我有一个 NSArray 对象,这些对象有 10 个属性。我想对这些对象进行文本搜索。
我知道如何一次搜索 1 个房产,但有没有一种简单的方法可以一次搜索所有房产?
以下是我的对象具有的属性列表:
@property (nonatomic, retain) NSString * name;
@property (nonatomic, retain) NSString * phone;
@property (nonatomic, retain) NSString * secondaryPhone;
@property (nonatomic, retain) NSString * address;
@property (nonatomic, retain) NSString * email;
@property (nonatomic, retain) NSString * url;
@property (nonatomic, retain) NSString * category;
@property (nonatomic, retain) NSString * specialty;
@property (nonatomic, retain) NSString * notes;
@property (nonatomic, retain) NSString * guid;
如果我搜索“医生”,我希望查看其中 1 个或多个属性中包含“医生”一词的所有结果。例如,如果 1 个对象的类别为“医生”,而另一个对象的电子邮件地址为“smit[电子邮件受保护] /cdn-cgi/l/email-protection”,它们都应该出现在结果中。
NSString *searchTerm = @"search this";
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF LIKE[cd] %@", searchTerm];
NSArray *filtered = [array filteredArrayUsingPredicate:predicate];
如果有特定属性,您可以将谓词更改为:
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF.propertyName LIKE[cd] %@", searchTerm];
要搜索所有属性,您必须使用逻辑运算符将它们绑定在一起
NSString *query = @"blah";
NSPredicate *predicateName = [NSPredicate predicateWithFormat:@"name contains[cd] %@", query];
NSPredicate *predicatePhone = [NSPredicate predicateWithFormat:@"phone contains[cd] %@", query];
NSPredicate *predicateSecondaryPhone = [NSPredicate predicateWithFormat:@"secondaryPhone contains[cd] %@", query];
NSArray *subPredicates = [NSArray arrayWithObjects:predicateName, predicatePhone, predicateSecondaryPhone, nil];
NSCompoundPredicate *predicate = [NSCompoundPredicate orPredicateWithSubpredicates:subPredicates];
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