![]() First click on the automation option on the top right corner of the screen. ![]() Now, let's implement the solution step by step. To make the example simpler, I will be keeping two statuses only: open and closed. That means, whatever the last record status is, should be for all the related discussion logs entries and projects as well. Now I would like to update the first entry status to be 'Closed' as well as the Project status to be 'Closed'. Let's assume that in the last discussion, I changed the status to 'Closed'. For example, in our case, we have two logs for one project open in our discussion logs table. The answer to the questions is, the status column is our trigger and we want to update all old statuses and also update the status of the project. If you have noticed, we have Status as a column in the project and discussion logs tables. Here we store discussion logs with clients.įor current scenarios what is our trigger and what do we want to update? The image below is of the Discussion logs table which is connected to the project's table. The image below is from the Project's table where we store Project's information and it is linked with the client's table and discussion logs table. The image below is of a Client's table where we store client's information and is linked with a project table. How we can update a previous discussion log's status and project status based on the last updated status for the same project when a certain condition is met? Basically, how do we update multiple record values with automation when a certain condition is met?Ĭurrent tables in airtable base. It's difficult to keep track of statuses for discussion logs and also for projects. Each time I the contact client, I open a new log in the discussion logs table. I might possibly contact the client more than once for the same or a new reason. At this point in time, I will create an entry in the discussion logs table which will be linked to the project's table. Now during the project development, I might require some information or want to discuss something with the client. First, I will create an entry in the client table and then one in the project table as well, linked to the client table. ![]() Let's assume I have a client and that client has a project. Each table is connected or linked with the other tables. Let's understand what we are going to do in this example. Understanding the problem that we will solve
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