Hello,
Writting a small script to make mozilla show the results of HTML code
written in vi, when vi saves, the script automaticly changes mozilla
to show that HTML.
I need to know if mozilla is running, if not I need to call it first
... seemed simple
if ! ps ax | grep "mozilla-bin" &>/dev/null; then
/usr/local/mozilla/mozilla -P web &
.....
This worked the first time then I get
web@debian:/usr/local/myfiles/dave/debian/sh files$ ./HTMLgui
7508 pts/0 S 0:00 grep mozilla-bin
Ie its picking up on the grep process, finding "mozilla-bin", saying
yep its running, dont call mozilla & bang, script fails.
Whatever I grep for, grep will find in ps ax as grep xxxxxx, a bit
frustrating !!
Help
Dave
PS still a newbe BASH scripter, If there is a better way of doing
this, let me know.
#!/bin/sh
# HTMLgui ... enable Mozilla to provide a GUI for vi
currentdir=/usr/local/myfiles/dave/websites/current
oldhtml="mismatch"
if ! ps ax | grep "mozilla-bin"; then
/usr/local/mozilla/mozilla -P web &
sleep 3s
fi
echo
while true; do
html=$(ls -t1 $currentdir/*.html | head -n 1)
if [ $oldhtml = $html ]; then
sleep 1s
else
echo -e "HTMLgui ... changing to ${html##/*/}"
oldhtml=$html
/usr/local/mozilla/mozilla -P web -remote "openFile($html)"
fi
done