Is it possible to echo a backspace in bash?
Something like
echo $'stack\b'
Shouldn't output stac? Or I'm missing something?
More specifically, I'd like to use that in:
ls | wc -l; echo $'\b items'
\b makes the cursor move left, but it does not erase the character. Output a space if you want to erase it.
For some distributions you may also need to use -e switch of echo:
-e enable interpretation of backslash escapes
So it will look like
echo -e 'stack\b '
Also, files=(*) ; echo "${#files[#]} items".
So to answer the actual question about backspaces this will simulate a backspace:
echo -e "\b \b"
It will move the character back one, then echo a space overwriting whatever character was there, then moving back again - in effect deleting the previous character. It won't go back up a line though so the output before that should not create a new line:
echo -n "blahh"; echo -e "\b \b"
It is not exactly what you're asking for, but also, in the line of Ignacio's answer, you could use for this case:
echo "$(ls | wc -l) items"
AFAIK you cannot print a character that deletes the one before, not even printing the char whose hexadecimanl number correspoonds to backspace. You can move back and print a blank space to delete, though. With cput you can do many things and print wherever you want in the screen.
Related
I have a simple bash script and I don't understand the return value.
My script
#!bin/bash
string=$(head -n 1 test.txt)
IFS=":"
read -r pathfile line <<< "$string"
echo "left"$line"right"
And my test.txt
filepath:file content
others lines
...
I have this return on the console.
rightfile content
The problem isn't when file only have 1 line.
I don't know why I don't have left value right to result.
Your input file has MSWin line ends (\x0D\x0A). Therefore, \x0D becomes part of $line and when printed, it moves the cursor back to the beginning, so $line"right" overwrites it.
Run dos2unix or fromdos on the input file to fix it.
BTW, you don't need to quote left and right. Quoting the variable might be needed, though.
echo left"$line"right
I only know what I have read so far, and I am confused about how to actually echo a variable as is.
echo "$var" might fail if var='-n'
printf '%s\n' "$var" might fail because of shell not implementig printf
echo -- "$var" might fail because it is a gnu extension
So if i would have to guess:
echo x"$var"|sed 's#^x##1' would be the only way, but I have never encountered that pattern. Why?
As a concrete question:
for source; do
target="$(echo "$source"|sed 's#[^a-z0-9]\+#.#')"
# do stuff with $source and $target
done
Does this work, or could someone "hack" / "break" my script by putting a file named '-n' somewhere, assuming my script is executed by some my_script * cron?
How do I write echo "$var" so it does not break?
Does this work, or could someone "hack" / "break" my script by putting
a file named '-n' somewhere?
There is nothing wrong with:
target="$(echo "$source"|sed 's#[^a-z0-9]\+#.#')"
What is happening:
"$(...)" is a command substitution which will substitute the results of the command within as the value -- in which case the result is assigned to target.
echo "$source"|sed 's#[^a-z0-9]\+#.#' simply pipes the output of echo (e.g. what is in source) to sed for the simple substitution of every character not lowercase or a digit followed by + with a period 1. Note: the quotes ".." around $source ARE proper within the command substitution.
There is no inherent reason assigning -n to a variable will cause any mischief. What you do with the variable is another question, but suffice it to say it is hard to see any problem.
"POSIX-shell's out there not implementing printf" -- Huh? Any shell not implementing printf would be more an exception rather than the rule. See printf - The Open Group Library that is POSIX.
If you are attempting to printf output that begins with '-' simply precede the output with "--" to indicate End-of-Options before the string your want to print and things will go fine. With your example of "-n", printf is about the only way you will output a variable beginning with the single '-', for example:
$ t="-n"
$ printf -- "%s\n" "$t"
-n
(note: you don't have to include "--" in printf "%s\n" "$var", the only time you must include it is with printf -- "-foo\n" or you will receive an "invalid option error".
For echo you can enable interpretation of backslash escapes with -e and include a backspace, e.g.
$ echo -e " \b$t"
-n
I think that has covered all issues. If not, let me know. Also, if you have any additional questions, drop a comment below or edit and add to your question.
footnotes:
note: + isn't part of basic regular expressions and it need not be escaped, but if there is any question, it is safer to include in a character class of its own, e.g. [^a-z0-9][+].
I'm having an issue in something that seems to be a rookie error, but I can't find a way to find a solution.
I have a bash script : log.sh
which is :
#!/bin/bash
echo $1 >> log_out.txt
And with a file made of filenames (taken from the output of "find" which names is filesnames.txt and contains 53 lines of absolute paths) I try :
./log.sh $(cat filenames.txt)
the only output I have in the log_out.txt is the first line.
I need each line to be processed separately as I need to put them in arguments in a pipeline with 2 softwares.
I checked for :
my lines being terminated with /n
using a simple echo without writing to a file
all the sorts of cat filenames.txt or (< filenames.txt) found on internet
I'm sure it's a very dumb thing, but I can't find why I can't iterate more than one line :(
Thanks
It is because your ./log.sh $(cat filenames.txt) is being treated as one argument.
while IFS= read -r line; do
echo "$line";
done < filenames.txt
Edit according to: https://mywiki.wooledge.org/DontReadLinesWithFor
Edit#2:
To preserve leading and trailing whitespace in the result, set IFS to the null string.
You could simplify more and skip using explicit variable and use the default $REPLY
Source: http://wiki.bash-hackers.org/commands/builtin/read
You need to quote the command substitution. Otherwise $1 will just be the first word in the file.
./log.sh "$(cat filenames.txt)"
You should also quote the variable in the script, otherwise all the newlines will be converted to spaces.
echo "$1" >> log_out.txt
If you want to process each word separately, you can leave out the quotes
./log.sh $(cat filenames.txt)
and then use a loop in the script:
#!/bin/bash
for word in "$#"
do
echo "$word"
done >> log_out.txt
Note that this solution only works correctly when the file has one word per line and there are no wildcards in the words. See mywiki.wooledge.org/DontReadLinesWithFor for why this doesn't generalize to more complex lines.
You can iterate with each line.
#!/bin/bash
for i in $*
do
echo $i >> log_out.txt
done
I have a big problem doing a script: basically, I read a line from files.
All lines are made of 3 to 8 characters contiguous (no space).
Then I used sed to replace those lines inside a pattern (aka "var" in my minimal script below)
var="iao"
for m in `more meshing/junction_names.txt`
do
echo $m
echo -n $m | xxd -ps | sed 's/[[:xdigit:]]\{2\}/\\x&/g'
echo $var |sed "s/a/b/"
echo $var |sed "s/a/$m/"
done
Now these are the first 3 record of my output (they are all the same anyway).
I am using linux. According kate, all files are encoded UTF-8. Very weird huh? Any idea why that is is welcome.
J_LEAK
\x4a\x5f\x4c\x45\x41\x4b\x0d
ibo
oJ_LEAK
JO_1
\x4a\x4f\x5f\x31\x0d
ibo
oJO_1
JPL2_F
\x4a\x50\x4c\x32\x5f\x46\x0d
ibo
oJPL2_F
JF_PL2
Your input file contains DOS carriage returns (or possibly, the absurd attempt to read it with more introduces them). The hex dump shows this clearly; every value ends with \x0d which translates to a control code which causes the terminal to jump the cursor back to the beginning of the line.
This is a massive FAQ and you can find many examples of how to troubleshoot this basic problem, including in the bash tag wiki.
Tangentially, you should always quote strings unless you specifically require the shell to perform wildcard expansion and whitespace tokenization on the value; and Bash has built-ins to avoid the inelegant and somewhat error-prone echo | sed. Finally, don't read lines with for.
var="iao"
tr -d '\015' <meshing/junction_names.txt |
while read -r m; do # don't use a for loop
echo "$m" # quote!
echo -n "$m" | xxd -ps | sed 's/[[:xdigit:]]\{2\}/\\x&/g'
echo "${var/a/b}" # quote; use Bash built-in substitution mechanism
echo "${var/a/$m}"
done
Perhaps you want to remove the carriage returns once and for all, and then just use while read .... done <fixed-file instead of the tr pipeline.
var=ab
echo -n "$var"
Output: ab
var=abc
echo "$var"
Output: ababc
I want to delete the first ab and replace it by abc
How would I do that?
Regards, intelinside
Perhaps you are looking for something like:
var=ab
echo -n "$var"
var=abc
echo -e "\r$var"
This doesn't actually delete anything, but merely moves the cursor to the beginning of the line and overwrites. If the text being written is too short, the content of the previous write will still be visible. You can either write spaces over the old text (very easy and portable):
printf "\r%-${COLUMNS}s" "$var"
or use some terminal escape sequences to delete the old text (not portable):
echo -e "\r$var\033[K"
to move to the beginning of the line, write new text, and then delete from the cursor to the end of the line. (This may not work, depending on the terminal.)