从mysql表中提取id答案

【问题标题】：extracting id's from a mysql table从mysql表中提取id
【发布时间】：2014-03-27 04:28:17
【问题描述】：

抱歉这个问题看起来很简单，但我是 php 新手，我正在尝试在索引页面中显示排名，但我不知道如何从列中提取 id一张桌子，有两张桌子：

- raking table -
    city = 'Boston'
    ranking_name = 'the best 5 places'
    business_ids = '67,43,1,6,78'

- business table -
    business_id = '67'
    business_name = 'planet pizza'

如何在前面显示这样的内容：

<ul>
<h2>the best 5 places</h2>
<li><span>1</span><?php echo $business_name ?></li>
<li><span>2</span><?php echo $business_name ?></li>
<li><span>3</span><?php echo $business_name ?></li>
<li><span>4</span><?php echo $business_name ?></li>
<li><span>5</span><?php echo $business_name ?></li>
</ul>

我已经在表格中找到了 id，所以我需要一些非常相似的东西，谁能帮助我。

感谢所有帮助问候

【问题讨论】：

如果您使用2个查询，您可以先选择business_ids，然后在php中使用explode，然后选择business_names

标签： php mysql

【解决方案1】：

尝试使用此查询

SELECT * FROM business_table WHERE business_id IN ( SELECT business_id FROM ranking_table WHERE city='Boston' );

见Demo

你的代码，

$con=mysqli_connect("HOSTNAEM","USERNAME","PASSWORD","DB") die('Could not connect: ' . mysql_error());

$query = "SELECT * FROM business_table WHERE business_id IN ( SELECT business_id FROM ranking_table WHERE city='Boston' )";

while($row = mysqli_fetch_array($query))
{
  echo '<li><span>1</span>'+$row['$business_name']+'></li></br>';
}

【讨论】：

【解决方案2】：

你可以使用类似......

<?php
$con=mysqli_connect("HOST","USER","PASS","DB");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = mysqli_query($con,"SELECT * FROM business_table WHERE business_id IN ( SELECT business_id FROM ranking_table WHERE city='Boston' )");

while($row = mysqli_fetch_array($result))
  {
  echo '<li><span>1</span>'+$row['$business_name']+'></li>';
  echo "<br>";
  }

mysqli_close($con);
?>

W3schools.com-Reference

【讨论】：