Autonumber function

[Anonymous]

Guys,

I have 15 columns in a table, but I want to generate keys using autonumber function for all the combinations of that 15 columns. Is it possible to do by script wise?

[sunny_talwar]

Should be possible...

AutoNumber(Field1&Field2&Field3&.....&Field15) as Key

[Kushal_Chawda]

Autonumber (key1&'|'&Key2&'|'&.........)

[Anonymous]

No, I want all the combinations. I mean

for example, for three columns itself I want these much of combinations,

AutoNumber(Field1&Field2&Field3) as Key

AutoNumber(Field1&Field3&Field2) as Key1

AutoNumber(Field3&Field1&Field2) as Key2

AutoNumber(Field2&Field1&Field3) as Key3

.

.

.

So, it goes on. Likewise, is it possible to generate key for all the combinations of 15 columns.

[sunny_talwar]

I think so... the only change I would do is this

AutoNumber(Field1&Field2&Field3, 'Key') as Key

AutoNumber(Field1&Field3&Field2, 'Key1') as Key1

AutoNumber(Field3&Field1&Field2, 'Key2') as Key2

AutoNumber(Field2&Field1&Field3, 'Key3') as Key3

.

.

.

[Anonymous]

Thank you. But do I have to write all the combinations of 15 columns or can I use loop function for that?

[Kushal_Chawda]

that menas 15 * 15 = 225 combinations?

[Anonymous]

yes exactly..

[sunny_talwar]

I guess you might be able to loop... let me see if there is an option

[sunny_talwar]

This is for 3 fields, but can be expanded to 15 fields

Table:

LOAD *,

RowNo() as UK;

LOAD * Inline [

ABC, DEF, GHI

1, 2, 3

92, 32, 54

3, 4, 2

];

FieldNames:

LOAD * Inline [

FieldName

ABC

DEF

GHI

];

FinalTable:

LOAD UK

Resident Table;

FOR i = 1 to FieldValueCount('FieldName')

LET vField1 = FieldValue('FieldName', $(i));

LET vField2 = FieldValue('FieldName', If($(i)+1 > 3, FieldValueCount('FieldName') - $(i) + 1, $(i)+1));

LET vField3 = FieldValue('FieldName', If($(i)+2 > 3, FieldValueCount('FieldName') - $(i) + 2, $(i)+2));

Left Join (FinalTable)

LOAD UK,

AutoNumber([$(vField1)]&'|'&[$(vField2)]&'|'&[$(vField3)], 'Key'&$(i)) as [Key$(i)]

Resident Table;

NEXT i