Hopefully this is a simple one, but I can't figure it out. I'm trying to work out why this command isn't finding any results:
find . -iname '*.cgi' -o -iname '*.txt' -o -iname '*.htm' -o -iname '*.html' -o -iname '*.php' -exec grep -l 'community.cgi' {} +
If I simplify it and just do:
find . -iname '*.cgi' -o -iname '*.txt' -o -iname '*.htm' -o -iname '*.html' -o -iname '*.php'
Then I get the list of files I'm expecting. For some reason the -exec part doesn't seem to be what I'm expecting. If I just run a basic grep on ALL files, I get the list of files as well:
grep -l 'community.cgi' .
Logical-and binds tighter than logical-or. Try:
find . \( -iname '*.cgi' -o -iname '*.txt' -o -iname '*.htm' -o -iname '*.html' -o -iname '*.php' \) -exec grep -l 'community.cgi' {} +
Here, parens (which have to be escaped to pass through the shell) are used to bind the -iname tests together so that -exec runs if any one of those tests is true.
Related
This works for me:
$> find . -name "*.log" -exec basename '{}' \;
20160114.log
20160115.log
20160116.log
20160117.log
20160118.log
Is the {}, \ and ';' mandatory when using -exec as any other syntax simply doesn't work?
The following, more complex example doesn't work:
$> find . -name "*.log" -exec echo $(basename '{}') \;
./log/20160114.log
./log/20160115.log
./log/20160116.log
./log/20160117.log
./log/20160118.log
echo here is just to demonstrate. I eventually plan to use something like rm $TARGET_DIR/$(basename '{}') in its place… It just doesn't work that way (nesting). Any ideas?
How can I find all emails in php files and replace with an email?
find . -iname '*.php' -exec grep -E -o "\b[a-zA-Z0-9.-_]+#[a-zA-Z0-9.-]+\.[a-zA-Z]+\b" -exec sed -i 's//email#domain.com/g' {} \;
Find emails in php files command is:
find . -iname '*.php' -exec grep -E -o "\b[a-zA-Z0-9.-_]+#[a-zA-Z0-9.-]+\.[a-zA-Z]+\b" {} \;
and replace command is:
find . -iname '*.php' -type f -exec sed -i 's/old-email#domain.com/new-email#domain.com/g' {} \;
But I don't know exactly how to join them in one command.
Any ideas?
I am writing a script that will saech for php files that contain a phrase and I would like that phrase replaced with a new one below is my little script but it is not working it searches ok but does not work with the search and replace section
find . -type f -name "*.php" -exec grep -H "define('DB_HOST', 'localhost');" {} \; | xargs sed -i "define('DB_HOST', 'localhost');/define('DB_HOST', '10.0.0.1');/g"
can someone explain to me what i am doing wrong
many thanks
Joe
did you forget the 's/' at the beggining of the sed expression? As in
sed 's/expression1/expression2/g'
You seem to have
sed 'espression1/expression2/g'
Edit
Another thing: You don't need to use xarg here. You can use multiple -exec flags - and it will to each only if all the previous succeeded:
find . -name '*.php' -exec grep 'whatever' {} \; -exec sed -i 's/whatever/you want/g' {} \;
This will work:
find . -type f -name "*.php" -exec grep -l "define('DB_HOST', 'localhost');" {} \; | xargs sed -i "s/define('DB_HOST', 'localhost');/define('DB_HOST', '10.0.0.1');/g"
Corrections
Missing s/ in sed search and replace command
use grep -l instead of grep -H
I have this working fine for me:
find Sources/$1-$2 -name '*' |xargs perl -pi -e "s/domain.com/$2/g"
But when I change it to the following it doesn't:
find Sources/$1-$2 -name '*.php,*.rb' |xargs perl -pi -e "s/domain.com/$2/g"
What wrong?
Here's some explanation behind the solution that others have provided.
The tests in a find command are combined with Boolean operators:
-a -and
-o -or
! -not
If you don't supply an operator, -and is the default.
find . -type f -name '*.rb' # The command as entered.
find . -type f -a -name '*.rb' # Behind the scenes.
Your search failed because it didn't find any matching files:
# Would find only files with bizarre names like 'foo.php,bar.rb'
find . -name '*.php,*.rb'
You need to supply the file extensions as separate -name tests, combined in an OR fashion.
find . -name '*.php' -o -name '*.rb'
you have to write it as:
find Sources/$1-$2 -name '*.php' -o -name '*.rb' ....
I'm guessing that you want all files then end in .php and .rb.
Try find Sources/$1-$2 \( -iname "*.php" -o -iname "*.rb" \) -print |xargs perl -pi -e "s/domain.com/$2/g"
It is much better filtering out find's result with [ef]grep. Why?
Because you can fed the grep pattern as an argument, or can read it from the config or soo. It is much easier to write: grep "$PATTERN" as constructing long find arguments with '-o'. (ofc, here are situations, where find args are better), but not in your case.
The cost is one more process. So, for you example is easy to write a script myscript.sh
find Sources/$1-$2 -print | egrep -i "$3" | xargs ...
you can call it
./myscript.sh aaa bbb ".(php|rb)$"
and the result will equivalent to more complicated
find Sources/$1-$2 \( -iname '*.php' -o -iname '*.rb' \) | xargs ...
but
why bother? If you have bash4+, (and shopt -s globstar in your .bashrc) you can simple write:
perl -pi -e '.....' Sources/aaa-bbb/**/*.{rb,php}
the ** is like a find -name.
By the way, xargs is not needed here.
find Sources/$1-$2 \( -name '*.php' -o -name '*.rb' \) \
-exec perl -i -pe "s/domain\.com/$2/g" {} +
Also notice the "." in /domain.com/ needs to be escaped.
This part " ( -name *txt -o -name *html ) " confuses me in the code:
find $HOME \( -name \*txt -o -name \*html \) -print0 | xargs -0 grep -li vpn
Can someone explain the the brackets and "-o"? Is "-o" a command or a parameter? I know the brackets are escaped by "\" , but why are they for?
By default, the conditions in the find argument list are 'and'ed together. The -o option means 'or'.
If you wrote:
find $HOME -name \*txt -o -name \*html -print0
then there is no output action associated with the file names end with 'txt', so they would not be printed. By grouping the name options with parentheses, you get both the 'html' and 'txt' files.
Consider the example:
mkdir test-find
cd test-find
cp /dev/null file.txt
cp /dev/null file.html
The comments below have an interesting side-light on this. If the command was:
find . -name '*.txt' -o -name '*.html'
then, since no explicit action is specified for either alternative, the default -print (not -print0!) action is used for both alternatives and both files are listed. With a -print or other explicit action after one of the alternatives (but not the other), then only the alternative with the action takes effect.
find . -name '*.txt' -print -o -name '*.html'
This also suggests that you could have different actions for the different alternatives.
You could also apply other conditions, such as a modification time:
find . \( -name '*.txt' -o -name '*.html' \) -mtime +5 -print0
find . \( -name '*.txt' -mtime +5 -o -name '*.html' \) -print0
The first prints txt or html files older than 5 days (so it prints nothing for the example directory - the files are a few seconds old); the second prints txt files older than 5 days or html files of any age (so just file.html). And so on...
Thanks to DevSolar for his comments leading to this addition.
The "-o" means OR. I.e., name must end in "txt" or "html". The brackets just group the two conditions together.
The ( and ) provide a way to group search parameters for the find command. The -o is an "or" operator.
This find command will find all files ending in "txt" or "html" and pass those as arguments to the grep command.