Table.PositionOf

语法

Table.PositionOf(table as table, row as record, optional occurrence as any, optional equationCriteria as any) as any

关于

返回 row 在指定 table 中第一个实例的位置。 如果未找到匹配项，则返回 -1。

• table：输入表。
• row：表中要查找其位置的行。
• occurrence：[可选] 指定要返回的行的匹配项。
• equationCriteria：[可选] 控制表行之间的比较。

示例 1

查找 [a = 2, b = 4] 在表 ({[a = 2, b = 4], [a = 6, b = 8], [a = 2, b = 4], [a = 1, b = 4]}) 中第一个实例的位置。

使用情况

Table.PositionOf(
    Table.FromRecords({
        [a = 2, b = 4],
        [a = 1, b = 4],
        [a = 2, b = 4],
        [a = 1, b = 4]
    }),
    [a = 2, b = 4]
)

输出

0

示例 2

查找 [a = 2, b = 4] 在表 ({[a = 2, b = 4], [a = 6, b = 8], [a = 2, b = 4], [a = 1, b = 4]}) 中第二个实例的位置。

使用情况

Table.PositionOf(
    Table.FromRecords({
        [a = 2, b = 4],
        [a = 1, b = 4],
        [a = 2, b = 4],
        [a = 1, b = 4]
    }),
    [a = 2, b = 4],
    1
)

输出

2

示例 3

查找 [a = 2, b = 4] 在表 ({[a = 2, b = 4], [a = 6, b = 8], [a = 2, b = 4], [a = 1, b = 4]}) 中所有实例的位置。

使用情况

Table.PositionOf(
    Table.FromRecords({
        [a = 2, b = 4],
        [a = 1, b = 4],
        [a = 2, b = 4],
        [a = 1, b = 4]
    }),
    [a = 2, b = 4],
    Occurrence.All
)

输出

{0, 2}