Sqlite数据库锁定错误-iOS
我使用sqlite来插入和更新表视图中的数据。我只能更新数据一次,下次尝试时显示数据库已锁定。即使我正在关闭数据库,请帮助。下面是代码Sqlite数据库锁定错误-iOS,ios,objective-c,sqlite,Ios,Objective C,Sqlite,我使用sqlite来插入和更新表视图中的数据。我只能更新数据一次,下次尝试时显示数据库已锁定。即使我正在关闭数据库,请帮助。下面是代码 -(void)saveUserCredential: (NSString *)email :(NSString *)userName :(NSString *)loginTime :(NSString *)source { NSDateFormatter *dateformate=[[NSDateFormatter alloc]init]; [dateforma
-(void)saveUserCredential: (NSString *)email :(NSString *)userName :(NSString *)loginTime :(NSString *)source
{
NSDateFormatter *dateformate=[[NSDateFormatter alloc]init];
[dateformate setDateFormat:@"yyyy-MM-dd HH:mm"]; // Date formater
NSString *todayDate = [dateformate stringFromDate:[NSDate date]];
const char *dbpath = [databasePath UTF8String];
if (sqlite3_open(dbpath, &database) == SQLITE_OK)
{
NSString * query = @"SELECT * from users";
int rc =sqlite3_prepare_v2(database, [query UTF8String], -1, &statement, NULL);
if(rc == SQLITE_OK)
{
if(sqlite3_step(statement) == SQLITE_ROW)
{
NSString *updateSQL = [NSString stringWithFormat:@"update users set email = '%@', username = '%@', source = '%@', created = '%@' where id = '%d'",email, userName, source, todayDate,1];
const char *update_stmt = [updateSQL UTF8String];
sqlite3_prepare_v2(database, update_stmt, -1, &statement, NULL );
//sqlite3_bind_int(statement, 1, 1);
if (sqlite3_step(statement) == SQLITE_DONE)
{
NSLog(@"successfully updated");
}
else{
NSLog(@"Error while updating. '%s'", sqlite3_errmsg(database));
}
}
else
{
NSString *insertSQL = [NSString stringWithFormat:@"insert into users (email,username,source,created) values('%@','%@','%@','%@')",email,userName,source,todayDate];
NSLog(@"INS SQL: %@", insertSQL);
const char *insert_stmt = [insertSQL UTF8String];
sqlite3_prepare_v2(database, insert_stmt,-1, &statement, NULL);
if (sqlite3_step(statement) == SQLITE_DONE)
{
NSLog(@"INSERTED");
}
else
{
NSLog(@"NOT INSERTED");
}
NSLog(@"hello ");
}
sqlite3_finalize(statement);
sqlite3_close(database);
}
}
}
你需要打电话
sqlite3_finalize(statement);
对于每个成功的sqlite\u prepare\u v2
调用。它应该是平衡的。对每个查询也使用不同的sqlite3\u stmt
因此,您的代码需要更改如下:
const char *dbpath = [databasePath UTF8String];
if (sqlite3_open(dbpath, &database) == SQLITE_OK)
{
NSString * query = @"SELECT * from users";
int rc =sqlite3_prepare_v2(database, [query UTF8String], -1, &statement, NULL);
if(rc == SQLITE_OK)
{
if(sqlite3_step(statement) == SQLITE_ROW)
{
NSString *updateSQL = [NSString stringWithFormat:@"update users set email = '%@', username = '%@', source = '%@', created = '%@' where id = '%d'",email, userName, source, todayDate,1];
const char *update_stmt = [updateSQL UTF8String];
sqlite3_stmt *upStmt;
sqlite3_prepare_v2(database, update_stmt, -1, &upStmt, NULL );
//sqlite3_bind_int(upStmt, 1, 1);
if (sqlite3_step(upStmt) == SQLITE_DONE)
{
NSLog(@"successfully updated");
}
else
{
NSLog(@"Error while updating. '%s'", sqlite3_errmsg(database));
}
sqlite3_finalize(upStmt);
}
else
{
NSString *insertSQL = [NSString stringWithFormat:@"insert into users (email,username,source,created) values('%@','%@','%@','%@')",email,userName,source,todayDate];
NSLog(@"INS SQL: %@", insertSQL);
const char *insert_stmt = [insertSQL UTF8String];
sqlite3_stmt *inStmt;
sqlite3_prepare_v2(database, insert_stmt,-1, &inStmt, NULL);
if (sqlite3_step(inStmt) == SQLITE_DONE)
{
NSLog(@"INSERTED");
}
else
{
NSLog(@"NOT INSERTED");
}
NSLog(@"hello ");
sqlite3_finalize(inStmt);
}
sqlite3_finalize(statement);
sqlite3_close(database);
}
}
因此,如果(sqlite3_step(statement)==SQLITE_DONE){NSLog(@“成功更新”);}或者{NSLog(@“更新时出错”。%s',sqlite3_errmsg(数据库));},我需要在这一行后面添加finalize语句它现在没有更新表。@Mac\u Play:我在代码中有一个输入错误,现在已经修复了,请现在检查您是否已经分别添加了2个sqlite3\u stmt来处理插入和更新,对吗?或者你做了其他的改变?