Question

是否有一个PHP代码可以让您将2个MySQL表相互比较并允许您将缺少的条目添加到一个中？

我有两张桌子。 hs_hr_employee和权利。我想从hs_hr_employee表中的某些列添加数据，以便它们在权限表中相同。

hs_hr_employee有多行，而权限表有5行。权限表从hs_hr_employee表中的4列获取信息，emp_number，employee_id，emp_firstname，emp_lastname

以下是代码：

    <?php

$connection = mysql_connect('localhost','admin','root');

if( isset($_POST['submit']) )
{
    if( isset( $_POST['cb_change'] ) && is_array( $_POST['cb_change'] ))
    {
        foreach( $_POST['cb_change']  as $emp_number => $permission)
        {
            $sql = "UPDATE `rights` SET Permission='".mysql_real_escape_string($permission)."' WHERE emp_number='".mysql_real_escape_string($emp_number)."'";
            echo __LINE__.": sql: {$sql}\n";
            mysql_query( $sql );
        }
    }
}
?>
<p style="text-align: center;">
    <span style="font-size:36px;"><strong><span style="font-family: trebuchet ms,helvetica,sans-serif;"><span style="color: rgb(0, 128, 128);">File Database - Administration Panel</span></span></strong></span></p>
<p style="text-align: center;">
    &nbsp;</p>

<head>
<style type="text/css">
table, td, th
{
border:1px solid #666;
font-style:Calibri;
}
th
{
background-color:#666;
color:white;
font-style:Calibri;
}
</style>
</head>

    <form method="post" action="admin.php">

    <?php 


        if (!$connection)
          {
          die('Could not connect: ' . mysql_error());
          }

        mysql_select_db('users', $connection);

        //mysql_query('INSERT into rights(Emp_num, ID, Name, Surname) SELECT emp_number, employee_id, emp_firstname, emp_lastname FROM hs_hr_employee');


        $result = mysql_query("SELECT emp_number, employee_id, emp_firstname, emp_lastname, Permissions FROM rights");

        mysql_query("INSERT INTO rights (emp_number, employee_id, emp_firstname, emp_lastname)
                    SELECT emp_number, employee_id, emp_firstname, emp_lastname
                    FROM hs_hr_employee
                    ON DUPLICATE KEY UPDATE employee_id = VALUES(employee_id), emp_number = VALUES(emp_number)
                    ");

        echo "<center>";

        echo "<table >
        <tr>
        <th>Employee Number</th>
        <th>ID</th>
        <th>Name</th>
        <th>Surname</th>
        <th>Permissions</th>
        <th>Change</th>
        </tr>";

        while($row = mysql_fetch_array($result))
          {
          echo "<tr>";
          echo "<td>" . $row['emp_number'] . "</td>";
          echo "<td>" . $row['employee_id'] . "</td>";
          echo "<td>" . $row['emp_firstname'] . "</td>";
          echo "<td>" . $row['emp_lastname'] . "</td>";
          echo "<td>" . $row['Permissions'] . "</td>";
          echo "<td> <select name='cb_change[]'><option value='all'>All</option> <option value='remote'>Remote Gaming</option> <option value='landbased'>Landbased Gaming</option> <option value='general'>General Gaming</option> </select> </td>"; 
          echo "</tr>" ;
          }


          #echo "<td>" . $row['Change'] . "</td>";


          echo "</table>";

          echo "</center>";


        #$_POST['cb_permissions'];


     mysql_close($connection);


    ?>

<p style="text-align: center;">
    &nbsp;</p>
<p style="text-align: center;">
    &nbsp;</p>

<p style="text-align: right;">
    <input name="Save_Btn" type="button" value="Save" />


    </p>


</form>

知道该怎么做吗？

谢谢，布赖恩

Answer 1

我会在这里找INSERT INTO SELECT。

假设您通过社会安全号码（ssn）识别员工，并使用此值更新名称和生育年份：

mysql_query("
INSERT INTO hs_hr_employee (ssn, name, birthyear)
SELECT ssn, name, birthyear
FROM hs_hr_rights
ON DUPLICATE KEY UPDATE name = VALUES(name), birthyear = VALUES(birthyear)
");

您还可以在WHERE和FROM之间添加ON DUPLICATE。像这样：

...
FROM hs_hr_rights
WHERE birthyear IS NULL
ON DUPLICATE KEY UPDATE name = VALUES(name), birthyear = VALUES(birthyear)
...

虽然，我认为不需要复制值，因为在大多数情况下，您可以通过JOIN获取它们。

Answer 2

select * from table a where a.common not in (select b.common from table b where a.common = b.common)

并运行它

select * from table b where b.common not in (select a.common from table a where a.common = b.common)

在PHP中不可能，但在mysql中它的ans如上所述

PHP - 比较两个MySQL表

2 个答案: