为什么这个PHP代码没有插入正确的方法

Question

在这里，当我尝试插入运行的第一个查询和插入的数据时，我的第二个查询不是Working.At，而是同时注释第一个查询，第二个很好。两者都意味着我被困住了

if($order_no!="" && $cus_name!="")
    {
        //fs_code: set primary key into auto means use this qry before inserting
        $query1 = "SET IDENTITY_INSERT order_creation ON";
        $stmt1 = sqlsrv_query($conn, $query1);

        $sql = "INSERT INTO order_creation (order_no,cus_name,cus_id,types,created_date,Status,id) VALUES('$order_no','$cus_name','$cus_id','$type',FORMAT(CURRENT_TIMESTAMP,'M/d/yyyy h:mm:sstt'),'$Status','$fmax')";

        if(sqlsrv_query($conn,$sql))
            {
                echo 'success-1';
            }
        else
            {
                echo 'failure-1';
            }
    }
for($i=0;$i<count($goods_name);$i++)
    {       
        if($goods_name[$i]!="")
            {                       
                $sql = "INSERT INTO order_Goods (order_no,goods_name,goods_qty,date,created_date,Balance_Qty,Supply_Qty,ProductPrice,id) VALUES('$order_no','$goods_name[$i]','$goods_qty[$i]',FORMAT(CURRENT_TIMESTAMP,'d/MM/yy h:mm:sstt'),'$created_date[$i]','$goods_qty[$i]','$Supply_Qty','$ProductPrice[$i]','$get_maxres')";

                if(sqlsrv_query($conn,$sql))
                    {
                        echo 'success-2';
                    }
                else
                    {
                        echo 'failure-2';
                    }
                $get_maxres++;
            }
    }

sqlsrv_close($conn);

Answer 1

您没有使用正确的语法将数组值替换为字符串。当数组索引包含变量时，需要使用“复合”语法，其中数组变量被{}包围。

$sql = "INSERT INTO order_Goods (order_no,goods_name,goods_qty,date,created_date,Balance_Qty,Supply_Qty,ProductPrice,id)
     VALUES('$order_no','{$goods_name[$i]}','{$goods_qty[$i]}',FORMAT(CURRENT_TIMESTAMP,'d/MM/yy h:mm:sstt'),'{$created_date[$i]}','{$goods_qty[$i]}','$Supply_Qty','{$ProductPrice[$i]}','$get_maxres')";

请参阅PHP文档中的变量解析一节。

您还需要在第二个表上启用IDENTITY_INSERT。将其放在for循环之前：

sqlsrv_query($conn, "SET IDENTITY_INSERT order_creation OFF");
sqlsrv_query($conn, "SET IDENTITY_INSERT order_Goods ON");

Answer 2

是的，我把这个修好了

SET IDENTITY_INSERT Table1 ON

INSERT INTO Table1
/*Note the column list is REQUIRED here, not optional*/
            (OperationID,
             OpDescription,
             FilterID)
VALUES      (20,
             'Hierachy Update',
             1)

SET IDENTITY_INSERT Table1 OFF