java - Use Parse.com and Android, how can I get a list of Games that a pair of Players share? -

java - Use Parse.com and Android, how can I get a list of Games that a pair of Players share? -

i making two-player android using parse.com. i'm determining 2 players in game introducing game table on parse.com, array of userids of players of game.

i need query parse.com list of games shared 2 given players. thought trick.

string[] userids = new string[] {player1, player2}; parsequery<parseobject> query = new parsequery<parseobject>("game"); query.whereequalto("players", userids);

the problem is, don't know order of array of players. half of time, order wrong won't right result query. i'll no game exists between 2 players.

so thought solve query reversed array, or 2 queries together.

string[] reverseduserids = new string[] {player2, player1}; parsequery<parseobject> query2 = new parsequery<parseobject>("game"); query2.whereequalto("players", reverseuserids); list<parsequery<parseobject>> querylist = new arraylist<parsequery<parseobject>>(); querylist.add(query); querylist.add(query2); parsequery<parseobject> mainquery = parsequery.or(querylist);

however, when run mainquery.find(), error looks this:

java.lang.illegalargumentexception: invalid type parseobject: class [ljava.lang.string;

instead of returning list of games both players playing each other.

you seek forcefulness ids pair ordered, illustration doing player1.compareto(player2). need query right order.

java android parse.com illegalargumentexception