Arangodb从文档中删除子项目

Question

How does one remove subitems from a document. So say I have a document called sales with each sale has a sale.item which contains {name,price,code}.

I want to remove each item which is not valid, by checking the code for blank or null.

Trying something like below fails with errors, am not sure if I need to use sub-query and how.

FOR sale in sales
FOR item in sale.items
    FILTER item.code == ""
REMOVE item IN sale.items

Another attempt

FOR sale in sales
    LET invalid = (
        FOR item in sale.items
            FILTER item.code == ""
        RETURN item
    )

REMOVE invalid IN sale.items LET removed = OLD RETURN removed

Answer 1

The following query will rebuild the items for each document in sales . It will only keep item whose code is not null and not the empty string:

FOR doc IN sales 
  LET newItems = (
    FOR item IN doc.items 
      FILTER item.code != null && item.code != '' 
      RETURN item
    ) 
  UPDATE doc WITH { items: newItems } IN sales

Here is the test data I used:

db.sales.insert({ 
  items: [ 
    { code: null, what: "delete-me" }, 
    { code: "", what: "delete-me-too" }, 
    { code: "123", what: "better-keep-me" }, 
    { code: true, what: "keep-me-too" }
  ]
});
db.sales.insert({ 
  items: [ 
    { code: "", what: "i-will-vanish" },
    { code: null, what: "i-will-go-away" },
    { code: "abc", what: "not me!" }
  ]
});
db.sales.insert({ 
  items: [ 
    { code: "444", what: "i-will-remain" },
    { code: null, what: "i-will-not" }
  ]
});

Answer 2

There's a better way to do this, without sub-queries. Instead, a function for removing an array element, will be used:

FOR doc IN sales
    FOR item IN doc.items 
        FILTER item.code == ""
            UPDATE doc WITH { items: REMOVE_VALUE( doc.items, item ) } IN sales

REMOVE_VALUE takes an array as the first argument, and an array item inside that array as the second argument, and returns an array that has all the items of the first argument, but without that specific item that was in the second argument.

Example:

REMOVE_VALUE([1, 2, 3], 3) = [1, 2]

Example with subdocuments being the values:

REMOVE_VALUE( [ {name: cake}, {name: pie, taste: delicious}, {name: cheese} ] , {name: cheese}) = [ {name: cake}, {name: pie, taste: delicious} ]

You cannot just use REMOVE_VALUE separately, the way you use the REMOVE command separately. You must use it as part of an UPDATE command not as part of a REMOVE command. Unfortunately, the way it works is to make a copy of the "items" list inside your one specific "doc" that you are currently dealing with, but the copy has the subdocument you don't like, removed from the "items" list. That new copy of the list replaces the old copy of the items list.

There is one more, most efficient way to remove subdocuments from a list - and that is by accessing the specific cell of the list with items[2] - and you have to use fancy array functions even fancier than the one I used here, to find out the specific cell in the list (whether it's [2] or [3] or [567]) and then to replace the contents of that cell with Null, using the UPDATE command, and then to set the options to KeepNull = false. That's the "most efficient" way to do it but it would be a monstrous looking complicated query ): I might write that query later and put it here but right now .. I would honestly suggest using the method I described above, unless you have a thousand subdocuments in each list.