How to use let statement scalar output in table function

Question

Hi,

I am using let statement where it gives tablename as a result based on some condition.

And I am trying use that value in table() function - table(tablename).

It gives me below error though I have my converted my let statement result to string using tostring and tried with toscalar function also.

table(): argument #1 must be a non-empty string literal.

Please help here to understand about table() function and how to solve this error.

Accepted Answer

Hi @Sangeetha Priya S ,

I have the same error when I use a dynamic variable for the table name, for example:

let MyTable = () {  iff( 1==1, "logsRaw","Telemetry") };
//
table(MyTable)
| take 50

When changing it to a constant variable, it will work:

let MyTable = () { "Telemetry" };
//
table(MyTable)
| take 50

According to the documentation:
A parameter, which isn't a scalar constant string, can't be passed as a parameter to the table() function.

The workaround:

let T1 = print x=1;
let T2 = print x=2;
let _choose = (_selector:string)
{
    union
    (T1 | where _selector == 'T1'),
    (T2 | where _selector == 'T2')
};
_choose('T2')

More info:
https://video2.skills-academy.com/en-us/azure/data-explorer/kusto/query/tablefunction?pivots=azuredataexplorer#use-table-with-non-constant-parameter

Kind Regards,

Wilko

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