从多个sql表中获取信息，并使用php将其放入一个表中

目前，此代码更新MYSQLTABLE1中的用户信息。我还希望将用户信息复制到WMYSQLTABLE，我可以这样做，但我也希望将MYSQLTABLE2中的代码复制到WMYSQLTABLE中的列中。以下是我需要更改的部分：

   $sql_insert2 = "INSERT INTO ".WMYSQLTABLE2."(ip,date,address)values
    ('','$ip','$date','$address')";      
    $res_insert2 = mysql_query($sql_insert2)or die(mysql_error());
//Need to also insert code from first row from      column named 'codes' in MYSQLTABLE2.

实际的代码，目前有点混乱，而if/else语句此时也做同样的事情。它可以工作，但我得到错误“列计数与第1行的值计数不匹配”，因为我无法用MYSQLTABLE2中的代码填充最后一列

<?php
include("includes/config.php");
include("includes/mysql.php");
include("amount.php");

if(isset($_POST['func']))

    {

        $address        = $_POST['address'];
        $ip             = $_POST['ip'];
        $date           = $_POST['date'];
        $time           = $_POST['time'];
        $price      = $_POST['price'];
        $increment              = $_POST['increment'];
        $id             = $_POST['id'];
        $num = 0;



        $sql_check_address = "SELECT * FROM ".MYSQLTABLE1." WHERE address='$address'";
        $res_check_address = mysql_query($sql_check_address)or die(mysql_error());
        $num               = mysql_num_rows($res_check_address);
        $row               = mysql_fetch_assoc($res_check_address);


        if($num > 0)
            {   
                $address        = $row['address'];
                $ip             = $row['ip'];
                $date           = $row['date'];
                $oldprice   = $row['price'];
                $id             = $row['id'];

                $newprice   = $oldprice - $payAmount*200;
                $newinc        = $increment - 200;

                $sql_update1 = "UPDATE ".MYSQLTABLE1." SET ip='$ip',date='$date',price='$newprice',increment='$newinc',address='$address' WHERE id='$id'";  
                $res_update1 = mysql_query($sql_update1)or die(mysql_error());

                /////////////////Insert user info and copy code from MYSQLTABLE2 to WMYSQLTABLE2

                $sql_insert2 = "INSERT INTO ".WMYSQLTABLE2."(ip,date,address)values
                ('','$ip','$date','$address')";
                $res_insert2 = mysql_query($sql_insert2)or die(mysql_error());

            }
        else{
            $address        = $row['address'];
            $ip             = $row['ip'];
            $date           = $row['date'];
            $oldprice   = $row['price'];
            $id             = $row['id'];

            $newprice   = $oldprice - $payAmount*200;
            $newinc        = $increment - 200;

            $sql_update = "UPDATE ".MYSQLTABLE1." SET ip='$ip',date='$date',price='$newprice',increment='$newinc',address='$address' WHERE id='$id'";   
            $res_update = mysql_query($sql_update)or die(mysql_error());

            /////////////////Insert user info and copy code from MYSQLTABLE2 to WMYSQLTABLE2

            $sql_insert2 = "INSERT INTO ".WMYSQLTABLE2."(ip,date,address)values
                ('','$ip','$date','$address')";
            $res_insert2 = mysql_query($sql_insert2)or die(mysql_error());

            e
                }

    }

---

查询的一般形式是：

$sql_insert = "INSERT INTO " . WMYSQLTABLE2 . "(code, ip, date, address)
               SELECT code, '$ip', '$date', '$address'
               FROM OtherTable
               WHERE <put something here to select the row>";

$sql\u insert=“插入”。WMYSQL2。“（代码、ip、日期、地址）
选择代码“$ip”、“$date”、“$address”
从其他表格
何处”；

您的代码易受SQL注入攻击。看见也。它们不再得到维护。相反，学习，并使用or-将帮助您决定哪一个。如果选择PDO，.

$sql_insert2=“插入到“.WMYSQLTABLE2.”（ip、日期、地址）值（“$ip”、“$date”、“$address”）请仔细查看，您正在为3列传递4个值。。。[''在“$ip”之前是额外的，用户输入是有限的，并且必须遵守规则，但是是的，我可能应该更改它，因为这对我来说是一个学习过程。sql注入可以用输入框以外的其他方式执行吗？看起来我还有很多工作要做^^