为什么不这样做!我已经把头发拉了几天试图让它起作用
$query2 = "SELECT * FROM temporders WHERE order_id=$id";
$result2 = mysql_query($query2);
$row = mysql_fetch_assoc($result2);
$osupplierid = $row['order_supplierid'];
$odescription = $row['order_description'];
$odate = $row['order_date'];
$oquoteid = $row['order_quoteid'];
$odetailid = $row['order_detailid'];
$otype = $row['order_type'];
$osupplierquote = $row['order_supplierquote'];
$osupplierquotecost = $row['order_supplierquotecost'];
$query = "INSERT INTO orders (order_supplierid, order_description, order_date, order_quoteid, order_detailid, order_type, order_supplierquote, order_supplierquotecost )
VALUES ($osupplierid, $odescription, $odate, $oquoteid, $odetailid, $otype, $osupplierquote, $osupplierquotecost )";
似乎没有得到行的值,因为没有任何输入,就像我用虚拟值替换它一样。
请帮忙!它的Xmas前夕我想早点完成工作。
伊恩
答案 0 :(得分:0)
INSERT查询中的一些参数是String,您必须使用单引号,即使它是PHP变量。
我为字段似乎是String-type添加了引号,请检查并再试一次!
$query = "INSERT INTO orders (order_supplierid, order_description, order_date, order_quoteid, order_detailid, order_type, order_supplierquote, order_supplierquotecost )
VALUES ($osupplierid, '$odescription', '$odate', $oquoteid, $odetailid, '$otype', '$osupplierquote', $osupplierquotecost )";
为了进一步了解,重要的是要注意PHP mysql_ *函数已被折旧,并且您应该使用PDO或mysqli_ *函数。
我建议您使用本指南快速忽略PDO:http://www.phpeveryday.com/articles/PHP-Data-Object/PDO-Tutorial-P842.html
答案 1 :(得分:0)
试试这个:
$query = "INSERT INTO orders (order_supplierid, order_description, order_date, order_quoteid, order_detailid, order_type, order_supplierquote, order_supplierquotecost )
VALUES ('$osupplierid', '$odescription', '$odate', '$oquoteid', '$odetailid', '$otype', '$osupplierquote, $osupplierquotecost ')";
答案 2 :(得分:0)
$query2 = "INSERT INTO orders (order_supplierid, order_description, order_date, order_quoteid, order_detailid, order_type, order_supplierquote, order_supplierquotecost ) SELECT order_supplierid, order_description, order_date, order_quoteid, order_detailid, order_type, order_supplierquote, order_supplierquotecost FROM temporders WHERE order_id=$id";
$result2 = mysql_query($query2);