Bash new line feed in results [duplicate] - bash

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Iterate over a list of files with spaces
(12 answers)
Closed 5 years ago.
Trying to create a mysql backup script.
However, I am finding that I am getting line feeds in the results:
#!/bin/bash
cd /home
for i in $(find $PWD -type f -name "wp-config.php" );
do echo "'$i'";
done
And the results show:
'/home/site1/public_html/folders/wp-config.php'
\'/home/site2/public_html/New'
'Website/wp-config.php'
'/home/site3/public_html/wp-config.php'
'/home/site4/public_html/old'
'website/wp-config.php'
'/home/site5/public_html/wp-config.php'
Do a ls from the command-line, we see for the folders in question:
New\ website
old\ website
and is treating the '\' as newline character.
OK.. Doing some research:
https://stackoverflow.com/a/5928254/175063
${foo/ /.}
Updating for what we may want:
${i/\ /}
The code now becomes:
#!/bin/bash
cd /home
for i in $(find $PWD -type f -name "wp-config.php" |${i/\ /});
do echo "'$i'";
done
Ref. https://tomjn.com/2014/03/01/wordpress-bash-magic/
Ultimately, I really want something like this:
!/bin/bash
# delete files older than 7 days
## find /home/dummmyacount/backups/ -type f -name '*.7z' -mtime +7 -exec rm {} \;
# set a date variable
DT=$(date +"%m-%d-%Y")
cd /home
for i in $(find $PWD -type f -name "wp-config.php" );
WPDBNAME=`cat $i | grep DB_NAME | cut -d \' -f 4`
WPDBUSER=`cat $i | grep DB_USER | cut -d \' -f 4`
WPDBPASS=`cat $i | grep DB_PASSWORD | cut -d \' -f 4`
do echo "$i";
#do echo $File;
#mysqldump...
done

Answer 1

You can do this
find . -type f -name "wp-config.php" -print0 | while read -rd $'\x00' f
do
printf '[%s]\n' "$f"
done
which uses the NUL character as the delimiter to avoid special chars