Cocoa-Touch-如何解析本地Json文件

一尘不染

Cocoa-Touch-如何解析本地Json文件

json

我是iOS开发人员中的新手，并且尝试解析本地Json文件，例如

{"quizz":[{"id":"1","Q1":"When Mickey was born","R1":"1920","R2":"1965","R3":"1923","R4","1234","response","1920"},{"id":"1","Q1":"When start the cold war","R1":"1920","R2":"1965","R3":"1923","rep4","1234","reponse","1920"}]}

这是我的代码：

 NSString *filePath = [[NSBundle mainBundle] pathForResource:@"data" ofType:@"json"];
NSString *myJSON = [[NSString alloc] initWithContentsOfFile:filePath encoding:NSUTF8StringEncoding error:NULL];
// Parse the string into JSON
NSDictionary *json = [myJSON JSONValue];

// Get all object
NSArray *items = [json valueForKeyPath:@"quizz"];

NSEnumerator *enumerator = [items objectEnumerator];
NSDictionary* item;
while (item = (NSDictionary*)[enumerator nextObject]) {
    NSLog(@"clientId = %@",  [item objectForKey:@"id"]);
    NSLog(@"clientName = %@",[item objectForKey:@"Q1"]);
    NSLog(@"job = %@",       [item objectForKey:@"Q2"]);
}

我在此站点上找到了一个示例，但出现以下错误

-JSONValue失败。错误是：对象键后不期望令牌“值分隔符”。

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2020-07-27

共1个答案

一尘不染

JSON具有严格的键/值表示法，用于R4和响应的键/值对不正确。试试这个：

NSString *jsonString = @"{\"quizz\":[{\"id\":\"1\",\"Q1\":\"When Mickey was born\",\"R1\":\"1920\",\"R2\":\"1965\",\"R3\":\"1923\",\"R4\":\"1234\",\"response\":\"1920\"}]}";

如果您从文件中读取字符串，则不需要所有的斜杠。
文件将如下所示：

{“ quizz”：[{“ id”：“ 1”，“ Q1”：“米奇出生时”，“ R1”：“ 1920”，“ R2”：“ 1965”，“ R3”：“
1923”， “ R4”：“ 1234”，“响应”：“ 1920”}，{“ id”：“ 1”，“ Q1”：“冷战开始时”，“ R1”：“ 1920”，“
R2”： “ 1965”，“ R3”：“ 1923”，“ R4”：“ 1234”，“响应”：“ 1920”}]}

我用以下代码进行了测试：

NSString *jsonString = @"{\"quizz\":[{\"id\":\"1\",\"Q1\":\"When Mickey was born\",\"R1\":\"1920\",\"R2\":\"1965\",\"R3\":\"1923\",\"R4\":\"1234\",\"response\":\"1920\"}, {\"id\":\"1\",\"Q1\":\"When start the cold war\",\"R1\":\"1920\",\"R2\":\"1965\",\"R3\":\"1923\",\"R4\":\"1234\",\"reponse\":\"1920\"}]}";
NSLog(@"%@", jsonString);
NSError *error =  nil;
NSDictionary *json = [NSJSONSerialization JSONObjectWithData:[jsonString dataUsingEncoding:NSUTF8StringEncoding] options:kNilOptions error:&error];

NSArray *items = [json valueForKeyPath:@"quizz"];

NSEnumerator *enumerator = [items objectEnumerator];
NSDictionary* item;
while (item = (NSDictionary*)[enumerator nextObject]) {
    NSLog(@"clientId = %@",  [item objectForKey:@"id"]);
    NSLog(@"clientName = %@",[item objectForKey:@"Q1"]);
    NSLog(@"job = %@",       [item objectForKey:@"Q2"]);
}

我的印象是，您复制了旧代码，因为您没有使用Apple的序列化和Enumerator而不是Fast
Enumeration。整个枚举内容可以写为

NSArray *items = [json valueForKeyPath:@"quizz"];
for (NSDictionary *item in items) {
    NSLog(@"clientId = %@",  [item objectForKey:@"id"]);
    NSLog(@"clientName = %@",[item objectForKey:@"Q1"]);
    NSLog(@"job = %@",       [item objectForKey:@"Q2"]);
}

甚至是基于块的枚举的爱好者，如果需要快速安全的枚举，还需要另外一个索引。

NSArray *items = [json valueForKeyPath:@"quizz"];
[items enumerateObjectsUsingBlock:^(NSDictionary *item , NSUInteger idx, BOOL *stop) {
    NSLog(@"clientId = %@",  [item objectForKey:@"id"]);
    NSLog(@"clientName = %@",[item objectForKey:@"Q1"]);
    NSLog(@"job = %@",       [item objectForKey:@"Q2"]);
}];

2020-07-27