- (UserInfo*)getCurrentUserInfo:(NSString*)userName
{
UserInfo *userInfo = [[UserInfo alloc]init];
sqlite3 *database;
sqlite3_stmt *selectstmt;
NSLog(@"userName:%@",userName);
if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
NSString *sqlString = [NSString stringWithFormat:@"SELECT USER_LEVEL FROM USER_INFO WHERE USER_NAME = '%@'" , userName];
NSLog(@"getCurrentUserInfo:%@",sqlString);
const char *SqlCommand = [sqlString UTF8String];
if (sqlite3_prepare_v2(database, SqlCommand, -1, &selectstmt, NULL) == SQLITE_OK) {
NSLog(@"success!");
while (sqlite3_step(selectstmt) == SQLITE_ROW) {
NSString *userInfoStr = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 0)];
NSLog(@"select result:%@",userInfoStr);
}
}
sqlite3_finalize(selectstmt);
}
sqlite3_close (database);
return userInfo;
}
以下是我的日志输出:
2012-03-06 17:50:11.556 MagicWords[508:f803] userName:Tan
2012-03-06 17:50:11.557 MagicWords[508:f803] getCurrentUserInfo:SELECT USER_LEVEL FROM USER_INFO WHERE USER_NAME = 'Tan'
它不会打印“成功”,所以sqlite3_prepare_v2不会返回yes.but我的数据库没问题:
我找不到问题?
答案 0 :(得分:2)
添加sqlite错误消息。它将深入了解内部情况。如果您的数据库的所有内容都正确,请检查表名。
UserInfo *userInfo = [[UserInfo alloc]init];
sqlite3 *database;
sqlite3_stmt *selectstmt;
NSLog(@"userName:%@",userName);
if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
NSString *sqlString = [NSString stringWithFormat:@"SELECT USER_LEVEL FROM USER_INFO WHERE USER_NAME = '%@'" , userName];
NSLog(@"getCurrentUserInfo:%@",sqlString);
const char *SqlCommand = [sqlString UTF8String];
if (sqlite3_prepare_v2(database, SqlCommand, -1, &selectstmt, NULL) == SQLITE_OK) {
NSLog(@"success!");
while (sqlite3_step(selectstmt) == SQLITE_ROW) {
NSString *userInfoStr = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 0)];
NSLog(@"select result:%@",userInfoStr);
}
}
//Add the error message here.
else
{
NSLog(@"%s",sqlite3_errmsg(database));
}
///////
sqlite3_finalize(selectstmt);
}
sqlite3_close (database);
return userInfo;
答案 1 :(得分:0)
删除''中的引号,并在sqlString的末尾使用Semicolon。
NSString *sqlString = [NSString stringWithFormat:@"SELECT USER_LEVEL FROM USER_INFO WHERE USER_NAME = %@;" , userName];
答案 2 :(得分:0)
在将sqlString转换为char
时尝试此操作const char * SqlCommand =(char *)[sqlString cStringUsingEncoding:NSUTF8StringEncoding];