion.
There are several ways to solve this. You could have bash trim the carriage return from the variable when you use it:
url="http://www.rcsb.org/pdb/files/${line%$'\r'}.pdb"
Or you could have read trim it by telling it that carriage return counts as whitespace (read will trim leading and trailing whitespace from what it reads):
while IFS=$'\r' read line
Or you could use a command like dos2unix (or whatever the equivalent is on your OS) to convert the id_numbers.txt file.
The -e echo option is used to output the desired content without inserting a new line, you do not need it here.
Also I suspect your file containing the ids to be malformed, on which OS did you create it?
Anyway, you can simplify your script this way:
!/bin/bash
while read line
do
wget "http://www.rcsb.org/pdb/files/$line.pdb"
done < id_numbers.txt
I was able to successfully test it with an id_numbers.txt file generated like so:
for i in $(0 9) ; do echo "$i" >> id_numbers.txt ; done
Try this:
url="http://www.rcsb.org/pdb/files/"$line
$url=$url".pdb"
For more info, check How to concatenate string variables in Bash?