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How you can create a column for counting number of entries in each cell of a column? Is there a formula for that?

  • September 22, 2021
  • 2 replies
  • 54 views

I am trying to count the number of values entered in each cell of a column, just so that I can perform certain automation. like for example, if the column name is 'Colleagues" and the first row in this column contains three names of colleagues. Then how can I automatically detect that there are 3 names/values entered in this cell?

Best answer by Ankit_Srivastav

Hi @Ankit_Srivastava ,

Welcome to Airtable community!

Do the values in the cell come from a "linked to " field? If yes you can create a new field with the type of “count” and this counts the values which are in the field.

If it doesn’t work please don’t hesitate to post back again :slightly_smiling_face:

Thanks


I have got the answer,
I tried out this one for counting number of entries(and yes it was linked using lookup for colleagues) ::
IF(
LEN({Colleagues})=0,0,
LEN({Colleagues})-LEN(SUBSTITUTE({
Colleagues},",",""))+1)

Thanks for the reply.

2 replies

Dimitris_Goudis
Forum|alt.badge.img+20

Hi @Ankit_Srivastava ,

Welcome to Airtable community!

Do the values in the cell come from a "linked to " field? If yes you can create a new field with the type of “count” and this counts the values which are in the field.

If it doesn’t work please don’t hesitate to post back again :slightly_smiling_face:

Thanks


  • Author
  • New Participant
  • Answer
  • September 23, 2021

Hi @Ankit_Srivastava ,

Welcome to Airtable community!

Do the values in the cell come from a "linked to " field? If yes you can create a new field with the type of “count” and this counts the values which are in the field.

If it doesn’t work please don’t hesitate to post back again :slightly_smiling_face:

Thanks


I have got the answer,
I tried out this one for counting number of entries(and yes it was linked using lookup for colleagues) ::
IF(
LEN({Colleagues})=0,0,
LEN({Colleagues})-LEN(SUBSTITUTE({
Colleagues},",",""))+1)

Thanks for the reply.