What is the most elegant way to remove a path from the $PATH variable in Bash?

A minute with awk:

# Strip all paths with SDE in them.
#
export PATH=`echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}'`

Edit: It response to comments below:

$ export a="/a/b/c/d/e:/a/b/c/d/g/k/i:/a/b/c/d/f:/a/b/c/g:/a/b/c/d/g/i"
$ echo ${a}
/a/b/c/d/e:/a/b/c/d/f:/a/b/c/g:/a/b/c/d/g/i

## Remove multiple (any directory with a: all of them)
$ echo ${a} | awk -v RS=: -v ORS=: '/a/ {next} {print}'
## Works fine all removed

## Remove multiple including last two: (any directory with g)
$ echo ${a} | awk -v RS=: -v ORS=: '/g/ {next} {print}'
/a/b/c/d/e:/a/b/c/d/f:
## Works fine: Again!

Edit in response to security problem: (that is not relevant to the question)

export PATH=$(echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}' | sed 's/:*$//')

This removes any trailing colons left by deleting the last entries, which would effectively add . to your path.

My dirty hack:

echo ${PATH} > t1
vi t1
export PATH=$(cat t1)