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我也跟着有關創建SQLite數據庫,插入數據,並從中尋找數據的教程和使用下面的方法來從數據庫中提取數據:如何從sqlite中提取名稱的數據搜索?
-(IBAction)findContacts{
const char *dbpath = [databasePath UTF8String];
sqlite3_stmt *statement;
if (sqlite3_open(dbpath, &contactDB)==SQLITE_OK) {
NSString *querySQL = [NSString stringWithFormat:@"SELECT adress, phone FROM CONTACTS WHERE name = \"%@\"", [name text]];
status.text = querySQL;
const char *query_stmt = [querySQL UTF8String];
if (sqlite3_prepare_v2(contactDB, query_stmt, -1, &statement, NULL) == SQLITE_OK) {
if (sqlite3_step(statement)==SQLITE_ROW) {
NSString *adressField = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 0)];
adress.text = adressField;
NSString *phoneField = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 1)];
phone.text = phoneField;
status.text = @"Match Found";
[adressField release];
[phoneField release];}
status.text = @"Match not Found";
adress.text = @"";
phone.text = @"";
sqlite3_finalize(statement);}
sqlite3_close(contactDB);}}
當我編譯,沒有錯誤!沒有注意!但不工作...
,我發現那是什麼,因爲
sqlite3_prepare_v2(contactDB, query_stmt, -1, &statement, NULL) == SQLITE_OK)
是不相等的,代碼不是如果statemens得到英寸我發現,當我扭轉的情況與
sqlite3_prepare_v2(contactDB, query_stmt, -1, &statement, NULL) != SQLITE_OK)
和鋸在status.text「匹配未找到」。但仍然無法找到任何真正的聯繫。
你知道代碼有什麼問題嗎?
在此先感謝...
解決你的幫助。感謝que que ... :) – Karaca 2012-02-03 16:24:38