超鏈接以交叉引用兩個表

Question

如何將這些超鏈接組合成一個可以交叉引用的引用或引文？

=IFERROR(HYPERLINK("#" & CELL("address", INDEX(INDIRECT("References[Reference]"), MATCH([@Resource], INDIRECT("References[Resource]"), 0))), "References"), "")

=IFERROR(HYPERLINK("#" & CELL("address", INDEX(INDIRECT("Citations[Citation]"), MATCH([@Resource], INDIRECT("Citations[Resource]"), 0))), "Citations"), "")

Answer 1

在資源表中，使用此鏈接到參考表。

=HYPERLINK(TEXT(ADDRESS(MATCH([@Resource], References[Reference], 0)+ROW(References[#Headers]), COLUMN(References[Reference]), 4, 1, CELL("filename", References[#All])), "\#@"), "References")

從資源表鏈接到引文表。

=HYPERLINK(TEXT(ADDRESS(MATCH([@Resource], Citations[Citation], 0)+ROW(Citations[#Headers]), COLUMN(Citations[Citation]), 4, 1, CELL("filename", Citations[#All])), "\#@"), "Citations")

結合。

=iferror(iferror(HYPERLINK(TEXT(ADDRESS(MATCH([@Resource], References[Reference], 0)+ROW(References[#Headers]), COLUMN(References[Reference]), 4, 1, CELL("filename", References[#All])), "\#@"), "References"), HYPERLINK(TEXT(ADDRESS(MATCH([@Resource], Citations[Citation], 0)+ROW(Citations[#Headers]), COLUMN(Citations[Citation]), 4, 1, CELL("filename", Citations[#All])), "\#@"), "Citations")), TEXT(,))

如果表位於不同的工作表上，則這些地址構造的確包含工作表名稱。