PHP - 比较两个 MySQL 表 [英] PHP - Compare two MySQL tables
问题描述
是否有 PHP 代码可以让您将 2 个 MySQL 表相互比较，并将缺失的条目添加到一个表中?
Is there a PHP code that let's you compare 2 MySQL tables with each other and lets you add missing entries into one?
我有两张桌子.hs_hr_employee 和权利.我想从 hs_hr_employee 表的某些列添加数据，以便它们在权限表中相同.
I have two tables. hs_hr_employee and rights. I want to add data from certain columns from the hs_hr_employee table so that they would be the same in the rights tables.
hs_hr_employee 有多行，而权限表有 5 行.权限表从 hs_hr_employee 表的 4 列中获取信息，emp_number、employee_id、emp_firstname、emp_lastname
hs_hr_employee has multiple rows, whereas the rights table has 5 rows. The rights table gets the info from 4 columns from the hs_hr_employee table, emp_number, employee_id, emp_firstname, emp_lastname
代码如下:
<?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( $sql );
}
}
}
?>
<p style="text-align: center;">
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<p style="text-align: center;">
</p>
<head>
<style type="text/css">
table, td, th
{
border:1px solid #666;
font-style:Calibri;
}
th
{
background-color:#666;
color:white;
font-style:Calibri;
}
</style>
</head>
<form method="post" action="admin.php">
<?php
if (!$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;">
</p>
<p style="text-align: center;">
</p>
<p style="text-align: right;">
<input name="Save_Btn" type="button" value="Save" />
</p>
</form>
知道怎么做吗?
谢谢，布莱恩
推荐答案
我会在这里使用 INSERT INTO SELECT.
假设您通过社会安全号码 (ssn) 识别出一名员工，并且您使用此值来更新姓名和出生年份:
Assume you recognize an employee by the Social Security Number (ssn), and you use this value to update, for instance, name and birthyear:
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FROM 和ON DUPLICATE 之间添加WHERE.像这样:
You can also add a WHERE in between FROM and ON DUPLICATE. Like so:
...
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.
Although, I don't see the need of copying values since in most cases you can fetch them through JOINs.
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