Question

例如，我有像@{@"firstName":@"aaa", @"lastName":@"bbb"}这样的对象。现在我尝试使用下一个谓词进行搜索：

NSPredicate *predicateFirstName = [NSPredicate predicateWithFormat:@"firstName contains [c] %@", searchString];
NSPredicate *predicateLastName = [NSPredicate predicateWithFormat:@"lastName contains [c] %@", searchString];
NSPredicate *orPredicate = [NSCompoundPredicate  orPredicateWithSubpredicates:@[predicateFirstName, predicateLastName]];
searchUsersArray = [usersArray filteredArrayUsingPredicate:orPredicate];

我需要一个谓词，它将返回搜索字符串@＆＃34; aa bb＆＃34;名字ENDED with＆＃34; aa＆＃34;和lastName BEGINS与@＆＃34; bb＆＃34;

NSPredicate *predicateLastName = [NSPredicate predicateWithFormat:@"(firstName AND lastName contains [c] %@)", searchString]

不能为我工作。

Answer 1

首先将搜索字符串分成末尾部分和开始部分：

NSArray *parts = [@"aa bb" componentsSeperatedByString:@" "];

然后使用BEGINSWITH和ENDSWITH：

[NSPredicate predicateWithFormat:@"(firstName ENDSWITH[c] %@) AND (lastName BEGINSWITH[c] %@", parts[0], parts[1]];

或者，您可以将计算属性添加到实体，并将名字和姓氏结合起来，然后在此属性中搜索。

Answer 2

尝试这样希望它可以帮助你

NSString *matchString =  [NSString stringWithFormat: @"(.*aa|bb.*)",searchText];
NSString *predicateString = @"keyword MATCHES[c] %@";
NSPredicate *predicate =[NSPredicate predicateWithFormat: predicateString, matchString];

由于

Answer 3

使用关键字BEGINSWITH而不是包含。

NSPredicate *predicateFirstName = [NSPredicate predicateWithFormat:@"firstName BEGINSWITH [c] %@", searchString];
NSPredicate *predicateLastName = [NSPredicate predicateWithFormat:@"lastName BEGINSWITH [c] %@", searchString];

如何用NSPredicate连接两个字段进行搜索？

3 个答案: