SQL 语法错误 MySQL 服务器版本在 'CompletedNo SELECT (OrderNo.) FROM `order` WHERE `OrderNo.` 附近使用正确的语法答案

【问题标题】：error in SQL syntax MySQL server version for the right syntax to use near 'CompletedNo SELECT (OrderNo.) FROM `order` WHERE `OrderNo.`SQL 语法错误 MySQL 服务器版本在 'CompletedNo SELECT (OrderNo.) FROM `order` WHERE `OrderNo.` 附近使用正确的语法
【发布时间】：2016-10-14 13:20:30
【问题描述】：

以下 php 脚本用于从数据库中检索一组数据（订单号）！单击删除链接时，我想删除该特定删除链接所在的行并将数据传输到“传输”表！

<?php
                require("includes/db.php");

                $sql="SELECT * FROM `order` ";
                $result=mysqli_query($db,$sql);
    echo"<head>";
    echo'
            <link rel="stylesheet" href="view.css">
                <head>


            ';
    echo"</head>";

    echo "<body >";
    echo "<table border=1 cellspacing=0 cellpadding=4 > " ;
    echo"<tr bgcolor=grey>";
    echo"<td align=center>";
    echo "<font size=4>";
    echo "<B>";
    echo "Order No.";
    echo "</B>";
    echo"</td>";
    echo"</tr>";


    while($row=mysqli_fetch_array($result))
    {

        echo"<tr>";
         echo"<td align=center>";
        echo $row["OrderNo."];
         echo "<br>";
         echo"</td>";
       echo "<td align=center>";
    echo "<a href='delete.php?del=";
    echo $row['OrderNo.'];
    echo "'>delete</a>";
    echo "<br>";
    echo"</td>";
        echo"</tr>";
    }
        echo"</table>";

?>

单击删除链接时会执行以下 php 脚本！当仅执行删除函数的查询时，它起作用了，但是在插入另一个查询以将值传输到“传输”表之后，编写了一个语法发生错误。

<?php
     include("includes/db.php");

    if( isset($_GET['del']) )
    {
        $id = $_GET['del'];
        $sql1="INSERT INTO transfer CompletedNo SELECT (OrderNo.) FROM `order` WHERE `OrderNo.` = '$id' ";
        $res1= mysqli_query($db,$sql1) or die("Failed".mysqli_error($db));
        $sql2= "DELETE FROM `order` WHERE `OrderNo.` = '$id' ";
        $res2= mysqli_query($db,$sql2) or die("Failed".mysqli_error($db));

    }
?>

请帮我改正错误只在线上

$sql1="INSERT INTO transfer CompletedNo SELECT (OrderNo.) FROM `order` WHERE `OrderNo.` = '$id' ";

【问题讨论】：

标签： php sql database syntax-error transfer

【解决方案1】：

您需要在Insert 列列表周围使用Parenthesis

INSERT INTO transfer (CompletedNo) 
SELECT `OrderNo.` FROM `order` WHERE `OrderNo.` = '$id'

【讨论】：