How to use For-loop in xquery?

Question

I'm having an url-pattern like "http://www.asdf.com/pages=1", now I want to operate for loop for this url-pattern upto 1547 i.e., ("http://www.asdf.com/pages=1547")

Christian Grün · Accepted Answer · 2012-05-16 11:49:44Z

5

This might do what you need:

for $i in (1 to 10, 1547, 1548)
return concat("http://www.asdf.com/pages=", $i)

answered May 16, 2012 at 11:49

Christian Grün

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6 Comments

Siddhu Over a year ago

Thanks and more query., if the url pattern is like this "asdf.com.?sid=1545148454855&page=1?sid=ABGDBSDGH58569ASDE" , now i have to change this thing?

Jens Erat Over a year ago

What about return concat("asdf.com.?sid=1545148454855&page=", $i, "?sid=ABGDBSDGH58569ASDE&q…")?

Siddhu Over a year ago

One more query adn how to replace "?" with "" [Null] in xquery.

Leo Wörteler Over a year ago

You'll have to escape the ampersand characters & as & too, so it should be more like concat(http://www.asdf.com./?sid=1545148454855&page=', $i, '?sid=ABGDBSDGH58569ASDE&q...').

Leo Wörteler Over a year ago

for replacing substrings, please look at fn:replace($string, $pattern, $replacement).

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Jens Erat · Accepted Answer · 2012-05-16 11:53:52Z

3

for $i in (1 to 1547)
return concat("http://www.asdf.com/pages=",$i)

answered May 16, 2012 at 11:53

Jens Erat

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