NSPredicate 通过对象数组过滤

Question

这是我的自定义 NSObject 类：

@interface personObject : NSObject

@property (nonatomic, copy) NSString *personName;
@property (nonatomic, copy) UIImage *personPhoto;

@property BOOL invited;

- (id) initWithName: (NSString *) personName Photo: (UIImage *) photo Invited: (BOOL *) invited;

@end

@implementation personObject
@synthesize personName = _personName;
@synthesize personPhoto = _personPhoto;
@synthesize invited = _invited;

 - (id) initWithName:(NSString *)personName Photo:(UIImage *)photo Invited:(BOOL *)invited{
    self = [super init];
    if (self) {
        self.personName = personName;
        self.personPhoto = photo;
        self.invited = invited;
    }
    return self;
}

@end

我在 viewDidLoad 中初始化了三个人对象。 现在我尝试了

[NSPredicate predicateWithFormat:@"personName beginswith[c], searchText];

但是我得到了一个错误——

-[personObject copyWithZone:]: unrecognized selector sent to instance 0x109684330'

所以我尝试了这个：

- (void) filterContententForSearchText: (NSString *) searchText scope:(NSString *) scope{
    NSPredicate *predicate = [NSPredicate predicateWithFormat:@"%K beginsWith[c]  %@",_currentPerson.personName, searchText ];
    self.searchArray = [self.peopleArray filteredArrayUsingPredicate:predicate];
}

所以我尝试设置

_currentPerson  = [self.peopleArray objectAtIndex.row];

在 tableViewcellForRowAtIndexPath 方法中并使用它来代替。但我得到一个不同的错误——原因：

'[<personObject 0x109524750> valueForUndefinedKey:]: this class is not key value coding-compliant for the key Paul Smith.'

当我开始在搜索栏中输入内容时，这两个错误都会发生。

这里是row方法...如何在搜索数组中将单元格文本设置为currentPerson.personName

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
    personObject *currentPerson = [self.peopleArray objectAtIndex:indexPath.row];

    self.personCell = (personCell *) [self.tableView dequeueReusableCellWithIdentifier:@"personCell"];

    if (tableView == self.searchDisplayController.searchResultsTableView) {
        self.personCell.nameLabel.text = [self.searchArray objectAtIndex:indexPath.row];
    }
    else{
        self.personCell.nameLabel.text = currentPerson.personName;
    }
    return _personCell;
}

Answer 1

您需要检查personName 密钥而不是_currentPerson.personName 作为密钥。当您提供_currentPerson.personName 时，它将作为密钥并将其视为应该是NSCodingcompliant 的整个密钥。但您没有需要它，您只能将@"personName" 作为字符串传递，它将通过keyValueCoding 工作。用下面的函数替换上面的函数

- (void) filterContententForSearchText: (NSString *) searchText scope:(NSString *) scope{
    NSPredicate *predicate = [NSPredicate predicateWithFormat:@"%K beginsWith[c]  %@",@"personName", searchText ];
   self.searchArray = [self.peopleArray filteredArrayUsingPredicate:predicate];
    NSLog(@"%@",seachArray);
}