1
我試圖從表中選擇數據:SQLite的選擇,同時插入數據到數據庫
NSMutableArray *tmpArray = [[NSMutableArray alloc] init];
if (sqlite3_open([path UTF8String], &database) == SQLITE_OK) {
const char *sql = "SELECT name FROM Artists ORDER BY name";
sqlite3_stmt *statement;
if (sqlite3_prepare_v2(database, sql, -1, &statement, NULL) == SQLITE_OK) {
while (sqlite3_step(statement) == SQLITE_ROW) {
char *nameChars = (char *)sqlite3_column_text(statement, 0);
NSString *name = [NSString stringWithUTF8String: nameChars];
[tmpArray addObject:name];
}
}else {
NSLog(@"Error");
}
}else {
NSLog(@"Error1");
}
我做這個選擇,並在我與這個數據插入到另一個數據庫中的表時間:
-(void)insertAutpPlaylist:(NSString*)playlistName withPlaylist:(NSMutableArray*)songsArray{
if (sqlite3_open([dataPath UTF8String], &database) == SQLITE_OK) {
for (int i = 0 ; i < [songsArray count]; i++) {
SongItem *song = [songsArray objectAtIndex:i];
sqlite3_stmt *insertStmt = nil;
NSString *name = song.name;
if(insertStmt == nil)
{
NSString *statement = [NSString stringWithFormat:@"INSERT INTO %@ (name) VALUES (?)",playlistName];
const char *insertSql = [statement UTF8String];
if(sqlite3_prepare_v2(database, insertSql, -1, &insertStmt, NULL) != SQLITE_OK){
NSLog(@"Error while creating insert statement.");
insertStmt = nil;
continue;
}
sqlite3_bind_text(insertStmt, 1, [name UTF8String], -1, SQLITE_TRANSIENT);
if(SQLITE_DONE != sqlite3_step(insertStmt)){
//NSAssert1(0, @"Error while inserting data. '%s'", sqlite3_errmsg(database));
NSLog(@"Error while inserting data.");
insertStmt = nil;
continue;
}
else{}
sqlite3_reset(insertStmt);
insertStmt = nil;
}
}
sqlite3_close(database);
}
}
此代碼工作正常,如果我不插入數據並在同一時間選擇,並插入和選擇是在不同的表上。
編輯
當我嘗試做一個選擇,我得到「錯誤」輸出:的NSLog(@「錯誤」);
編輯2
當我添加sqlite3_errmsg(database)
select方法我在控制檯中看到:
reason: 'Error while inserting data. 'database is locked'
你打開數據庫之前,火災查詢(sqlite3_open([/ *數據庫ase path */UTF8String],&database)!= SQLITE_OK) – priyanka 2012-04-19 13:10:44
即使兩個不同線程同時存在兩個不同的表,也無法訪問數據庫。 – coder 2012-04-19 13:13:27
如果我只做選擇? – MTA 2012-04-19 13:40:12