Shell Script with following awk command pls help

Shell Programming and Scripting
1

Hi

I want to create a shell script with the following awk command & also get the filenames in output.

awk '/<catetcsecuretty0>/ {p=1} /<catvarlogmessages0>/ {p=0} p' *.xml

As there will be multiple outputs related to many xml files I cannot identify which output belongs to which file

can anyone help?

2

Hello sharp488,

Welcome to forums, could you please try following and let me know if this helps.

for file in *.xml
do
    echo "File name which is processing currently: " $file
    awk '/<catetcsecuretty0>/ {p=1} /<catvarlogmessages0>/ {p=0} p' $file 
    echo "******************************"
    echo ""
done

Thanks,
R. Singh

3

Its giving me error

$ ./roottelnet.sh
./roottelnet.sh: line 2: syntax error near unexpected token `$'do\r''
'/roottelnet.sh: line 2: `do
4

Hello sharp488,

Seems you may have carriage characters in your script which you are using, could you please use following command to confirm the same.

cat -v  Input_file

Where Input_file is your script name. If you are seeing control M characters there like as an example as follows.

again test test1 test1 test2^M
test test test1 test chumma.^M

If you find above control M characters into your script then you could try following to remove them.

awk '{gsub(/\r/,X,$0);print}' roottelnet.sh > tmp_roottelnet.sh
mv tmp_roottelnet.sh  roottelnet.sh

Then you could try to run your script. Kindly do let us know if you have any queries on same.

Thanks,
R. Singh

5

Try (untested):

awk '/<catetcsecuretty0>/ {p=1; print FILENAME ":"} /<catvarlogmessages0>/ {p=0} p' *.xml
6

i did a vi on the file its showing

for file in *.xml
do
    echo "File name which is processing currently: " $file
    awk '/<catetcsecuretty0>/ {p=1} /<catvarlogmessages0>/ {p=0} p' $file
    echo "******************************"
    echo ""
done
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---------- Post updated at 07:47 PM ---------- Previous update was at 07:41 PM ----------

Thank You.
Its working :b: