无法（使用迅速Ç源码）打开数据库

Question

override func viewDidLoad() { 
    super.viewDidLoad() 
    let path = NSSearchPathForDirectoriesInDomains(.documentDirectory, .userDomainMask, true).first! 
    let db = "\(path)/ishop.sqlite3" 

    let name: [CChar]? = db.cString(using: String.Encoding.utf8) 
    getInput(UnsafeMutablePointer<Int8>(mutating: name!)) 

} 

INT getInput（烧焦DBNAME []） {

sqlite3 *db; 
sqlite3_stmt *res; 

int rc = sqlite3_open_v2("dbname", &db, SQLITE_OPEN_CREATE | SQLITE_OPEN_READWRITE | SQLITE_OPEN_URI |SQLITE_OPEN_MAIN_DB , NULL); 

if (rc != SQLITE_OK) { 


    fprintf(stderr, "Cannot open database: %s\n", sqlite3_errmsg(db)); 
    sqlite3_close(db); 

    return 1; 
} 

Answer 1

sqlite3_open_v2("dbname", 

第一个参数是一个字符串，由六个字母dbname的。

如果您想使用dbname参数，请不要将它用引号引起来。