我通常会像这样加入:
$query = "SELECT table1.ColA, table2.ColC ".
"FROM table1, table2".
"WHERE table1.uid = table2.uid";
但在这种情况下,WHERE部分已经用于使用PHP变量指定记录.在这种情况下如何加入表2?
SELECT sql_CALC_FOUND_ROWS ColA, ColB FROM Table1 WHERE value = $value
解决方法:
请试试这个
$query = "SELECT table1.ColA, table2.ColC FROM table1,
table2 WHERE table1.uid = table2.uid and table1.value=$value";
要么
$query = "SELECT table1.ColA, table2.ColC FROM table1 INNER JOIN table2
ON (table1.uid = table2.uid) WHERE table1.value=$value";