Challenges level status for unlocking should look on previous level waiver amount, not on its value.
Also, if second level is marked as locked, then all next levels should also be marked as locked.
This commit is contained in:
BONNe 2020-06-17 08:32:15 +03:00
parent 679fe48871
commit dd8834f1df
1 changed files with 9 additions and 3 deletions

View File

@ -1107,6 +1107,7 @@ public class ChallengesManager
// The first level is always unlocked and previous for it is null.
ChallengeLevel previousLevel = null;
int doneChallengeCount = 0;
boolean previousUnlocked = true;
// For each challenge level, check how many the storageDataID has done
for (ChallengeLevel level : challengeLevelList)
@ -1116,20 +1117,25 @@ public class ChallengesManager
// remove waiver amount to get count of challenges that still necessary to do.
int challengesToDo = previousLevel == null ? 0 :
(previousLevel.getChallenges().size() - doneChallengeCount - level.getWaiverAmount());
(previousLevel.getChallenges().size() - doneChallengeCount - previousLevel.getWaiverAmount());
// As level already contains unique ids of challenges, just iterate through them.
doneChallengeCount = (int) level.getChallenges().stream().filter(playerData::isChallengeDone).count();
// Mark if level is unlocked
boolean unlocked = previousUnlocked && challengesToDo <= 0;
result.add(new LevelStatus(
level,
previousLevel,
challengesToDo,
level.getChallenges().size() == doneChallengeCount,
challengesToDo <= 0));
unlocked));
previousLevel = level;
previousUnlocked = unlocked;
}
return result;
}
@ -1160,7 +1166,7 @@ public class ChallengesManager
ChallengeLevel previousLevel = levelIndex < 1 ? null : challengeLevelList.get(levelIndex - 1);
int challengesToDo = previousLevel == null ? 0 :
(previousLevel.getChallenges().size() - level.getWaiverAmount()) -
(previousLevel.getChallenges().size() - previousLevel.getWaiverAmount()) -
(int) previousLevel.getChallenges().stream().filter(playerData::isChallengeDone).count();
// As level already contains unique ids of challenges, just iterate through them.