NSPredicate:将CONTAINS与IN结合使用

时间:2012-11-20 18:16:46

标签: core-data nspredicate

我在CoreData中有一组用户,在我的应用程序中有一个搜索字段。用户具有属性名字和名称。

目前我有一个谓词“user.name CONTAINS [c]%@ OR user.firstname CONTAINS [c]%@”

这一直有效,直到用户输入“john smith”这样的全名。即使他输入“john sm”,也应找到John Smith-Object。

将搜索词的数组(IN)与CONTAINS组合起来的谓词是什么?

4 个答案:

答案 0 :(得分:36)

我认为你不能在谓词中将“IN”和“CONTAINS”结合起来。但是您可以将搜索字符串拆分为单词,并创建“复合谓词”:

NSString *searchString = @"John  Sm ";
NSArray *words = [searchString componentsSeparatedByString:@" "];
NSMutableArray *predicateList = [NSMutableArray array];
for (NSString *word in words) {
    if ([word length] > 0) {
        NSPredicate *pred = [NSPredicate predicateWithFormat:@"user.name CONTAINS[c] %@ OR user.firstname CONTAINS[c] %@", word, word];
        [predicateList addObject:pred];
    }
}
NSPredicate *predicate = [NSCompoundPredicate andPredicateWithSubpredicates:predicateList];
NSLog(@"%@", predicate);

此示例生成谓词

(user.name CONTAINS[c] "John" OR user.firstname CONTAINS[c] "John") AND
(user.name CONTAINS[c] "Sm" OR user.firstname CONTAINS[c] "Sm")

哪个匹配“John Smith”,但不匹配“John Miller”。

答案 1 :(得分:14)

2。5年后,我可以用swift中的一个更复杂的例子回答我的问题:

var predicateList = [NSPredicate]()

let words = filterText.componentsSeparatedByString(" ")

for word in words{

     if count(word)==0{
           continue
     }

     let firstNamePredicate = NSPredicate(format: "firstName contains[c] %@", word)
     let lastNamePredicate = NSPredicate(format: "lastName contains[c] %@", word)
     let departmentPredicate = NSPredicate(format: "department contains[c] %@", word)
     let jobTitlePredicate = NSPredicate(format: "jobTitle contains[c] %@", word)

     let orCompoundPredicate = NSCompoundPredicate(type: NSCompoundPredicateType.OrPredicateType, subpredicates: [firstNamePredicate, lastNamePredicate,departmentPredicate,jobTitlePredicate])

     predicateList.append(orCompoundPredicate)
}

request.predicate = NSCompoundPredicate(type: NSCompoundPredicateType.AndPredicateType, subpredicates: predicateList)

答案 2 :(得分:0)

我遇到了同样的问题并解决了这个问题:

在我的NSManagedObject类(用户)中,我添加了以下方法,将这两个值合并为一个字符串:

- (NSString *)name
{
    return [NSString stringWithFormat:@"%@ %@", self.firstname, self.lastname];
}

然后你只需要以下几行来匹配你的列表,在我的例子中是一个带有用户对象的NSArray:

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"name contains[c] %@", searchString];
NSArray *filteredUsers = [users filteredArrayUsingPredicate:predicate];

答案 3 :(得分:0)

Swift 4更新:

let firstName = NSPredicate(format: "firstName CONTAINS[c] %@", searchText)
let lastName = NSPredicate(format: "lastName CONTAINS[c] %@", searchText)

let orCompoundPredicate = NSCompoundPredicate(orPredicateWithSubpredicates: 
[firstNamePredicate,lastNamePredicate]

在获取数据时使用或CompoundPredicate