通过NSString过滤字典数组

问题描述 投票:3回答:4

在实现搜索功能时,我需要过滤字典数组。我正在使用自动完整的文本字段方法搜索栏并将其存储到字符串中。我能够解析数组,但面对下面的json

[{"CertProfID":"4","Name":"Dodge","Location":"loc4","City":"city4","State":"state4","Zip":"zip5","Website":"http:\/\/cnn.com","Phone":"phone4","Email":"email4"},
{"CertProfID":"5","Name":"cat","Location":"loc5","City":"city5","State":"State5","Zip":"zip5","Website":"web5","Phone":"phone5","Email":"email5"}]

在这里,我需要过滤字典以使其完成

我尝试使用下面的代码但其返回的数组具有空值:(

 NSString *substring = [NSString stringWithString:textField.text];
  NSLog(@"substring %@",substring);
  NSMutableArray *arr2Filt= [arraylist valueForKey:@"Name"];
  NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF  contains[c] %@",substring];
  filteredarr = [NSMutableArray arrayWithArray:[arr2Filt filteredArrayUsingPredicate:predicate]];
ios objective-c json nsarray nsdictionary
4个回答
8
投票

此代码将解决您的问题,它将返回一个字典数组

NSString *substring = [NSString stringWithString:textField.text];
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"Name contains[c] %@",searchString];
NSArray *filteredArry=[[arrayOfDict filteredArrayUsingPredicate:predicate] copy];

arrayOfDict是您原始的字典数组

Swift 4.2版:::

let namePredicate = NSPredicate(format: "Name contains[c] %@",searchString)

let filteredArray = arrayOfDict.filter { namePredicate.evaluate(with: $0) }

print("names = \(filteredArray)")

希望它会对你有所帮助


2
投票

这里有一个观察结果是filteredArrayUsingPredicateNSArray的一种方法,你使用的是NSMutableArray

使用临时NSMutableArray对象更改NSArray作为谓词。

例如:

NSString *substring = [NSString stringWithString:textField.text];

NSArray *tempArray = [arraylist valueForKey:@"Name"];

// If [arraylist valueForKey:@"Name"]; line returns NSMutableArray than use below line
// NSArray *tempArray = [NSArray arrayWithArray:[arraylist valueForKey:@"Name"]];

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF contains[c] %@",substring];

filteredarr = [[tempArray filteredArrayUsingPredicate:predicate] mutableCopy];

2
投票

你可以改用块。在比赛条件方面,他们不那么手艺了。

NSString *substring = [NSString stringWithString:textField.text];
NSLog(@"substring %@",substring);
NSPredicate *predicate = [NSPredicate predicateWithBlock:^BOOL(id evaluatedObject, NSDictionary *bindings) {
  return [evaluatedObject[@"Name"] containsString:substring];
}];
filteredarr = [NSMutableArray arrayWithArray:[arraylist filteredArrayUsingPredicate:predicate]];

1
投票

嗨我知道有很多答案,在我的情况下我存储了值为JSON MODEL对象,这是代码,使用JSON MODEL

 NSString *searchString = searchBar.text;
    NSPredicate *predicate = [NSPredicate predicateWithBlock:^BOOL(id evaluatedObject, NSDictionary *bindings) {
        NSDictionary *temp=[evaluatedObject toDictionary];
        return [temp[@"user"][@"firstName"] localizedCaseInsensitiveContainsString:searchString];
}];
    filteredContentList = [NSMutableArray arrayWithArray:[searchListValue filteredArrayUsingPredicate:predicate]];
    [tableViews reloadData];

searchListValue是一个NSMutableArray,它包含一些json模型对象。如果有人需要帮助,请点击我

Swift版本高于2.2

var customerNameDict = ["firstName":"karthi","LastName":"alagu","MiddleName":"prabhu"];
var clientNameDict = ["firstName":"Selva","LastName":"kumar","MiddleName":"m"];
var employeeNameDict = ["firstName":"karthi","LastName":"prabhu","MiddleName":"kp"];
var attributeValue = "ka";

var arrNames:Array = [customerNameDict,clientNameDict,employeeNameDict];

//var namePredicate =
//    NSPredicate(format: "firstName like %@",attributeValue);  
//uncomment above line to search particular word
let namePredicate =
    NSPredicate(format: "firstName contains[c] %@",attributeValue);
let filteredArray = arrNames.filter { namePredicate.evaluateWithObject($0) };
print("names = ,\(filteredArray)");

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