Rendering different values based on count function

Question

I have a constant that stores three different values. My table has a list of items, where the constant can be either pass, fail, or na. 
 
 I want to have three different sections on a new interface where it will show one section as all items that are in value 1 from the constant. All items that are in value 2 and all that are in value 3. I am not sure exactly how to go about this?

Omkar Ambardekar · Accepted Answer

Hi Jorge, I am not sure how the data in your tables is structured. But you could take a look at the sample code I have posted below and build your interface similarly: 
 with( local!difficulty: {"HIGH", "MEDIUM", "LOW"}, local!randomItems: { {task: "Build a pyramid", difficulty: "HIGH"}, {task: "Run a marathon", difficulty: "MEDIUM"}, {task: "Space travel", difficulty: "HIGH"}, {task: "Make coffee", difficulty: "LOW"} }, a!forEach( local!difficulty, a!richTextDisplayField( value: { with( local!indices: wherecontains(tostring(fv!item), touniformstring(local!randomItems.difficulty)), { a!richTextItem( text: length(local!indices)&" "&fv!item&" items:", ), char(10), a!richTextBulletedList( items: a!forEach(local!indices, a!richTextItem( text: index(index(local!randomItems, fv!item, {}), "task", {}) ) ) ) } ) } ) ) )

Omkar Ambardekar · Answer

You would just be replacing local!difficulty with your status constant (or a db query fetching ref data) and local!randomItems with a queryEntity that returns the items from the db.

Peter Lewis · Answer

It might be even easier if you have the data coming from a database. You can set up your query to aggregate on the field you want to compare to the constant. Then, you can use wherecontains to find the count of the associated data. Here's an example: a!localVariables( local!query:a!queryEntity( entity: cons!, query: a!query( pagingInfo: a!pagingInfo( startIndex: 1, batchSize: -1 ), aggregation: a!queryAggregation( aggregationColumns: { a!queryAggregationColumn(field: "", isGrouping: true), a!queryAggregationColumn(field: "id", aggregationFunction: "COUNT", alias: "count") } ) ) ), a!textField( label: "Count of Constant Value 1", readOnly: true, value: index( local!query.data.count, wherecontains(cons![1], touniformstring(local!query.data.)), 0 ) ) ) In this example, I group by the field in the database to create the correct count. Then, I find whether that value exists in the output using wherecontains(). I can then find the associated count using the index() function.

Rendering different values based on count function

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