Hi,
I want to insert a formula into a label that when a Building name is selected from a dropdown, the formula will look up the building from a Sharepoint list and return the value in a column determined by another separate label.
Below is my Sharepoint List Data.
Building Name | 2017 | 2018 | 2019 |
Building 1 | 100 | 400 | 700 |
Building 2 | 200 | 500 | 800 |
Building 3 | 300 | 600 | 900 |
I have set up the dropdown to show the options Building 1, 2 and 3, and the below formula works fine
ComboBox1.Selected.'2018'
However, I want to change “’2018’” to the value in Label 1, which for now say is 2019. I have tried the below formula, however this causes an error.
ComboBox1.Selected.'Label1'
Please can someone advise me what to change my formula to instead of “’Label1’”.
Example below
Data
Drop down value = Building 2
Label = 2019
Answer
Label with Formula = 800
Thanks
Solved! Go to Solution.
Hey @SG1990
Reference of column name is not possible using a label directly, in such a case, you can use:
Switch(Label1.Text, "2018", ComboBox1.Selected.'2018', "2019", ComboBox1.Selected.'2019', "2020", ComboBox1.Selected.'2020')
Hope this Helps!
If this reply has answered your question or solved your issue, please mark this question as answered. Answered questions helps users in the future who may have the same issue or question quickly find a resolution via search. If you liked my response, please consider giving it a thumbs up. THANKS!
Hey @SG1990
Reference of column name is not possible using a label directly, in such a case, you can use:
Switch(Label1.Text, "2018", ComboBox1.Selected.'2018', "2019", ComboBox1.Selected.'2019', "2020", ComboBox1.Selected.'2020')
Hope this Helps!
If this reply has answered your question or solved your issue, please mark this question as answered. Answered questions helps users in the future who may have the same issue or question quickly find a resolution via search. If you liked my response, please consider giving it a thumbs up. THANKS!
Thanks for your comment.
Unfortunately this just causes an error for me. Additionally I was hoping someone might have a different solution to this as in my example I have only 3 columns, however what I am working on has over 100 columns and using that formula would be very long winded for 100+ columns.
Thanks again for your help though.
@yashag2255 , apologies but the formula did work, I must have made a mistake first time. Even though this is more long winded than I hoped, this method does work. Thanks.
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