Table.PositionOfAny
Syntax
Table.PositionOfAny(table as table, rows as list, optional occurrence as nullable number, optional equationCriteria as any) as any
About
Returns the row(s) position(s) from the table of the first occurrence of the list of rows. Returns -1 if no occurrence is found.
table: The input table.rows: The list of rows in the table to find the positions of.occurrence: [Optional] Specifies which occurrences of the row to return.- `equationCriteria: [Optional] Controls the comparison between the table rows.
Example 1
Find the position of the first occurrence of [a = 2, b = 4] or [a = 6, b = 8] in the table ({[a = 2, b = 4], [a = 6, b = 8], [a = 2, b = 4], [a = 1, b = 4]}).
Usage
Table.PositionOfAny(
Table.FromRecords({
[a = 2, b = 4],
[a = 1, b = 4],
[a = 2, b = 4],
[a = 1, b = 4]
}),
{
[a = 2, b = 4],
[a = 6, b = 8]
}
)
Output
0
Example 2
Find the position of all the occurrences of [a = 2, b = 4] or [a = 6, b = 8] in the table ({[a = 2, b = 4], [a = 6, b = 8], [a = 2, b = 4], [a = 1, b = 4]}.
Usage
Table.PositionOfAny(
Table.FromRecords({
[a = 2, b = 4],
[a = 6, b = 8],
[a = 2, b = 4],
[a = 1, b = 4]
}),
{
[a = 2, b = 4],
[a = 6, b = 8]
},
Occurrence.All
)
Output
{0, 1, 2}