我有一个包含Employees Name的
XML文件和由他们完成的Job.
XML文件的结构是 –
XML文件的结构是 –
<Employee>AAA@A#B#C#D</Employee> <Employee>BBB@A#B#C#D</Employee> <Employee>CCC@A#B#C#D</Employee> <Employee>DDD@A#B#C#D</Employee>
有成千上万的记录,我必须改变结构 –
<Employee> <Name>AAA</Name> <Jobs> <Job>A</Job> <Job>B</Job> <Job>C</Job> <Job>D</Job> </Jobs> </Employee>
如何在BaseX中使用XQuery完成此操作？
解决方法
3个XQuery函数,substring-before,substring-after和tokenize用于获取
所需的输出.
所需的输出.
substring-before用于获取Name.
类似地,substring-after用于获取Job部分.
然后,tokenize函数用于拆分作业.
let $data :=
<E>
<Employee>AAA@A#B#C#D</Employee>
<Employee>BBB@A#B#C#D</Employee>
<Employee>CCC@A#B#C#D</Employee>
<Employee>DDD@A#B#C#D</Employee>
</E>
for $x in $data/Employee
return
<Employee>
{<Name>{substring-before($x,"@")}</Name>}
{<Jobs>{
for $tag in tokenize(substring-after($x,"@"),'#')
return
<Job>{$tag}</Job>
}</Jobs>
}</Employee>
HTH …
