Get the index of a Row by a value

Hi,
I want get the index of a row by giving a value so i used the function :
row(nameOrIndex1, nameOrIndex2…, caseSensitive=True)
op(‘table1’).row(‘item2’)
that return me a list, but i can’t extract in the list the index of the row :frowning:
The list look at this :
[type:Cell cell:(1, 0) owner:/project1/table1 value:item2, type:Cell cell:(1, 1) owner:/project1/table1 value:, type:Cell cell:(1, 2) owner:/project1/table1 value:, type:Cell cell:(1, 3) owner:/project1/table1 value:]
Image Code

So how i get the value : 1 ?
Thx for the help

hey anodin, see the docs about the row method:

It says it returns a list of cells from the row matching the name/index.

So to get the row index of the first cell of that list of cells:
op('table1').row('item2')[0].row

Oh Thank you Nettoyeur,

I didn’t understand that i need a .row at this end…
Thx for the exemple :raised_hands: