ios中的程序流问题

我正在解析json以在tableview中显示内容。我在getReceivedData中填充了包含解析json的数组，该数组在UITAbleView委托方法之后调用。因此，填充tableview是一个问题，因为当编译器尝试填充tableview时，数组尚未初始化

- (void)getReceivedData:(NSMutableData *)data sender:(RestAPI *)sender{

NSError * error=nil;
NSArray *receivedData = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:&error];
NSString *dictionaryKey=@"department";
NSString *predicateString=@"software";
NSPredicate *predicate=[NSPredicate predicateWithFormat:@" %K == %@ ", dictionaryKey,predicateString];
NSArray *shortlisted=[receivedData filteredArrayUsingPredicate:predicate];
for(int i = 0; i<shortlisted.count; i++)
{
    NSDictionary *detailItems=[shortlisted objectAtIndex:i];
    NSString *name=[detailItems objectForKey:@"emp_name"];
    NSString *designation=[detailItems objectForKey:@"designation"];
    NSString *email=[detailItems objectForKey:@"email"];
    NSString *phone_no=[detailItems objectForKey:@"phone_no"];
  //  NSString *image=[detailItems objectForKey:@"url_path"];
    dictionary1=[NSMutableDictionary dictionaryWithObjectsAndKeys:
               name, @"keyname",
                designation, @"keydesignation",
                email, @"keyid",
                phone_no, @"keyphone",

               nil];
     [myObject1 addObject:dictionary1];
   }

 }

-(NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:
(NSInteger)section{
   if(isfiltered==YES){
       return [filteredArray count];
  }
   else{
  return [myObject1 count];
   }

 }
-(UITableViewCell *)tableView:(UITableView *)tableView 
cellForRowAtIndexPath:(NSIndexPath *)indexPath{

MyTableCell *cell=[tableView dequeueReusableCellWithIdentifier:@"myCell"];
if(!cell){
    [tableView registerNib:[UINib nibWithNibName:@"MyTableCell" bundle:nil] 
forCellReuseIdentifier:@"myCell"];
    cell=[tableView dequeueReusableCellWithIdentifier:@"myCell"];
}


if(isfiltered==NO)
{
  NSDictionary * tmpdict= [myObject objectAtIndex:indexPath.row];
    cell.nameLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keyname"]];
    cell.designationLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keydesignation"]];
    cell.idLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keyid"]];
    cell.phoneLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keyphone"]];
    cell.mainImg.image = [UIImage imageNamed:[tmpdict objectForKeyedSubscript:@"keyimage"]];

}
else{
    NSDictionary * tmpdict= [filteredArray objectAtIndex:indexPath.row];
    cell.nameLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keyname"]];
    cell.designationLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keydesignation"]];
    cell.idLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keyid"]];
    cell.phoneLabel.text=[NSMutableString stringWithFormat:@"%@",[tmpdict objectForKeyedSubscript:@"keyphone"]];
             cell.mainImg.image = [UIImage imageNamed:[tmpdict objectForKeyedSubscript:@"keyimage"]];
      }
   return cell;


}

-（void）getReceivedData:（NSMutableData*）数据发送方：（RestAPI*）发送方{
n错误*错误=nil；
NSArray*receivedData=[NSJSONSerialization JSONObjectWithData:数据选项：NSJSONReadingAllowFragments错误：&错误]；
NSString*dictionaryKey=@“部门”；
NSString*谓词字符串=@“软件”；
NSPredicate*谓词=[NSPredicate谓词格式：@“%K==%”，字典键，谓词字符串]；
NSArray*入围=[receivedData filteredArrayUsingPredicate:predicate]；
对于（int i=0；i实际上，这不是像您那样初始化字典的正确方法，@“%@”被使用%@是任何对象的格式字符串中的占位符

dictionary1=[NSMutableDictionary dictionaryWithObjectsAndKeys:
                name,@"keyname",
                designation,@"keydesignation",
                email,@"keyid",
                phone_no,@"keyphone",
               nil];

实际上，这不是像您那样初始化字典的正确方法，@“%@”被使用%@是任何对象的格式字符串中的占位符

dictionary1=[NSMutableDictionary dictionaryWithObjectsAndKeys:
                name,@"keyname",
                designation,@"keydesignation",
                email,@"keyid",
                phone_no,@"keyphone",
               nil];

在易变字典中，首先你必须给对象，然后给它的键。你做错了

for(int i = 0; i<shortlisted.count; i++)
{
    NSDictionary *detailItems=[shortlisted objectAtIndex:i];
    NSString *name=[detailItems objectForKey:@"emp_name"];
    NSString *designation=[detailItems objectForKey:@"designation"];
    NSString *email=[detailItems objectForKey:@"email"];
    NSString *phone_no=[detailItems objectForKey:@"phone_no"];


dictionary1=[NSMutableDictionary dictionaryWithObjectsAndKeys:
                name,@"keyname",
                designation,@"keydesignation",
                email,@"keyid",
                phone_no,@"keyphone",
               nil];
 [myObject1 addObject:dictionary1];
}

for（inti=0；i在易变字典中，首先必须给对象赋值，然后才是它的键。你做错了

for(int i = 0; i<shortlisted.count; i++)
{
    NSDictionary *detailItems=[shortlisted objectAtIndex:i];
    NSString *name=[detailItems objectForKey:@"emp_name"];
    NSString *designation=[detailItems objectForKey:@"designation"];
    NSString *email=[detailItems objectForKey:@"email"];
    NSString *phone_no=[detailItems objectForKey:@"phone_no"];


dictionary1=[NSMutableDictionary dictionaryWithObjectsAndKeys:
                name,@"keyname",
                designation,@"keydesignation",
                email,@"keyid",
                phone_no,@"keyphone",
               nil];
 [myObject1 addObject:dictionary1];
}

for（int i=0；i
@“%@”，name，@“keyname”，
=>
[NSString stringWithFormat:@“%@”，name]，@“keyname”
，等等。或者干脆
NSString*name=[NSString stringWithFormat:@“%@”，“detailItems objectForKey:@“emp_name”]；和
name，@“keyname”`（不带“%@”）。只需像
dictionary1一样更改即可=[NSMutableDictionary Dictionary With Objects and Keys:名称、@“keyname”、名称、@“keydesignation”、电子邮件、@“keyid”、电话号码、@“keyphone”、无]；
@Larme非常感谢much@Anbu.Karthik非常感谢你，它成功了again@AashimaAnand-您是否获得了正确的字典数组，那么为什么再次更改，在需要
@“%@”，name，@“keyname”，name，
=>
[NSString stringWithFormat:@“%@”，name]，@“keyname”的地方直接使用字典数组
，等等。或者简单地
NSString*name=[NSString stringWithFormat:@“%@”、[detailItems objectForKey:@“emp_name”]；和
name、@“keyname”`（不带“%@”）。只需像
dictionary1=[NSMutableDictionary dictionary dictionary with objectsandkeys:name、@“keyname”、designation、@“keynesignation”这样更改即可，电子邮件，@“keyid”，电话号码，@“keyphone”，无]；
@Larme非常感谢much@Anbu.Karthik非常感谢你，它成功了again@AashimaAnand-你得到的字典数组是否正确，那么你为什么再次更改，在需要的地方直接使用你的字典数组