在Xpath查询中使用联合静态值

Question

如何使用Xpath查询合并静态值

以下是我的XML数据

<breakfast_menu>
<food>
    <name>Belgian Waffles</name>
    <price>$5.95</price>
    <description>Two of our famous Belgian Waffles with plenty of real maple syrup</description>
    <calories>650</calories>
</food>
<food>
    <name>Strawberry Belgian Waffles</name>
    <price>$7.95</price>
    <description>Light Belgian waffles covered with strawberries and whipped cream</description>
    <calories>900</calories>
</food>
</breakfast_menu>

通过下面的Xpath查询，我可以实现Node Name的值

1) breakfast_menu/food[not(name=preceding-sibling::food/name)]/name

输出：

Belgian Waffles
Strawberry Belgian Waffles

我的意图是将值“ ALL”附加到输出。

Xpath查询下面是我的工作，使用Union Operator实现

1) breakfast_menu/food[not(name=preceding-sibling::food/name)]/name | //food[name="ALL"]

Xpath的认沽权除外：

ALL
Belgian Waffles
Strawberry Belgian Waffles

Answer 1

假设您确实拥有XPath 2.0：

string-join(('ALL', distinct-values(breakfast_menu/food/name)), '\n')

将产生带有行的单个字符串

ALL
Belgian Waffles
Strawberry Belgian Waffles

如果您想要一个字符串序列，则只需使用

'ALL', distinct-values(breakfast_menu/food/name)