如果我有這樣的SELECT語句:PHP/MySQL:如何在多個表中調用MySQL結果?
SELECT T1.TEXT, T2.FIELD FROM TABLE1 T1, TABLE2 T2 WHERE T1.ID = T2.ID
然後,我用的mysql_query使用它與mysql_fetch_object獲取它。
我怎樣才能訪問現場? $ fetched-> T1.TEXT不工作...
求助THX
UPDATE 2013:這個問題是很老,但正確的,最簡單的方式就是給每一個領域的別名select語句,所以它們是唯一的。示例:
SELECT T1.TEXT AS MYTEXT, T2.FIELD AS MYFIELD FROM TABLE1 T1, TABLE2 T2 WHERE T1.ID = T2.ID
PHP:
mysql_connect('host','user','pass');
mysql_select_db('mydb');
$resource = mysql_query(/* the query again */);
$firstrow = mysql_fetch_object($resource);
$text = $firstrow->MYTEXT;
$field = $firstrow->MYFIELD;
嘗試$ fetched-> TEXT – cristian 2011-01-09 21:22:54
我認爲它不會工作,因爲如果我選擇T1.TEXT和T2.TEXT,這將不起作用。 – 2011-01-09 21:24:01
爲什麼不給你的字段添加一個別名,例如`SELECT T1.TEXT as t1_text,T2.FIELD as t2_field FROM TABLE1 T1,TABLE2 T2 WHERE T1.ID = T2.ID`,然後嘗試$ fetched-> t1_text或$ fetched - > t2_field – cristian 2011-01-09 21:28:15