Hi Rechell,
At 01:42 PM 1/8/2004, you wrote:
Thanks, Wendell, Vasu, Michael, and David for your helpfule responses.
You are very welcome.
Am I correct that:
1)not(Tag[normalize-space()]) will treat all blanks as if it were empty
text, whereas not(string(Tag)) will not?
Basically, but...
There's an important difference between
[a] Tag[normalize-space()] or Tag[string()]
(short for child::Tag[normalize-space(self::node())] and
child::Tag[string(self::node())] )
and
[b] string(Tag) or normalize-space(Tag)
(short for string(child::Tag) or normalize-space(child::Tag) ) ...
The [a] expressions test true if there exists a Tag child with a non-null
string value (string()) or a non-null normalized-string value
(normalize-space()).
The [b] expressions perform the same tests, except only on the first Tag
child of the context node: since the string() and normalize-space()
functions require a single node as operand, the first node in the set in
document order is used.
So, the answer is basically "yes", with the caveat that the two expressions
are also not alike in that crucial other respect.
normalize-space() collapses white space and trims leading and trailing
white space, so boolean(normalize-space(' ')) is false since
normalize-space(' ') = ''.
2)If I had multiple sets of a given element, not(Tag[normalize-space()])
will return true for the following xml document because of the second
Tag element, while not(string(Tag)) will return false since the first
tag element has text?
<Xml>
<Tag>somevalue</Tag>
<Tag></Tag>
<Tag>Another value</Tag>
</Xml>
Not quite. Tag[normalize-space()] will test true by virtue of the existence
of the first and third Tag elements (which have text content even when
trimmed), so not(Tag[normalize-space()]) tests false.
You are correct, however, that not(string(Tag)) will test false since the
first Tag child has text content besides whitespace.
If you wanted to test whether there exists a Tag child without text
content, that would be test="Tag[not(normalize-space())]", which would test
true in the above example because the second node passes the test in the
predicate, therefore the node set returned is not empty.
Remember a node-set casts to Boolean "true" if there exists a node in the
set, false if not.
I hope this helps--
Wendell
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