如何获得正确的输出格式
how to get the correct output format
xquery version "1.0-ml";
let $letter := ("A","B")
let $number := (1,2,3)
for $i in $letter
for $j in $number
return ($i,$j)
我怎样才能得到输出格式呢。谢谢
A
1
2
3
B
1
2
3
尝试:
for $i in $letter
return ($i,$number)
或
for $i in $letter
return ($i,
for $j in $number
return $j)
输出(除了格式化):
A
1
2
3
B
1
2
3
如果您打算在输出中使用缩进和回车 returns:
let $letter := ("A","B")
let $number := (1,2,3)
for $i in $letter
let $line := (
$i,
for $j in $number
return " "||$j
)
string-join($line, " ")
更简洁,使用 simple map operator !
而不是 for 循环:
let $letter := ("A","B")
let $number := (1,2,3)
return
string-join($letter ! (., $number ! (" "||.)), " ")
xquery version "1.0-ml";
let $letter := ("A","B")
let $number := (1,2,3)
for $i in $letter
for $j in $number
return ($i,$j)
我怎样才能得到输出格式呢。谢谢
A
1
2
3
B
1
2
3
尝试:
for $i in $letter
return ($i,$number)
或
for $i in $letter
return ($i,
for $j in $number
return $j)
输出(除了格式化):
A
1
2
3
B
1
2
3
如果您打算在输出中使用缩进和回车 returns:
let $letter := ("A","B")
let $number := (1,2,3)
for $i in $letter
let $line := (
$i,
for $j in $number
return " "||$j
)
string-join($line, " ")
更简洁,使用 simple map operator !
而不是 for 循环:
let $letter := ("A","B")
let $number := (1,2,3)
return
string-join($letter ! (., $number ! (" "||.)), " ")