删除与图案匹配的线

Question

I want to search for patterns in a file and remove the lines containing the pattern. To do this, am using:

originalLogFile='sample.log'
outputFile='3.txt'
temp=$originalLogFile 

 while read line
 do
    echo "Removing" 
    echo $line
    grep -v "$line" $temp > $outputFile
    temp=$outputFile
done <$whiteListOfErrors

This works fine for the first iteration. For the second run, it throws :

grep: input file ‘3.txt’ is also the output

Any solutions or alternate methods?

Answer 1

以下应等效

grep -v -f  "$whiteListOfErrors" "$originalLogFile" > "$outputFile"

Answer 2

originalLogFile='sample.log'
outputFile='3.txt'
tmpfile='tmp.txt'
temp=$originalLogFile 
while read line
do
   echo "Removing" 
   echo $line
   grep -v "$line" $temp > $outputFile
   cp $outputfile $tmpfile
   temp=$tmpfile
done <$whiteListOfErrors

Answer 3

Use sed for this:

sed '/.*pattern.*/d' file

If you have multiple patterns you may use the -e option

sed -e '/.*pattern1.*/d' -e '/.*pattern2.*/d' file

If you have GNU sed (typical on Linux) the -i option is comfortable as it can modify the original file instead of writing to a new file. (But handle with care, in order to not overwrite your original)

Answer 4

Used this to fix the problem:

while read line
do
    echo "Removing" 
    echo $line
    grep -v "$line" $temp | tee $outputFile 
    temp=$outputFile
done <$falseFailures

Answer 5

Trivial solution might be to work with alternating files; eg

idx=0
while ...
    let next='(idx+1) % 2'
    grep ... $file.$idx > $file.$next
    idx=$next

A more elegant might be the creation of one large grep command

args=( )
while read line; do args=( "${args[@]}" -v "$line" ); done < $whiteList
grep "${args[@]}" $origFile