sqlite3_prepare_v2失败(Xcode sqlite 3)

Question

我是如此想要连接sqlite但不能.有人能帮我吗？我在这里看不到任何错误,我在谷歌和这里搜索了这个问题的高低,但我找不到任何方法来解决这个问题.

我在这里按照教程

帮助我,我对这个错误感到疯狂.=(

[super viewDidLoad];
// Do any additional setup after loading the view, typically from a nib.

sqlite3 *contactDB;  //declare pointer to database structurery
const char *dbPath = [databasePath UTF8String];  //convert NSString to UTF8String

//open database file. TRUE = success, FALSE = failed
if (sqlite3_open(dbPath, &contactDB)== SQLITE_OK){     
    NSLog(@"Opened");

sqlite3_stmt *statement;
NSString *querySQL =  @"SELECT address, phone FROM contacts";
const char *query_stmt = [querySQL UTF8String];

if (sqlite3_prepare_v2(contactDB, query_stmt, 1, &statement, NULL) == SQLITE_OK) {
    NSLog(@"Statement is OK");
}else{
    NSLog(@"Statement FAILED");
}

}else{
    NSLog(@"Failed");
}

日志只显示"已打开",然后显示"语句失败"

Answer 1

我就是这样做的.(我的数据库类的编译部分)也
显示sql错误.

sqlite3 *_database;
sqlite3_open([databasePath UTF8String], &_database);

NSString *sqlStatement = @"SELECT address, phone FROM contacts";
const char *sql = [sqlStatement UTF8String];
static sqlite3_stmt *compiledStatement;
int result = sqlite3_prepare_v2(_database, sql, -1, &compiledStatement, NULL);
if(result != SQLITE_OK) {
    NSLog(@"Prepare-error #%i: %s", result, sqlite3_errmsg(_database));
}

result = sqlite3_step(compiledStatement);
if (result != SQLITE_DONE) {
    NSLog(@"Step-error #%i for '%@': %s", result, sqlStatement, sqlite3_errmsg(_database));
}

sqlite3_finalize(compiledStatement);