iOS-更新表失败?
我的桌子在插入过程中没有上升。在tableview中也不显示,但单击“添加新记录”后,消息会显示“已添加记录”,因此没有任何错误 在表函数中插入:iOS-更新表失败?,ios,objective-c,xcode,sqlite,Ios,Objective C,Xcode,Sqlite,我的桌子在插入过程中没有上升。在tableview中也不显示,但单击“添加新记录”后,消息会显示“已添加记录”,因此没有任何错误 在表函数中插入: -(int) insert:(NSString *)filePath withBrand:(NSString *)brand { sqlite3* db = NULL; int rc=0; rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncod
-(int) insert:(NSString *)filePath withBrand:(NSString *)brand
{
sqlite3* db = NULL;
int rc=0;
rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncoding], &db, SQLITE_OPEN_READWRITE , NULL);
if (SQLITE_OK != rc)
{
sqlite3_close(db);
NSLog(@"Failed to open db connection");
}
else
{
NSString * query = [NSString
stringWithFormat:@"INSERT INTO brands (bikeBrandName) VALUES (?)"];
char * errMsg;
rc = sqlite3_exec(db, [query UTF8String] ,NULL,NULL,&errMsg);
if(SQLITE_OK != rc)
{
NSLog(@"Failed to insert record rc:%d, msg=%s",rc,errMsg);
}
sqlite3_close(db);
}
return rc;
}
我也试过这个:
-(int) insert:(NSString *)filePath withBrand:(NSString *)brand
{
sqlite3* db = NULL;
int rc=0;
rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncoding], &db, SQLITE_OPEN_READWRITE , NULL);
if (SQLITE_OK != rc)
{
sqlite3_close(db);
NSLog(@"Failed to open db connection");
}
else
{
NSString * query = [NSString
stringWithFormat:@"INSERT INTO brands (bikeBrandName) VALUES (\"%@\")",brand];
char * errMsg;
rc = sqlite3_exec(db, [query UTF8String] ,NULL,NULL,&errMsg);
if(SQLITE_OK != rc)
{
NSLog(@"Failed to insert record rc:%d, msg=%s",rc,errMsg);
}
sqlite3_close(db);
}
return rc;
}
不是吗
应该是
INSERT INTO brands (bikeBrandName) VALUES ('%@')",brand
^ ^
要进一步调试,请使用check
当然会有帮助的谢谢你的回答。。但是不起作用仍然不更新直到不起作用…尝试了教程,但很难实现到我的代码中,因为使用其他方法插入到数据库中
INSERT INTO brands (bikeBrandName) VALUES ('%@')",brand
^ ^