Printf adding extra space in shell script - bash

I am writing a shell script to create file and write multiple lines to it.
#!/bin/bash
CURRENTHOST="$(hostname -I)"
touch genesis_start.sh
printf '%s\n\t''--http-server-address '"$CURRENTHOST"':8888 \' >> genesis_start.sh
There are more lines but I am just showing one.
The above code looks like in genesis_start.sh file:
--http-server-address 192.168.110.164 :8888 \
The extra space is added between 164 and 8888. How to display it without space?

There will always be space after the IP address, however, you can remove it like this :
CURRENTOST=${CURRENTHOST::-1}
Reference: https://unix.stackexchange.com/questions/144298/delete-the-last-character-of-a-string-using-string-manipulation-in-shell-script

Your code quality would really benefit from the help of the shellcheck.net linter.
There are many issues bugs with your script:
Misuse of printf (constant parts and variable formatter should go to the format string, there must be at least as many variable arguments as variable formatters)
Invalid escaping of single quote.
Wrong parsing of hostname -I which can return multiple space delimited entries.
Fixed your code:
#!/usr/bin/env bash
# Get the first IP address only
read -r CURRENTHOST _ < <(hostname -I)
printf '\t--http-server-address %s:8888\n' "$CURRENTHOST" > genesis_start.sh

Related

How to create new names for files with problematic characters for use in an existing bash scripted environment?

The goal is to get rid of (by changing) filenames that give headaches for scripting by translating them to something else. The reason is that in this nearly 30 year Unix / Linux environment, with a lot of existing scripts that may not be "written correctly", a new, large and important cache of files arrived that have to be managed, and so, a colleague has asked me to write a script to help with "problematic filenames" and translate them. They've got a list of chars to turn into dots, such as the comma, and another list to turn into underscores, such as whitespace, as but two examples and ran into problems which I asked about over here.
I was using tr to do it, but commenters to it said I should perhaps ask just about this instead of how to get tr to work. So, I have!
Parameter expansion can do this for you.
Note that unlike when using tr (as requested on your other question), when using parameter expansion you don't need to use backslashes inside your character class definitions: put the expansion in double quotes and bash will treat the results of that expansion as literal.
#!/usr/bin/env bash
toDots='\,;:|+##$%^&*~'
toUnderscores='}{]['"'"'="()`!'
# requires bash 5+: if debug=1, then print what we would do instead of doing it
runOrDebug() {
if (( debug )); then
printf '%s\n' "${*#Q}"
else
"$#"
fi
}
renameFiles() {
local name subDots subBoth
for name; do
subDots=${name//["$toDots"]/.}
subBoth=${subDots//["$toUnderscores"]/_}
if [[ $subBoth != "$name" ]]; then
runOrDebug mv -- "$name" "$subBoth"
fi
done
}
debug=1 renameFiles '[/a],/;[p:r|o\b+lem#a#t$i%c]/#(%$^!/(e^n&t*ry)~='
Note that toUnderscores is (except for the single quote in the middle) in single quotes, so all the backslashes in it are part of the variable's data rather than being syntax; because globs use character class syntax from REs, they're parsed as POSIX regular expression character class syntax.
See a demonstration of the technique running at https://ideone.com/kKE7IJ

Using space-separated arguments from a field in a tab-separated file

I'm writing a shell script intended to edit audio files using the sox command. I've been running into a strange problem I never encountered in bash scripting before: When defining space separated effects in sox, the command will work when that effect is written directly, but not when it's stored in a variable. This means the following works fine and without any issues:
sox ./test.in.wav ./test.out.wav delay 5
Yet for some reason the following will not work:
IFS=' ' # set IFS to only have a tab character because file is tab-separated
while read -r file effects text; do
sox $file.in.wav $file.out.wav $effects
done <in.txt
...when its in.txt is created with:
printf '%s\t%s\t%s\n' "test" "delay 5" "other text here" >in.txt
The error indicates this is causing it to see the output file as another input.
sox FAIL formats: can't open input file `./output.wav': No such file or directory
I tried everything I could think of: Using quotation marks (sox "$file.in.wav" "$file.out.wav" "$effects"), echoing the variable in-line (sox $file.in.wav $file.out.wav $(echo $effects)), even escaping the space inside the variable (effects="delay\ 5"). Nothing seems to work, everything produces the error. Why does one command work but not the other, what am I missing and how do I solve it?
IFS does not only change the behavior of read; it also changes the behavior of unquoted expansions.
In particular, unquoted expansions' content are split on characters found in IFS, before each element resulting from that split is expanded as a glob.
Thus, if you want the space between delay and 5 to be used for word splitting, you need to have a regular space, not just a tab, in IFS. If you move your IFS assignment to be part of the same simple command as the read, as in IFS=$'\t' read -r file effects text; do, that will stop it from changing behavior in the rest of the script.
However, it's not good practice to use unquoted expansions for word-splitting at all. Use an array instead. You can split your effects string into an array with:
IFS=' ' read -r -a effects_arr <<<"$effects"
...and then run sox "$file.in.wav" "$file.out.wav" "${effects_arr[#]}" to expand each item in the array as a separate word.
By contrast, if you need quotes/escapes/etc to be allowed in effects, see Reading quoted/escaped arguments correctly from a string

Renaming the file Directory which contains Space based on CSV in Shell

I need to rename the files inside the folder that has a space in it eg(Deco/main library/file1.txt )
code:
while IFS="," read orig new pat
do
mv -v $pat$new $pat$orig
done < new.csv
csv file:
newname,file1.txt,Deco/main\\\ library/
error:
mv: invalid option -- '\'
Welcome to Stackoverflow!
First: Use quotes around the use of variables. That means except in very rare occasions, you always should use "$foo" instead of $foo because if you are using the latter, the shell is supposed (and will) interpret spaces in the variables as word delimiters which you rarely want. Especially in your case you do not want it.
Second: Your CSV file seems to contain backslashes to quote the spaces. And some additional step seems to have added another level of quotation so than now you end up with three backslashes and a space for each original space. If this really is the case (please double check if what you wrote in your question is correct, otherwise my answer doesn't fit), you need to unquote this before you can use it.
There are security issues involved in using eval, so do not use it lightly (this disclaimer is necessary whenever proposing to use eval), but if you have trust in the input you are handling to not contain any nastinesses, then you can do this using this code:
while IFS="," read orig new pat
do
eval eval mv -v "$pat$new" "$pat$orig"
done < new.csv
Using this, two levels of quotation are evaluated (that's what eval does) before the mv command is executed.
I strongly suggest to do a dry run by adding echo before the mv first. Then instead of executing your commands they are merely printed first.

Printf splits a string at spaces using Bash [duplicate]

This question already has answers here:
Why a variable assignment replaces tabs with spaces
(2 answers)
Closed 7 years ago.
I'm having some troubles with the printf function in bash.
I wrote a little script on which I pass a name and two letters (such as "sh", "py", "ht") and it creates a file in the current working directory named "name.extension".
For instance, if I execute seed test py a file named test.py is created in the current working dir with the shebang #!/usr/bin/python3.
So far, so good, nothing fancy: I'm learning shell scripting and I thought this could be a simple exercise to test the knowledge gained so far.
The problem is when I want to create an HTML file. This is the function that I use:
creaHtml(){
head='<!--DOCTYPE html-->\n<html>\n\t<head>\n\t\t<meta charset=\"UTF-8\">\n\t</head>\n\t<body>\n\t</body>\n</html>'
percorso=$CARTELLA_CORRENTE/$NOME_FILE.html
printf $head>>$percorso
chmod 755 $percorso
}
If I run, for instance, seed test ht the correct function (creaHtml) is called, test.html is created but if I try to look into it I only see:
<!--DOCTYPE
And nothing else.
This is the trace for that function:
[sviluppo:~/bin]$ seed test ht
+ creaHtml
+ head='<!--DOCTYPE html-->\n<html>\n\t<head>\n\t\t<meta charset=\"UTF-8\">\n\t</head>\n\t<body>\n\t</body>\n</html>'
+ percorso=/home/sviluppo/bin/test.html
+ printf '<!--DOCTYPE' 'html-->\n<html>\n\t<head>\n\t\t<meta' 'charset=\"UTF-8\">\n\t</head>\n\t<body>\n\t</body>\n</html>'
+ chmod 755 /home/sviluppo/bin/test.html
+ set +x
However, if I try to run printf '<!--DOCTYPE html-->\n<html>\n\t<head>\n\t\t<meta charset=\"UTF-8\">\n\t</head>\n\t<body>\n\t</body>\n</html>' from the terminal, I see the correct output: the "skeleton" of an HTML file neatly displayed with indentation and everything. What am I missing here?
Try echo -e instead of printf. printf is for printing formatted strings. Since you didn't protect $head with quotes, bash splits the string to form the command. The first word (before first white space) forms the format string. The rest are just arguments for things you didn't specify to print.
echo -e "$head" > "$percorso"
The -e evaluates your \n into newlines. I changed your >> to > since it looks like you want this to be the whole file, rather than append to any existing file you might have.
You have to be careful with quotes in bash. One thing can become many things. This actually makes it more powerful, but it can be confusing for people learning. Notice that I also put the file name "$percorso" in double quotes too. This evaluates the variable and makes sure that it ends up as one thing. If you use single quotes, it will be one word, but not evaluated. Unlike Python, there is a big difference between single and double quotes.
If you want to use printf for compatibility as #chepner pointed out, just be sure to quote it:
printf "$head" > "$percorso"
Actually that is much simpler anyway.

Bash curl and variable in the middle of the url

I would need to read certain data using curl. I'm basically reading keywords from file
while read line
do
curl 'https://gdata.youtube.com/feeds/api/users/'"${line}"'/subscriptions?v=2&alt=json' \
> '/home/user/archive/'"$line"
done < textfile.txt
Anyway I haven't found a way to form the url to curl so it would work. I've tried like every possible single and double quoted versions. I've tried basically:
'...'"$line"'...'
"..."${line}"..."
'...'$line'...'
and so on.. Just name it and I'm pretty sure that I've tried it.
When I'm printing out the URL in the best case it will be formed as:
/subscriptions?v=2&alt=jsoneeds/api/users/KEYWORD FROM FILE
or something similar. If you know what could be the cause of this I would appreciate the information. Thanks!
It's not a quoting issue. The problem is that your keyword file is in DOS format -- that is, each line ends with carriage return & linefeed (\r\n) rather than just linefeed (\n). The carriage return is getting read into the line variable, and included in the URL. The giveaway is that when you echo it, it appears to print:
/subscriptions?v=2&alt=jsoneeds/api/users/KEYWORD FROM FILE"
but it's really printing:
https://gdata.youtube.com/feeds/api/users/KEYWORD FROM FILE
/subscriptions?v=2&alt=json
...with just a carriage return between them, so the second overwrites the first.
So what can you do about it? Here's a fairly easy way to trim the cr at the end of the line:
cr=$'\r'
while read line
do
line="${line%$cr}"
curl "https://gdata.youtube.com/feeds/api/users/${line}/subscriptions?v=2&alt=json" \
> "/home/user/archive/$line"
done < textfile.txt
Your current version should work, I think. More elegant is to use a single pair of double quotes around the whole URL with the variable in ${}:
"https://gdata.youtube.com/feeds/api/users/${line}/subscriptions?v=2&alt=json"
Just use it like this, should be sufficient enough:
curl "https://gdata.youtube.com/feeds/api/users/${line}/subscriptions?v=2&alt=json" > "/home/user/archive/${line}"
If your shell gives you issues with & just put \&, but it works fine for me without it.
If the data from the file can contain spaces and you have no objection to spaces in the file name in the /home/user/archive directory, then what you've got should be OK.
Given the contents of the rest of the URL, you could even just write:
while read line
do
curl "https://gdata.youtube.com/feeds/api/users/${line}/subscriptions?v=2&alt=json" \
> "/home/user/archive/${line}"
done < textfile.txt
where strictly the ${line} could be just $line in both places. This works because the strings are fixed and don't contain shell metacharacters.
Since you're code is close to this, but you claim that you're seeing the keywords from the file in the wrong place, maybe a little rewriting for ease of debugging is in order:
while read line
do
url="https://gdata.youtube.com/feeds/api/users/${line}/subscriptions?v=2&alt=json"
file="/home/user/archive/${line}"
curl "$url" > "$file"
done < textfile.txt
Since the strings may end up containing spaces, it seems (do you need to expand spaces to + in the URL?), the quotes around the variables are strongly recommended. You can now run the script with sh -x (or add a line set -x to the script) and see what the shell thinks it is doing as it is doing it.

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