從mysql表中提取ID

Question

抱歉，這個問題似乎很簡單，但是我是php的新手，我正試圖在索引頁中顯示排名，但是我不知道如何從表的列中提取ID，有2張桌子：

- 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'

我如何在前面顯示如下內容：

<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>

我的桌子上已經有身份證了，所以我需要非常相似的東西，有人可以幫我嗎。

感謝所有的幫助

Answer 1

嘗試使用此查詢

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

見演示

您的代碼，

$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>';
}

Answer 2

您可以使用類似...的方法。

<?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);
?> 

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