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我使用下面的函数在我的应用程序，我已经使用SQ-精简版最近才开始和我想获得的是我与那个正确或不打算你的意见。sqlite的敲定和Db锁定问题
由于搜索时我发现我需要使用sqlite3的敲定声明我面对我的应用程序数据库锁定问题。
什么我不知道是我需要放置一个敲定语句给每个SQLITE3准备语句
请让我知道
- (BOOL) addNewCate:(NSString*)dbPath:(NSString*)title:(NSString*)tierOneID:(NSString*)tierTwoID{
BOOL returnVal = NO;
if (sqlite3_open([dbPath UTF8String], &database) == SQLITE_OK)
{
const char *sql = "insert into storyboard_phrases(phrase) Values(?)";
sqlite3_stmt *addStmt;
if(sqlite3_prepare_v2(database, sql, -1, &addStmt, NULL) == SQLITE_OK){
sqlite3_bind_text(addStmt, 1, [title UTF8String], -1, SQLITE_TRANSIENT);
}
if(sqlite3_step(addStmt) != SQLITE_DONE) {
NSLog(@"Error: %s", sqlite3_errmsg(database));
} else {
NSLog(@"Insert into row id = %d", sqlite3_last_insert_rowid(database));
int ph_id = sqlite3_last_insert_rowid(database);
int sub_category_id = [tierTwoID intValue];
int main_category_id = [tierOneID intValue];
addStmt = nil;
sql = "insert into phrase_reference(phrase_id, sub_category_id,main_category_id) Values(?,?,?)";
if(sqlite3_prepare_v2(database, sql, -1, &addStmt, NULL) == SQLITE_OK){
sqlite3_bind_int(addStmt, 1, ph_id);
sqlite3_bind_int(addStmt, 2, sub_category_id);
sqlite3_bind_int(addStmt, 3, main_category_id);
}
if(sqlite3_step(addStmt) != SQLITE_DONE) {
NSLog(@"Error: %s", sqlite3_errmsg(database));
} else {
NSLog(@"Insert into row id = %d", sqlite3_last_insert_rowid(database));
returnVal = YES;
}
}
sqlite3_finalize(addStmt);
}
sqlite3_close(database);
return returnVal;
}