-1
任何人都可以請幫助我對每個不同的packnumber的序列號的邏輯如下。 輸入:XSLT 1.0中循環序列號的條件
<Move-Afile>
<Afile>
<Item>
<PackNumber>1234</PackNumber>
</Item>
<Item>
<PackNumber>1234</PackNumber>
</Item>
<Item>
<PackNumber>1234</PackNumber>
</Item>
<Item>
<PackNumber>126</PackNumber><!-- different PackNumber counter start from 1 -->
</Item>
<Item>
<PackNumber>126</PackNumber>
</Item>
<Item>
<PackNumber>135</PackNumber><!-- different PackNumber counter start from 1 -->
</Item>
<Item>
<PackNumber>135</PackNumber>
</Item>
</Afile>
</Move-Afile>
<Item>
是無界的element.for每個Item PackNumber是present.Here有像對於每個唯一<PackNumber>
計數器應從1開始並遞增得到像下面輸出的條件。
輸出:
<?xml version="1.0"?>
<A>
<target>
<counter>1</counter>
<PNumber>1234</PNumber>
<counter>2</counter>
<PNumber>1234</PNumber>
<counter>3</counter>
<PNumber>1234</PNumber>
<counter>1</counter><!-- different PackNumber counter start from 1 -->
<PNumber>126</PNumber>
<counter>2</counter>
<PNumber>126</PNumber>
<counter>1</counter><!-- different PackNumber counter start from 1 -->
<PNumber>135</PNumber>
<counter>2</counter>
<PNumber>135</PNumber>
</target>
</A>
這是工作fine.Thanks很多關於你的幫助。 – sum