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linux shell script that executes sed substitution based on mysql query

hard coded sed substitution
cat $OUTFILE_1 |sed -e 's/ABC1/123/g' -e 's/DEF1/456/g'

Is there a way to look up the values from a mysql table.  something like select field2 where field1 like '%1';  that could be put in a result set array to dynamically substitute all the field1 with field2?
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mcgilljd
Asked:
mcgilljd
1 Solution
 
medveddCommented:
If you can get values from mysql and put them into shell variables, you can do something like this:

sed -e 's/'"$field1"'/'"$field2"'/'
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brb6708Commented:
#! /bin/bash

tmpfile=/tmp/some_unlikely_double_filename.7455
rm -f $tmpfile
touch $tmpfile
for i in `mysql -BrN -ubrb  -p test -e"select object_id  from testad where object_id  like '%$1'"`; do
    echo "-e s/$1/$i/g" >> $tmpfile
done

cat $OUTFILE_1 |sed `tr '\n' ' ' < $tmpfile`

should do the job
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brb6708Commented:
.... sorry replace sqlstatement according your needs:

-e"select field2  from your_table where field1  like '%$1'"

tried it in a test database on my system and forgot to replace names.....
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mankowitzCommented:
You might be able to do something like this:

% mysql.exe -uxxxx -pYYYYYY -e "SELECT concat('sed s/', field1, '/', field2, '/;') FROM schema.table" | sh
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mcgilljdAuthor Commented:
That works, but I was wondering if it could be done without writing to a tempfile.
Maybe
echo "cat $OUTFILE_1 |sed";for i in `mysql -BrN -ubrb  -p test -e"select object_id  from testad where object_id  like '%$1'"`; do
    echo " -e s/$1/$i/g " > $outfile
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mcgilljdAuthor Commented:
forgot the done.

echo "cat $OUTFILE_1 |sed";for i in `mysql -BrN -ubrb  -p test -e"select object_id  from testad where object_id  like '%$1'"`; do
    echo " -e s/$1/$i/g ";done > $outfile
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