如何在 mySQL 中连接更多表？答案

【问题标题】：How to connect more tables in mySQL?如何在 mySQL 中连接更多表？
【发布时间】：2015-05-07 00:10:01
【问题描述】：

有人可以告诉我如何将更多连接到我设置的表格只有一个并且我需要覆盖按钮可以打开更多...用户应该选择按下按钮您将选择表格...

    <?php



//include("connection.php");

$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="vaktija";
$tbl_name="table 1";

// Connect to server and select databse.

$con=mysql_connect("$host", "$username", "$password") or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");



$sql = "SELECT * FROM `table 1`";
$result=mysql_query($sql);

?>

<html>


<body bgcolor="#f5fff5">
<div align="center" </div>
<br>
<br>
<b>Vaktija za SBK</b>
<br>
<br>

<div align="left" </div>
&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;
<strong><a href="main_page.php" >  Back to main </a> </strong>
<br>
<br>


</html>

<?php


echo "<table width='1100' border='0' align='center' cellpadding='0' cellspacing='1' bgcolor='#99FF33'>

<tr>
<th>Dan</th>
<th>Zora</th>
<th>Izlazak Sunca</th>
<th>Podne</th>
<th>Ikindija</th>
<th>Aksam</th>
<th>Jacija</th>


</tr>";

$color="1";

while($rows = mysql_fetch_array($result)) 
  {
  if($color==1){

  echo "<tr bgcolor='#FFFFFF'>

<td>".$rows['Dan']."</td><td>".$rows['Zora']."</td><td>".$rows['Izlazak Sunca']."</td><td>".$rows['Podne']."</td><td>".$rows['Ikindija']."</td><td>".$rows['Aksam']."</td><td>".$rows['Jacija']."</td>
</tr>";

  $color="2";
}


else {
echo "<tr bgcolor='#FFFFCC'>


<td>".$rows['Dan']."</td><td>".$rows['Zora']."</td><td>".$rows['Izlazak Sunca']."</td><td>".$rows['Podne']."</td><td>".$rows['Ikindija']."</td><td>".$rows['Aksam']."</td><td>".$rows['Jacija']."</td>
</tr>";

$color="1";
}
}

echo "</table>";


mysql_close($con);

?>



<html>


<br>
<br>
<br>


</html>

【问题讨论】：

要查询一个不同的表，构造'$sql = "SELECT * FROM table 1";'基于用户输入。要同时查询多个表，请了解JOIN。

标签： html mysql visual-studio xampp

【解决方案1】：

请阅读JOIN

一个简单的例子可以是：你有

Table 1
+----+-------+----------+
| id | name  | lastname |
+----+-------+----------+
|  1 | Peter | Smith    |
|  2 | Susan | Johnson  |
+----+-------+----------+

还有表2

+----+--------------+-------------+-------+
| id |   address    |    city     | state |
+----+--------------+-------------+-------+
|  1 | 111 NE 11st. | Manhattan   | NY    |
|  2 | 222 NW 22st. | Long Island | NY    |
+----+--------------+-------------+-------+

如果您运行此查询

SELECT 
    *
FROM
    table1 as t1
        JOIN
    table2 as t2 ON t1.id = t2.id

你会得到Table1+Table2

+----+-------+-------------+---------+--------------+-------------+----+
| id | name  | lastname id | address |     city     |    state    |    |
+----+-------+-------------+---------+--------------+-------------+----+
|  1 | Peter | Smith       |       1 | 111 NE 11st. | Manhattan   | NY |
|  2 | Susan | Johnson     |       2 | 222 NW 22st. | Long Island | NY |
+----+-------+-------------+---------+--------------+-------------+----+

注意这只是一个虚拟的例子，只是为了说明答案

【讨论】：