Attempting to Assign to an Object That is not an L-Value

[#1]

Hi, I'm getting this error.

"attempting to assign to an object that is not an L-value{1} in access or evaluation of 'Assign' , Row() & dt:sequence[Row() + 1] = Row() + 1"

Not quite sure what L-value means. Think it might be to do with the fact that sequence is an inbulit function. Any help would be appreciated.

[#2]

The Assign() function takes two arguments, the left and the right side as they appear using the operator. An example of an valid assignment is x = Log(5). An example of an invalid assignment is Log(5) = x. 

[#3]

Okay thanks, I changed the order of the statement to: Row() = dt:sequenceSteps[Row()];

but I'm still getting the same error. Should I assign Row() to a variable? Still don't quite understand what the issue is.

[#4]

an L-value is a value that can appear on the left-hand side of an assignment statement. Here's an example that might be similar to what you are trying to do:

dt = New Table( "Untitled",
  Add Rows( 5 ),
  New Column( "sequenceSteps",
    Numeric,
    "Continuous",
    Format( "Best", 12 ),
    Set Values( [5, 3, 1, 2, 4] )
  )
);

Row() = 1;

For( i = 1, i <= 20, i++,
  Row() = dt:sequenceSteps[Row()];
  dt << clearSelect;
  dt << selectrows( Row() );
  Print( Row() );
  Wait( 1 );
);

5
4
2
3
1
5
4
2
3
1
5
4
2
3
1
5
4
2
3
1

[#5]

I have a very simple example of this:  I have a table with a column named "Electrolyte Solution".  When I did this:

for each row(
	:Electrolye Solution = "Lot 5";
);

I got the error "attempting to assign to an object that is not an L-value".  Of course, I had left the "t" out of the column name.  You know what would have been super helpful?  An error message that said:

"Electrolye Solution" is not a valid column name.

Then, I could have fixed my script in 5 seconds instead of 20 minutes.

Thanks!

Eric