I work on a script compressing files. I want to do an 'until loop' til' the content of variable matches the pattern. The script is using zenity. This is the major part:
part="0"
pattern="^([0-9]{1}[0-9]*([km])$"
until `grep -E "$pattern" "$part"` ; do
part=$(zenity --entry \
--title="Zip the file" \
--text "Choose the size of divided parts:
(0 = no division, *m = *mb, *k = *kb)" \
--entry-text "0");
if grep -E "$pattern" "$part" ; then
zenity --warning --text "Wrong text entry, try again." --no-cancel;
fi
done
I want it to accept string containing digits ended with 'k' or 'm' (but not both of them) and don't accept string started with '0'.
Is the pattern ok?
$ grep -w '^[1-9][0-9]*[km]$' <<< 45k
45k
$ grep -w '^[1-9][0-9]*[km]$' <<< 001023m
$ grep -w '^[1-9][0-9]*[km]$' <<< 1023m
1023m
Don't forget the <<< in your expression, you're not grep'ing a file, but a string. To be more POSIX-compliant, you can also use:
echo 1023m | grep -w '^[1-9][0-9]*[km]$'
But it is kinda ugly.
Edit:
Longer example:
initmessage="Choose the size of divided parts:\n(0 = no division, *m = *mb, *k = *kb)"
errmessage="Wrong input. Please re-read carefully the following:\n\n$initmessage"
message="$initmessage"
while true ; do
part=$(zenity --entry \
--title="Zip the file" \
--text "$message")
if grep -qw '^[1-9][0-9]*[km]$' <<< "$part" ; then
zenity --info --text 'Thank you !'
break
else
message="$errmessage"
fi
done
Also, this is not directly related to the question, but you may want to have a look at Yad, which does basically the same things Zenity does, but has more options. I used it a lot when I had to write Bash scripts, and found it much more useful than Zenity.
You don't want the back-quotes in the until line. You might write:
until grep -E "$pattern" "$part"
do
...body of loop...
done
Or you might add arguments to grep to suppress the output (or send the output to /dev/null). As written, the script tries to execute the output of the grep command and use the success/failure status of that (not the grep per se) as an indication of whether to continue the loop or not.
Additionally, your pattern needs some work. It is:
pattern="^([0-9]{1}[0-9]*([km])$"
There is an unmatched open parenthesis in there. It also looks to me as though it is trying to allow a leading zero. You probably want:
pattern='^[1-9][0-9]*[km]$'
Single quotes are generally safer than double quotes for things like regular expressions.
I just want to check if my variable called part is well-formed after writing it in Zenity entry dialog. I just realised that grep needs a file, but my part is a variable initialised in this script. How to get along now?
In bash, you can use the <<< operator to redirect from a string:
until grep -E "$pattern" <<< "$part"
In most other shells, you'd write:
until echo "$part" | grep -E "$pattern"
This also works in bash, of course.
Related
I'm writing a shell script that greps for $foo then counts the number of occurrences then runs a command. Each time that command is run, there is one less instance of $foo in that file. Uncertain on how to continuously read that file and reduce the value in the variable I set.
$count= `grep -o $foo /some/file |wc -w`
until [ $count -eq 0 ]
do
some_command_that_deletes_foo_in_file
done
However I realize that $count is set once at runtime and is not updated. What I want is $count to be updated to the current count in /some/file while the script is looping through /some/file until there is 0 instances of the phrase I'm grepping for. Uncertain to what the best approach is.
Unless you have additional code that you haven't showed us that depends on $count, you don't actually need to count occurrences; you just need to know whether the string appears in the file. For that, you can write:
while grep -q -- "$foo" /some/file ; do
some_command_that_deletes_foo_in_file
done
(using the fact that grep returns success when it finds the value, and failure when it does not, and using the -q flag to suppress its console output).
You could add the grep command inside the loop:
count=$(grep -o "$foo" /some/file |wc -w)
until (( count == 0 ))
do
some_command_that_deletes_foo_in_file
count=$(grep -o "$foo" /some/file |wc -w)
done
You simply want to delete the string "$foo"? Use sed:
sed "/$foo/d" /some/file > /some/other/file
The sed command is an editor. The /$foo/ is taking a regular expression (whatever the value of $foo), finding it in the file. The d tells it to delete the line.
sed doesn't usually do an in place edit. You usually have to write to another file and then to a move. However, some sed commands may have such a parameter. You can check your manage.
Second Try
I think it must take some action or perform some processing or something, and one of its effects is that one of the $foos is gone. (But I could be wrong.) – ruakh yesterday
This is what I get answering these questions at night.
You can take advantage of the fact that grep returns true (i.e. exit value = 0) if it can find your regular expression and false (i.e. exit value not equal to 0) otherwise:
while grep -o -q "$foo" "/some/file"
do
some_command_that_deletes_foo_in_file
done
The -q tells grep not to output anything. Instead, the exit value of grep will be true if the string is found and false if it isn't.
Read your manpage on grep. Some grep commands don't have the -q parameter. In that case, you'l need to redirect both STDOUT and STDERR to /dev/null.
Another tact may be to do your count of the number of lines, and then use that as a counter:
count=$(grep -o "$foo" "/some/file" | wc -w) # $(...) is the preferred syntax over `...`
for loop in {1..$count}
do
some_command_that_deletes_foo_in_file
done
The advantage is that you only grep through the file once (which maybe a long operation). However, your count maybe incorrect if $foo is on more than one line.
A few notes:
The $(...) is preferred over backticks
When you set a variable, you don't use the dollar sign in front of that variable. Note I have count= and not $count=.
Watch spaces. count= $(...) is wrong because there's a space after the equals sign.
I have a loop in my script that will append a list of email address's to a file "$CRN". If this script is executed again, it will append to this old list. I want it to overwrite with the new list rather then appending to the old list. I can submit my whole script if needed. I know I could test if "$CRN" exists then remove file, but I'm interested in some other suggestions? Thanks.
for arg in "$#"; do
if ls /students | grep -q "$arg"; then
echo "${arg}#mail.ccsf.edu">>$CRN
((students++))
elif ls /users | grep -q "$arg$"; then
echo "${arg}#ccsf.edu">>$CRN
((faculty++))
fi
Better do this :
CRN="/path/to/file"
:> "$CRN"
for arg; do
if printf '%s\n' /students/* | grep -q "$arg"; then
echo "${arg}#mail.ccsf.edu" >> "$CRN"
((students++))
elif printf '%s\n'/users/* | grep -q "${arg}$"; then
echo "${arg}#ccsf.edu" >> "$CRN"
((faculty++))
fi
done
don't parse ls output ! use bash glob instead. ls is a tool for interactively looking at file information. Its output is formatted for humans and will cause bugs in scripts. Use globs or find instead. Understand why: http://mywiki.wooledge.org/ParsingLs
"Double quote" every expansion, and anything that could contain a special character, eg. "$var", "$#", "${array[#]}", "$(command)". See http://mywiki.wooledge.org/Quotes http://mywiki.wooledge.org/Arguments and http://wiki.bash-hackers.org/syntax/words
take care to false positives like arg=foo and glob : foobar, that will match. You need grep -qw then if you want word boundaries. UP2U
The output is blank fr the below script. What is it missing? I am trying to grep a string
#!/bin/ksh
file=$abc_def_APP_13.4.5.2
if grep -q abc_def_APP $file; then
echo "File Found"
else
echo "File not Found"
fi
In bash, use the <<< redirection from a string (a 'Here string'):
if grep -q abc_def_APP <<< $file
In other shells, you may need to use:
if echo $file | grep -q abc_def_APP
I put my then on the next line; if you want your then on the same line, then add ; then after what I wrote.
Note that this assignment:
file=$abc_def_APP_13.4.5.2
is pretty odd; it takes the value of an environment variable ${abc_def_APP_13} and adds .4.5.2 to the end (it must be an env var since we can see the start of the script). You probably intended to write:
file=abc_def_APP_13.4.5.2
In general, you should enclose references to variables holding file names in double quotes to avoid problems with spaces etc in the file names. It is not critical here, but good practices are good practices:
if grep -q abc_def_APP <<< "$file"
if echo "$file" | grep -q abc_def_APP
Yuck! Use the shell's string matching
if [[ "$file" == *abc_def_APP* ]]; then ...
#!/bin/bash
fname=$2
rname=$1
echo "$(<$fname)" | while read line ; do
result=`echo "$(<$rname)" | grep "$line"; echo $?`
if [ $result != 0 ]
then
sed '/$line/d' $fname > newkas
fi 2> /dev/null
done
Hi all, i am new to bash.
i have two lists one older than another. I wish to compare the names on 'fname' against 'rname'. 'Result' is the standard out put which i will get if the name is still available in 'rname'. if is not then i will get the non-zero output.
Using sed to delete that line and re route it to a new file.
I have tried part by part of the code and it works until i add in the while loop function. sed don't seems to work as the final output of 'newkas' is the same as the initial input 'fname'.
Is my method wrong or did i miss out any parts?
Part 1: What's wrong
The reason your sed expression "doesn't work" is because you used single quotes. You said
sed '/$line/d' $fname > newkas
Supposing fname=input.txt' and line='example text' this will expand to:
sed '/$line/d' input.txt > newkas
Note that $line is still literally present. This is because bash will not interpolate variables inside single quotes, thus sed sees the $ literally.
You could fix this by saying
sed "/$line/d/" $fname > newkas
Because inside double quotes the variable will expand. However, if your sed expression becomes more complicated you could run into difficulty in cases where bash interprets things which you intended to be interpreted by sed. I tend to use the form
sed '/'"$line"'/d/' $fname > newkas
Which is a bit harder to read but, if you look carefully, single-quotes everything I intend to be part of the sed expression and double quotes the variable I want to expand.
Part 2: How to improve it
Your script contains a number things which could be improved.
echo "$(<$fname)" | while read line ; do
:
done
In the first place you're reading the file with "$(<$fname)" when you could just redirect the stdin of the while loop. This is a bit redundant, but more importantly you're piping to while, which creates an extra subshell and means you can't modify any variables from the enclosing scope. Better to say
while IFS= read -r line ; do
:
done < "$fname"
Next, consider your grep
echo "$(<$rname)" | grep "$line"
Again you're reading the file and echoing it to grep. But, grep can read files directly.
grep "$line" "$rname"
Afterwards you echo the return code and check its value in an if statement, which is a classic useless construct.
result=$( grep "$line" "$rname" ; echo $?)
Instead you can just pass grep directly to if, which will test its return code.
if grep -q "$line" "$rname" ; then
sed "/$line/d" "$fname" > newkas
fi
Note here that I have quoted $fname, which is important if it might ever contain a space. I have also added -q to grep, which suppresses its output.
There's now no need to suppress error messages from the if statement, here, because we don't have to worry about $result containing an unusual value or grep not returning properly.
The final result is this script
while IFS= read -r line ; do
if grep -q "$line" "$rname" ; then
sed "/$line/d" "$fname" > newkas
fi
done < "$fname"
Which will not work, because newkas is overwritten on every loop. This means that in the end only the last line in $fname was used. Instead you could say:
cp "$fname" newkas
while IFS= read -r line ; do
if grep -q "$line" "$rname" ; then
sed -i '' "/$line/d" newkas
fi
done < "$fname"
Which, I believe, will do what you expect.
Part 3: But don't do that
But this is all tangential to solving your actual problem. It appears to me that you want to simply create a file newkas which contains the all the lines of $fname except those that appear in $rname. This is easily done with the comm utility:
comm -2 -3 <(sort "$fname") <(sort "$rname") > newkas
This also changes the sort order of the lines, which may not be good for you. If you want to do it without changing the ordering then using the method #fge suggests is best.
grep -F -v -x -f "$rname" "$fname"
If I understand your need correctly, you want a file newaks which contains the lines in $fname which are also in $rname.
If this is what you want, using sed is overkill. Use fgrep:
fgrep -x -f $fname $rname > newkas
Also, there are problems with your script:
you capture the output of grep in result, which means it will never be exactly 0; what you want is executing the command and simply check for $?
your echoes are convoluted, just do grep whatever thefilename, or while...done <thefile;
finally, you take the line as is from the source file: the line can potentially be a regex, which means you will try and match a regex in $rname, which may yield to unexpected results.
And others.
I am working on a bash script which execute a command depending on the file type. I want to use the the "file" option and not the file extension to determine the type, but I am bloody new to this scripting stuff, so if someone can help me I would be very thankful! - Thanks!
Here the script I want to include the function:
#!/bin/bash
export PrintQueue="/root/xxx";
IFS=$'\n'
for PrintFile in $(/bin/ls -1 ${PrintQueue}) do
lpr -r ${PrintQueue}/${PrintFile};
done
The point is, all files which are PDFs should be printed with the lpr command, all others with ooffice -p
You are going through a lot of extra work. Here's the idiomatic code, I'll let the man page provide the explanation of the pieces:
#!/bin/sh
for path in /root/xxx/* ; do
case `file --brief $path` in
PDF*) cmd="lpr -r" ;;
*) cmd="ooffice -p" ;;
esac
eval $cmd \"$path\"
done
Some notable points:
using sh instead of bash increases portability and narrows the choices of how to do things
don't use ls when a glob pattern will do the same job with less hassle
the case statement has surprising power
First, two general shell programming issues:
Do not parse the output of ls. It's unreliable and completely useless. Use wildcards, they're easy and robust.
Always put double quotes around variable substitutions, e.g. "$PrintQueue/$PrintFile", not $PrintQueue/$PrintFile. If you leave the double quotes out, the shell performs wildcard expansion and word splitting on the value of the variable. Unless you know that's what you want, use double quotes. The same goes for command substitutions $(command).
Historically, implementations of file have had different output formats, intended for humans rather than parsing. Most modern implementations have an option to output a MIME type, which is easily parseable.
#!/bin/bash
print_queue="/root/xxx"
for file_to_print in "$print_queue"/*; do
case "$(file -i "$file_to_print")" in
application/pdf\;*|application/postscript\;*)
lpr -r "$file_to_print";;
application/vnd.oasis.opendocument.*)
ooffice -p "$file_to_print" &&
rm "$file_to_print";;
# and so on
*) echo 1>&2 "Warning: $file_to_print has an unrecognized format and was not printed";;
esac
done
#!/bin/bash
PRINTQ="/root/docs"
OLDIFS=$IFS
IFS=$(echo -en "\n\b")
for file in $(ls -1 $PRINTQ)
do
type=$(file --brief $file | awk '{print $1}')
if [ $type == "PDF" ]
then
echo "[*] printing $file with LPR"
lpr "$file"
else
echo "[*] printing $file with OPEN-OFFICE"
ooffice -p "$file"
fi
done
IFS=$OLDIFS