How can i convert string into XPATH, below is the code
let $ti := "item/title"
let $tiValue := "Welcome to America"
return db:open('test')/*[ $tiValue = $ti]/base-uri()
Here is one way to solve it:
let $ti := "item/title"
let $tiValue := "Welcome to America"
let $input := db:open('test')
let $steps := tokenize($ti, '/')
let $process-step := function($input, $step) { $input/*[name() = $step] }
let $output := fold-left($input, $steps, $process-step)
let $test := $output[. = $tiValue]
return $test/base-uri()
The path string is split into single steps (item, title). With fold-left, all child nodes of the current input (initially db:open('test')) will be matched against the current step (initially, item). The result will be used as new input and matched against the next step (title), and so on. Finally, only those nodes with $tiValue as text value will be returned.
Your question is very unclear - the basic problem is that you've shown us some code that doesn't do what you want, and you're asking us to work out what you want by guessing what was going on in your head when you wrote the incorrect code.
I suspect -- I may be wrong -- that you were hoping this might somehow give you the result of
db:open('test')/*[item/title = $ti]/base-uri()
and presumably $ti might hold different path expressions on different occasions.
XQuery 3.0/3.1 doesn't have any standard way to evaluate an XPath expression supplied dynamically as a string (unless you count the rather devious approach of using fn:transform() to invoke an XSLT transformation that uses the xsl:evaluate instruction).
BaseX however has an query:eval() function that will do the job for you. See https://docs.basex.org/wiki/XQuery_Module
XQuery
Input: (1,2,3,4,5,6,7,14,15,16,17,24,25,26,27,28)
Output: (1,7,14,17,24,28)
I tried to remove consecutive numbers from the input sequence using the XQuery functions but failed doing so
xquery version "1.0" encoding "utf-8";
declare namespace ns1="http://www.somenamespace.org/types";
declare variable $request as xs:integer* external;
declare function local:func($reqSequence as xs:integer*) as xs:integer* {
let $nonRepeatSeq := for $count in (1 to count($reqSequence)) return
if ($reqSequence[$count+1] - $reqSequence) then
remove($reqSequence,$count+1)
else ()
return
$nonRepeatSeq
};
local:func((1,2,3,4,5,6,7,14,15,16,17,24,25,26,27,28))
Please suggest how to do so in XQuery functional language.
Two simple ways to do this in XQuery. Both rely on being able to assign the sequence of values to a variable, so that we can look at pairs of individual members of it when we need to.
First, just iterate over the values and select (a) the first value, (b) any value which is not one greater than its predecessor, and (c) any value which is not one less than its successor. [OP points out that the last value also needs to be included; left as an exercise for the reader. Or see Michael Kay's answer, which provides a terser formulation of the filter; DeMorgan's Law strikes again!]
let $vseq := (1,2,3,4,5,6,7,14,15,16,17,24,25,26,27,28)
for $v at $pos in $vseq
return if ($pos eq 1
or $vseq[$pos - 1] ne $v - 1
or $vseq[$pos + 1] ne $v + 1)
then $v
else ()
Or, second, do roughly the same thing in a filter expression:
let $vseq := (1,2,3,4,5,6,7,14,15,16,17,24,25,26,27,28)
return $vseq[
for $i in position() return
$i eq 1
or . ne $vseq[$i - 1] + 1
or . ne $vseq[$i + 1] - 1]
The primary difference between these two ways of performing the calculation and your non-working attempt is that they don't say anything about changing or modifying the sequence; they simply specify a new sequence. By using a filter expression, the second formulation makes explicit that the result will be a subsequence of $vseq; the for expression makes no such guarantee in general (although because for each value it returns either the empty sequence or the value itself, we can see that here too the result will be a subsequence: a copy of $vseq from which some values have been omitted.
Many programmers find it difficult to stop thinking in terms of assignment to variables or modification of data structures, but its worth some effort.
[Addendum] I may be overlooking something, but I don't see a way to express this calculation in pure XPath 2.0, since XPath 2.0 seems not to have any mechanism that can bind a variable like $vseq to a non-singleton sequence of values. (XPath 3.0 has let expressions, so it's not a challenge there. The second formulation above is itself pure XPath 3.0.)
In XSLT this can be done as:
<xsl:for-each-group select="$in" group-adjacent=". - position()">
<xsl:sequence select="current-group()[1], current-group()[last()]"/>
</xsl:for-each-group>
In XQuery 3.0 you can do it with tumbling windows, but I'm too lazy to work out the detail.
An XPath 2.0 solution (assuming the input sequence is in $in) is:
for $i in 1 to count($in)
return $in[$i][not(. eq $in[$i - 1]+1 and . eq $in[$i+1]-1)]
There are several logic and XQuery usage errors in your solution, but the main problem with it is that variables in XQuery are immutable, so you cannot reassign a value to one once assigned. Therefore, it's often easier to think about these types of problems in terms of recursive solutions:
declare function local:non-consec(
$prev as xs:integer?,
$rest as xs:integer*
) as xs:integer*
{
if (empty($rest)) then ()
else
let $curr := head($rest)
let $next := subsequence($rest, 2, 1)
return (
if ($prev eq $curr - 1 and $curr eq $next - 1)
then () (: This number is part of a consecutive sequence :)
else $curr,
local:non-consec(head($rest), tail($rest))
)
};
local:non-consec((), (1,2,3,4,5,6,7,14,15,16,17,24,25,26,27,28))
=>
1
7
14
17
24
28
For me when making a recursive method. I always need to spend a lot of time to do it, because I will make some test cases and to see whether my recursive case works and to draw a stack diagram. However, when I ask other about it, they just say that I need to believe myself it will work. How am I suppose to believe that if you don't know what is going on in the recursive case?
You define what is going on in the recursive case, just as you define the rest of the method. Imagine someone else wrote a method to do what the one you are writing does; you wouldn't have a problem calling that, would you? The only difference is that you are that method's author, and it just happens to be the one being written.
For example: I am writing the following method:
// Sort array a[i..j-1] in ascending order
method sort_array( a, i, j ) {
..
}
The base case is easy:
if ( i >= j-1 ) // there is at most one element to be sorted
return; // a[i..j-1] is already sorted
Now, when that isn't true, I could do the following:
else {
k = index_of_max( a, i, j );
swap( a, j-1, k );
At this point, I know that a[j-1] has the correct value, so I just need to sort what comes before it -- fortunately, I have a method to do just that:
sort_array( a, i, j-1 );
}
No leap of faith is required; I know that recursive call will work because I wrote the method to do just that.
I'm wondering whether in XQuery it is possible to access some elements in a variable from within the variable itself.
For instance, if you have a variable with several numbers and you want to sum them all up inside the variable itself. Can you do that with only one variable? Consider something like this:
let $my_variable :=
<my_variable_root>
<number>5</number>
<number>10</number>
<sum>{sum (??)}</sum>
</my_variable_root>
return $my_variable
Can you put some XPath expression inside sum() to access the value of the preceding number elements? I've tried $my_variable//number/number(text()), //number/number(text()), and preceding-sibling::number/number(text()) - but nothing worked for me.
You cannot do that. The variable is not created, till everything in it is constructed.
But you can have temporary variables in the variable
Like
let $my_variable :=
<my_variable_root>{
let $numbers := (
<number>5</number>,
<number>10</number>
)
return ($numbers, <sum>{sum ($numbers)}</sum>)
} </my_variable_root>
Or (XQuery 3):
let $my_variable :=
<my_variable_root>{
let $numbers := (5,10)
return (
$numbers ! <number>{.}</number>,
<sum>{sum ($numbers)}</sum>)
} </my_variable_root>
This is not possible, neither by using the variable name (it is not defined yet), nor using the preceding-sibling axis (no context item bound).
Construct the variable's contents in a flwor-expression instead:
let $my_variable :=
let $numbers := (
<number>5</number>,
<number>10</number>
)
return
<my_variable_root>
{ $numbers }
<sum>{ sum( $numbers) }</sum>
</my_variable_root>
return $my_variable
If you have similar patterns multiple times, consider writing a function; using XQuery Update might also be an alternative (but does not seem to be the most reasonable one to me, both in terms of readability and probably performance).
Im new on this project and am going to write, what i thought was a simple thing. A recursive function that writes nested xml elements in x levels (denoted by a variable). So far I have come up with this, but keeps getting a compile error. Please note that i have to generate new xml , not query existing xml:
xquery version "1.0";
declare function local:PrintTest($amount)
{
<test>
{
let $counter := 0
if ($counter <= $amount )
then local:PrintTest($counter)
else return
$counter := $counter +1
}
</test>
};
local:PrintPerson(3)
My error is:
File Untitled1.xquery: XQuery transformation failed
XQuery Execution Error!
Unexpected token - " ($counter <= $amount ) t"
I never understood xquery, and cant quite see why this is not working (is it just me or are there amazingly few resources on the Internet concerning XQuery?)
You have written this function in a procedural manner, XQuery is a functional language.
Each function body can only be a single expression; it looks like you are trying to write statements (which do not exist in XQuery).
Firstly, your let expression must be followed by a return keyword.
return is only used as part of a FLWOR expression, a function always evaluates to a value. As you have written it return is equivalent to /return and so will return a node called return.
The line $counter := $counter + 1 is not valid XQuery at all. You can only set a variable like this with a let expression, and in this case it would create a new variable called counter which replaced the old one, that would be in scope only in the return expression of the variable.
The correct way to do what you are trying to do is to reduce the value of $argument each time the function recurses, and stop when you hit 0.
declare function local:Test($amount)
{
if ($amount == 0)
then ()
else
<test>
{
local:Test($amount - 1)
}
</test>
};
local:Test(3)
Note that I have changed the name of the function to Test. The name "PrintTest" was misleading, as this implies that the function does something (namely, printing). The function in fact just returns a node, it does not do any printing. In a purely functional langauge (which XQuery is quite close to) a function never has any side effects, it merely returns a value (or in this case a node).
The line $counter := $counter + 1 is valid XQuery Scripting.