Shell Programming and Scripting

Hello All,


I am trying to write a script to search in my current directory to look for all files that end with HTML and look for any HTML tags that are in upper case. for example if I were to grep test.html and test.html has a tag <P> instead of <p> then it would print the file name. This is what I have:


#!/bin/sh

for x in *.html

do
echo $x | grep \<[A-Z]+\>
if [ $? -ge "1" ]
then
echo $x
fi
done




thank you in advance.

Hi Raw,

Try this one..hope this should work..

#!/bin/sh
for x in *.html
do
echo $x | grep -l "<[A-Z]>"
if [ $? -ge "0" ]
then
echo $x
fi
done

oR

You can use like ..

grep -l "<[A-Z]>" *.html

this will print all the file name which has upper case tab inside...

Thanks
Sha

Hello Shahul,

Thank you for your response and help.

With the edited changes it does not work.

Lets say I have a directory of:

uppercase.sh
test_0.html --- File has <P> capital tag
test_1.html --- File has no capital tags
test_2.html --- File has no capital tags

The output result is as follows:

test_0.html
test_1.html
test_2.html

Expected Result: ( I could be wrong)

test_0.html test_0.html
test_1.html
test_2.html

Please try now and let me know

#!/bin/sh
for x in *.html
do
echo $x | grep "<[A-Z]>"
if [ $? -eq "0" ]
then
echo $x
fi
done

Thanks
Sha

I think you mean

cat $x | grep "<[A-Z]>"
not
echo $x | grep "<[A-Z]>"

Thank you again . Same result

however: grep -l "<[A-Z]>" *.html
works flawlessly

Not sure why:
#!/bin/sh

for x in *.html

do
echo $x | grep "<[A-Z]>"
if [ $? -ge "0" ]
then
echo $x
fi
done


Doesn't work

Try this

grep -l "\<[A-Z]\>" *.html