Function CASE projects one or multiple data columns to a new column. Starting from the first WHEN, this function checks a data row against the condition claimed in each WHEN, once a condition is met the projection result in the correspondent THEN will be returned, and the rest of WHEN and ELSE will be skipped.

When projecting multiple columns to a table, please make sure the values in the same table column have same data type; all keywords involved in this function are case insensitive.

Syntax 1:

case
  when <condition1> then <projection1>
  when <condition2> then <projection2>
  ...
  else <other>
end

Syntax 2:

case <expression>
  when <value1> then <projection1>
  when <value2> then <projection2>
  ...
  else <other>
end

• <projectionN> is the expression to compute when <conditionN> is met in Syntax 1, or when <expression> equals <valueN> in Syntax 2
• <other> is the expression to compute when none of the conditions is met; a 0 or empty string will be generated if ELSE is omitted
• Projecting to NODE/EDGE will be processed as projecting to the UUID of NODE/EDGE, while projecting to PATH will be processed as projecting to the row number of PATH

If multiple columns are non-homologous, Cartesian Product will be applied when the function is called in a WITH clause, or column length will be trimmed when called in a RETURN clause.

Constant

Example: divide @customer nodes into 4 age groups based on birth years: 1980s, 1990s, and 2000s, count the number of people in each age group

find().nodes({@customer}) as nodes
with CASE
when year(nodes.birthday) <=> [1980, 1989] then "1980s"
when year(nodes.birthday) <=> [1990, 1999] then "1990s"
when year(nodes.birthday) <=> [2000, 2009] then "2000s"
else "others" END as ageGroup
group by ageGroup
return ageGroup, count(ageGroup)

Example: abbreviate the example above

find().nodes({@customer}) as nodes
with CASE floor(year(nodes.birthday)/10)
when 198 then "1980s"
when 199 then "1990s"
when 200 then "2000s"
else "others" END as ageGroup
group by ageGroup
return ageGroup, count(ageGroup)

Function

Example: calculate the actual monthly paydays in 2022, with 15th each month as planned payday and payday will be put off to the nearest Monday if it is a non-workday.

uncollect ["2022-1-15 0:0:0","2022-2-15 0:0:0","2022-3-15 0:0:0","2022-4-15 0:0:0","2022-5-15 0:0:0","2022-6-15 0:0:0","2022-7-15 0:0:0","2022-8-15 0:0:0","2022-9-15 0:0:0","2022-10-15 0:0:0","2022-11-15 0:0:0","2022-12-15 0:0:0"] as plan
return CASE day_of_week(plan)
when 1 then date_add(plan, 1, "day")
when 7 then date_add(plan, 2, "day")
else date_add(plan, 0, "day")
END

Analysis: the first condition in case adds 1 day to a planned payday that happens to be a Sunday, the second condition adds 2 days to a planned payday on Saturday , ELSE keeps the planned payday as it is by adding 0 day. Note: ELSE can not be followed by plan, for plan's structure type is string, date_add() can only be filled with time, so to keep data structure type consistent plan should be transformed to a time format.

Alias

Example: find one 1-step path from each card CA001, CA002, and CA005, count each path as [<transferCardID>, <amount>,<receivercardID>]; if a path do not exist, count it as [<CardID>, 0, "none"]

find().nodes({_id in ["CA001","CA002","CA005"]}) as cards
optional n(cards).le({@transfer} as e).n({@card} as n).limit(1)
return CASE n
when 0 then [cards._id, 0, "none"]
else [cards._id, e.amount, n._id] END

Analysis: 1) when using optional, if one of the template query does not return result, then conduct a psyeudo-path 0-0-0; 2）use 'CASE' to find the psyeudo-path 0-0-0 and map it as [<CardID>, 0, "none"], map the rest of paths as a table [<TransferorCardID>,<amount>,<ReceiverCardID>], note that data type should be consistent in each column.