[__NSCFString objectForKey:]：无法识别的选择器发送到实例 0xa6a2750答案

【问题标题】：[__NSCFString objectForKey:]: unrecognized selector sent to instance 0xa6a2750[__NSCFString objectForKey:]：无法识别的选择器发送到实例 0xa6a2750
【发布时间】：2014-04-26 18:45:45
【问题描述】：

我将我的数据库值转换为 JSON 格式。它在 URL 中显示为 JSON。然后我从 JSON URL 获取值然后存储到本地数据库。但我遇到以下错误：[__NSCFString objectForKey:]: unrecognized selector sent to instance 0xa6a2750 和 *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[__NSCFString objectForKey:]: unrecognized selector sent to instance 0xa6a2750'

URL 中的 JSON 值：

[{"cat_id":"196","category_name":"Performance Parts"},{"cat_id":"212","category_name":"Car Care"}]

获取值并插入到本地：

NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory,NSUserDomainMask, YES);
 NSString *documentsDirectory = [paths objectAtIndex:0];

 NSString *path = [documentsDirectory stringByAppendingPathComponent:@"Mdb.sqlite"];
 NSLog(@"filepath %@",path);



 //NSDictionary *jsonDict = [stringFromFileAtURL JSONValue];
 //array

 NSArray *userData = [str JSONValue];

 //  NSLog(@"userData is %@", userData);


 BOOL notExist = TRUE;
 sqlite3_stmt *statement, *addStmt;



 for (NSArray *skarray in userData) {

 NSLog(@"test");


 for (NSDictionary *tuser in skarray) {

 NSLog(@"test1");

 //write all this in the table
 //if already exists in data base id then overwrite the name

 //category table
 //NSLog(@"CategoryId is %@",[tuser objectForKey:@"cat_id"]);
 //NSLog(@"CategoryName is %@",[tuser objectForKey:@"cat_name"]);

 if (sqlite3_open([path UTF8String], &database) == SQLITE_OK) {

 NSLog(@"test2");


 const char *sql = [[NSString stringWithFormat:@"SELECT cat_id FROM categories where cat_id = '%@'",[tuser objectForKey:@"cat_id"]] cStringUsingEncoding:NSUTF8StringEncoding];

 NSLog(@"test3");


 NSLog(@"check category is %s", sql);

 if (sqlite3_prepare_v2(database, sql, -1, &statement, NULL) == SQLITE_OK) {

 notExist = TRUE;

 while (sqlite3_step(statement) == SQLITE_ROW) {

 notExist = FALSE;

 }

 }


 if(notExist){
 //NSLog(@"cat id does not exist");

 const char *sqlInsert = [[NSString stringWithFormat:@"insert into categories (cat_id, category_name) values ('%@', '%@')", [tuser objectForKey:@"cat_id"], [tuser objectForKey:@"category_name"]] cStringUsingEncoding:NSUTF8StringEncoding];


 NSLog(@"Insert category is %s", sqlInsert);


 if(sqlite3_prepare_v2(database, sqlInsert, -1, &addStmt, NULL) != SQLITE_OK)
 NSAssert1(0, @"Error while creating add statement. '%s'", sqlite3_errmsg(database));

 if(SQLITE_DONE != sqlite3_step(addStmt))
 NSAssert1(0, @"Error while inserting data. '%s'", sqlite3_errmsg(database));





 }

 }



 }
 }

【问题讨论】：

标签： ios objective-c json sqlite

【解决方案1】：

删除for (NSArray *skarray in userData) {循环并使用

for (NSDictionary *tuser in userData) {

    NSLog(@"test1");
    if (sqlite3_open([path UTF8String], &database) == SQLITE_OK) {

     const char *sql = [[NSString stringWithFormat:@"SELECT cat_id FROM categories where cat_id = '%@'",[tuser objectForKey:@"cat_id"]] cStringUsingEncoding:NSUTF8StringEncoding];


     if (sqlite3_prepare_v2(database, sql, -1, &statement, NULL) == SQLITE_OK) {

    notExist = TRUE;

    while (sqlite3_step(statement) == SQLITE_ROW) {

       notExist = FALSE;

    }
}

【讨论】：

【解决方案2】：

您正在尝试从NSDictionary（或可变）获取数据，但在您的情况下，它以某种方式转换为NSString（NSCFString）并且没有objectForKey 属性。查看使用objectForKey 的编译步骤。

【讨论】：