Bash separation of line with newlines instead of spaces - linux

I did two following commands:
With the first one I listed content of directory and stored it in variable.
Second one shows content of variable.
Now I decided that I want to separate listing not with spaces but with newlines, I do the following:
I get a mess. Why?
It's worth to note that when I changed command so, it worked as I wanted:
Could someone please explain, why 0x20 or 32 ( I tried this number too) is not treated Bash as space in this case?

tr simply doesn't recognize hex but octal. This would work:
tr '\040' '\n'
And the easier way to show your files is
shopt -s nullglob ## Optional.
printf '%s\n' *

The problem with tr '\0x20' is, tr is treating all the character sequence as literal characters. And the characters are 0, x, 2. Note all of theese characters were replaced in the output by \n. That's why you have .t instead of txt. Also 2 didn't appear too.
This is not bash, its tr which is making you unhappy. If you really want to iterate over file names there are better ways to do that.
for f in *; do
# do work with $f. But always use quotes. Like `"$f"`
done

Related

How to add character at the end of specific line in UNIX/LINUX?

Here is my input file. I want to add a character ":" into the end of lines that have ">" at the beginning of the line. I tried seq -i 's|$|:|' input.txt but ":" was added to all the ending of each line. It is also hard to call out specific line numbers because, in each of my input files, the line contains">" present in different line numbers. I want to run a loop for multiple files so it is useless.
>Pas_pyrG_2
AAAGTCACAATGGTTAAAATGGATCCTTATATTAATGTCGATCCAGGGACAATGAGCCCA
TTCCAGCATGGTGAAGTTTTTGTTACCGAAGATGGTGCAGAAACAGATCTGGATCTGGGT
>Pas_rpoB_4
CAAACTCACTATGGTCGTGTTTGTCCAATTGAAACTCCTGAAGGTCCAAACATTGGTTTG
ATCAACTCGCTTTCTGTATACGCAAAAGCGAATGACTTCGGTTTCTTGGAAACTCCATAC
CGCAAAGTTGTAGATGGTCGTGTAACTGATGATGTTGAATATTTATCTGCAATTGAAGAA
>Pas_cpn60_2
ATGAACCCAATGGATTTAAAACGCGGTATCGACATTGCAGTAAAAACTGTAGTTGAAAAT
ATCCGTTCTATTGCTAAACCAGCTGATGATTTCAAAGCAATTGAACAAGTAGGTTCAATC
TCTGCTAACTCTGATACTACTGTTGGTAAACTTATTGCTCAAGCAATGGAAAAAGTAGGT
AAAGAAGGCGTAATCACTGTAGAAGAAGGCTCAGGCTTCGAAGACGCATTAGACGTTGTA
Here is experted output file:
>Pas_pyrG_2:
AAAGTCACAATGGTTAAAATGGATCCTTATATTAATGTCGATCCAGGGACAATGAGCCCA
TTCCAGCATGGTGAAGTTTTTGTTACCGAAGATGGTGCAGAAACAGATCTGGATCTGGGT
>Pas_rpoB_4:
CAAACTCACTATGGTCGTGTTTGTCCAATTGAAACTCCTGAAGGTCCAAACATTGGTTTG
ATCAACTCGCTTTCTGTATACGCAAAAGCGAATGACTTCGGTTTCTTGGAAACTCCATAC
CGCAAAGTTGTAGATGGTCGTGTAACTGATGATGTTGAATATTTATCTGCAATTGAAGAA
>Pas_cpn60_2:
ATGAACCCAATGGATTTAAAACGCGGTATCGACATTGCAGTAAAAACTGTAGTTGAAAAT
ATCCGTTCTATTGCTAAACCAGCTGATGATTTCAAAGCAATTGAACAAGTAGGTTCAATC
TCTGCTAACTCTGATACTACTGTTGGTAAACTTATTGCTCAAGCAATGGAAAAAGTAGGT
AAAGAAGGCGTAATCACTGTAGAAGAAGGCTCAGGCTTCGAAGACGCATTAGACGTTGTA
Do seq have more option to modify or the other commands can solve this problem?
sed -i '/^>/ s/$/:/' input.txt
Search the lines of input for lines that match ^> (regex for "starts with the > character). Those that do substitute : for end-of-line (you got this part right).
/ slashes are the standard separator character in sed. If you wish to use different characters, be sure to pass -e or s|$|:| probably won't work. Since / characters, unlike | characters, are not meaningful character within the shell, it's best to use them unless the pattern also contains slashes, in which case things get unwieldy.
Be careful with sed -i. Make a backup - make sure you know what's changing by using diff to compare the files.
On OSX -i requires an argument.
Using ed to edit the file:
printf "%s\n" 'g/^>/s/$/:/' w | ed -s input.txt
For every line starting with >, add a colon to the end, and then write the changed file back to disk.

Remove lines with japanese characters from a file

First question on here- I've searched around to put together an answer to this but have come up empty thus far.
I have a multi-line text file that I am cleaning up. Part of this is to remove lines that include Japanese characters. I have been using sed for my other operations but it is not working in this instance.
I was under the impression that using the -r switch and the \p{Han} regular expression would work (from looking at other questions of this kind), but it is not working in this case.
Here is my test string - running this returns the full string, and does not filter out the JP characters as I was expecting.
echo 80岁返老还童的处女: 第3话 | sed -r "s/\\p\{Han\}//g"
Am I missing something? Is there another command I should be using instead?
I think this might work for you:
echo "80岁返老还童的处女: 第3话" | tr -cd '[:print:]\n'
sed doesn't support unicode classes AFAIK, and nor support multibyte ranges.
-d deletes characters in SET1, and -c reverses it.
[:print:] matches all printable characters including space.
\n is a newline
The above will not only remove Japanese characters but all multibyte characters, including control characters.
Perl can also be used:
PERLIO=:utf8 perl -pe 's/\p{Han}//g' file
PERLIO=:utf8 tells Perl to tread input and output as UTF-8

Change all non-ascii chars to ascii Bash Scripting

I am trying to write a script that take people names as an arguments and create a folder with their names. But in folder names, the non-ascii chars and whitespaces can sometimes make problem so I want to remove or change them to ascii chars.
I can remove the whitespace between name and surname but I can not figure out how can I change ş->s, ç->c, ğ->g, ı->i, ö->o.
Here is my code :
#!/bin/bash
ARRAY=("$#")
ELEMENTS=${#ARRAY[#]}
for (( i=0;i<$ELEMENTS;i++))
do #C-like for loop syntax
echo ${ARRAY[$i]} | grep "[^ ]*\b" | tr -d ' '
done
I run my script like that myscript.sh 'Çişil Aksoy' 'Cem Dalgıç'
It should change the arguments like : CisilAksoy CemDalgic
Thanks in advance
EDIT :
I found this solution, this does not look very pretty but it works.
sed 's/ş/s/gI; s/ç/c/gI; s/ü/u/gI; s/ö/o/gI; s/ı/i/gI;'
EDIT2 : SOLVED
#!/bin/bash
ARRAY=("$#")
ELEMENTS=${#ARRAY[#]}
for (( i=0;i<$ELEMENTS;i++))
do #C-like for loop syntax
v=$(echo ${ARRAY[$i]} | grep "[^ ]*\b" | tr -d ' ' | sed 's/ş/s/gI; s/ç/c/gI; s/ü/u/gI; s/ö/o/gI; s/ı/i/gI;')
mkdir $v
done
Anything that converts from UTF-8 to ASCII is going to be a compromise.
The iconv program does what was requested (not necessarily satisfying everyone, as in Transliterate any convertible utf8 char into ascii equivalent). Given
Çişil Aksoy' 'Cem Dalgıç
in "foo.txt", and the command
iconv -f UTF8 -t ASCII//TRANSLIT <foo.txt
that would give
Cisil Aksoy' 'Cem Dalg?c
The lynx browser has a different set of ASCII approximations. Using this command
lynx -display_charset=us-ascii -force_html -nolist -dump foo.txt
I get this result:
C,isil Aksoy' 'Cem Dalgic,
Simply put, you can't. ASCII only supports 128 characters.
International characters typically use some variation of Unicode, which can store a much much greater number of characters.
I think your best bet is to identify WHY your folder creation fails when using these characters. Does the method or function not support Unicode? If it does, figure out how to specify that instead of ASCII. If not, you might be stuck with sed and/or tr, which is probably not sustainable.
[UPDATED]
You should be able to substitute multiple characters via tr like follows:
echo şğıö | tr şçğıö scgio
sgio
(I removed my comment from earlier. I tried it on a different server and it worked fine.)

Understanding sed expression 's/^\.\///g'

I'm studying Bash programming and I find this example but I don't understand what it means:
filtered_files=`echo "$files" | sed -e 's/^\.\///g'`
In particular the argument passed to sed after '-e'.
It's a bad example; you shouldn't follow it.
First, understanding the sed expression at hand.
s/pattern/replacement/flags is the a sed command, described in detail in man sed. In this case, pattern is a regular expression; replacement is what that pattern gets replaced with when/where found; and flags describe details about how that replacement should be done.
In this case, the s/^\.\///g breaks down as follows:
s is the sed command being run.
/ is the sigil used to separate the sections of this command. (Any character can be used as a sigil, and the person who chose to use / for this expression was, to be charitable, not thinking about what they were doing very hard).
^\.\/ is the pattern to be replaced. The ^ means that this replaces anything only at the beginning; \. matches only a period, vs . (which is regex for matching any character); and \/ matches only a / (vs /, which would go on to the next section of this sed command, being the selected sigil).
The next section is an empty string, which is why there's no content between the two following sigils.
g in the flags section indicates that more than one replacement can happen each line. In conjunction with ^, this has no meaning, since there can only be one beginning-of-the-line per line; further evidence that the person who wrote your example wasn't thinking much.
Using the same data structures, doing it better:
All of the below are buggy when handling arbitrary filenames, because storing arbitrary filenames in scalar variables is buggy in general.
Still using sed:
# Use printf instead of echo to avoid bugginess if your "files" string is "-n" or "-e"
# Use "#" as your sigil to avoid needing to backslash-escape all the "\"s
filtered_files=$(printf '%s\n' "$files" | sed -e 's#^[.]/##g'`)
Replacing sed with a bash builtin:
# This is much faster than shelling out to any external tool
filtered_files=${files//.\//}
Using better data structures
Instead of running
files=$(find .)
...instead:
files=( )
while IFS= read -r -d '' filename; do
files+=( "$filename" )
done < <(find . -print0)
That stores files in an array; it looks complex, but it's far safer -- works correctly even with filenames containing spaces, quote characters, newline literals, etc.
Also, this means you can do the following:
# Remove the leading ./ from each name; don't remove ./ at any other position in a name
filtered_files=( "${files[#]#./}" )
This means that a file named
./foo/this directory name (which has spaces) ends with a period./bar
will correctly be transformed to
foo/this directory name (which has spaces) ends with a period./bar
rather than
foo/this directory name (which has spaces) ends with a periodbar
...which would have happened with the original approach.
man sed. In particular:
-e script, --expression=script
add the script to the commands to be executed
And:
s/regexp/replacement/
Attempt to match regexp against the pattern space. If success-
ful, replace that portion matched with replacement. The
replacement may contain the special character & to refer to that
portion of the pattern space which matched, and the special
escapes \1 through \9 to refer to the corresponding matching
sub-expressions in the regexp.
In this case, it replaces any occurence of ./ at the beginning of a line with the empty string, in other words removing it.

Replacing strings with special characters with linux sed

I've read lots of posts to understand how to correctly escape white spaces and special characters inside strings using sed, but still i can't make it, here's what i'm trying to achieve.
I have a file containing the some strings like this one:
JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=some_value"
and i'm trying to replace 'some_value' using the following:
sed -i "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1$DORG_APACHE_CATALINA_JSESSIONID/" $JBOSS_CONFIGURATION/jboss.configuration
$JBOSS_CONFIGURATION is a variable containing an absolute Linux path.
jboss.configuration is a file i'm pointing as the target for replace
operations.
$DORG_APACHE_CATALINA_JSESSIONID contains the value i want instead
of 'some_value'.
Please note that the pattern:
JAVA_OPTS="$JAVA_OPTS -D
Is always present, and org.apache.catalina.jsessionid is an example of a variable value i'm trying to replace with this script.
What's missing/wrong ? i tried also escaping whitespaces using \s without success,
and echoing the whole gives me the following:
echo "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1$DORG_APACHE_CATALINA_JSESSIONID/"
s/^\(JAVA_OPTS="$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*$/\1/
is echo interpreting the search pattern as sed does ?
any info/help/alternative ways of doing it are highly welcome,
thank you all
echo 'JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=some_value"' | (export DORG_APACHE_CATALINA_JSESSIONID=FOO/BAR/FOOBAR; sed "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1${DORG_APACHE_CATALINA_JSESSIONID////\/}\"/")
Note the bash expansion (in order to escape any / that may trip up sed) and the extra \" after $DORG_APACHE_CATALINA_JSESSIONID in order to properly close the double quote. Other than that your sed expression works for me and the above command outputs the follwoing result:
JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=FOO/BAR/FOOBAR"
You can use sed like this:
sed -r '/\$JAVA_OPTS -D/{s/^(.+=).*$/\1'"$DORG_APACHE_CATALINA_JSESSIONID"'/;}' $JBOSS_CONFIGURATION/jboss.configuration
You can specify a pattern that'll match the desired string rather than trying to specify it exactly.
The following should work for you:
sed -i 's#^\(JAVA_OPTS.*Dorg.apache.catalina.jsessionid\)=\([^"]*\)"#\1='"$DORG_APACHE_CATALINA_JSESSIONID"'"#' $JBOSS_CONFIGURATION/jboss.configuration
sed 's/=\w.*$/='"$DORG_APACHE_CATALINA_JSESSIONID"'/' $JBOSS_CONFIGURATION/jboss.configuration

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