我在ios中编写了一个代码,我正在使用quickblox api进行ios视频聊天: -
这是我的代码: -
//获取用户结果
- (void)completedWithResult:(Result *)result{
AppDelegate *appDelegate = (AppDelegate *)[UIApplication sharedApplication].delegate;
if(result.success && [result isKindOfClass:[QBUUserPagedResult class]]){
QBUUserPagedResult *usersResult = (QBUUserPagedResult *)result;
NSLog(@"Users=%@", usersResult.users);
NSLog(@"Page parameters: currentPage %lu, totalPages %lu, perPage %lu, totalEntries %lu",
(unsigned long)usersResult.currentPage,
(unsigned long)usersResult.totalPages,
(unsigned long)usersResult.perPage,
(unsigned long)usersResult.totalEntries);
[noOfuser removeAllObjects];
[noOfuser addObjectsFromArray:usersResult.users];
for (int i=0; i<[noOfuser count];i++)
{
QBUUser *tmpUser=(QBUUser *)[noOfuser objectAtIndex:i];
NSLog(@"bbbbb%@",tmpUser);
if (appDelegate.currentUser==i) {
[noOfuser removeObject:noOfuser[i]];
}
}
[self.tblTeamList reloadData];
}else{
NSLog(@"errors=%@", result.errors);} }
这里我要获取添加到我的Quickblox应用程序中的用户列表,但我不想显示列表中登录的人员的姓名,我想只显示除了登录用户。你能否检查一下代码并纠正我的错误。
实施例: -
如果我在一个组中有四个名为a,b,c,d的用户,那么当a登录时,b,c,d应显示为a。
答案 0 :(得分:0)
QBUUser *currentUser = ...
...
if(result.success && [result isKindOfClass:[QBUUserPagedResult class]]){
QBUUserPagedResult *usersResult = (QBUUserPagedResult *)result;
NSMutableArray *allUsers = [NSMutableArray arrayWithArray:usersResult.users];
NSMutableArray *allUsersCopy = [NSMutableArray arrayWithArray:allUsers];
for(QBUUser *user in allUsersCopy){
if(user.ID == currentUser.ID){
[allUsers removeObject:user];
break;
}
}
}
下一步 - 使用 allUsers 数组 - 它不包含您当前的用户