Php 从mysql表中提取id

很抱歉，这个问题看起来很简单，但我对php很陌生，我试图在索引页面中显示排名，但我不知道如何从表中的列中提取id，共有2个表：

- 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'

如何在前面显示如下内容：

<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>

我已经把身份证放在桌子上了，所以我需要一些非常相似的东西，有人能帮我吗

谢谢你的帮助 请尝试使用此查询

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

看

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>';
}

尝试使用此查询

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

看

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>';
}

尝试使用此查询

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

看

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>';
}

尝试使用此查询

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

看

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>';
}

---

你可以用像

<?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);
?> 

---

你可以用像

<?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);
?> 

---

你可以用像

<?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);
?> 

---

你可以用像

<?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);
?> 

---

如果使用2个查询，则可以先选择业务\u ID，然后在php中使用explode，然后选择业务\u名称如果使用2个查询，则可以先选择业务\u ID，然后在php中使用explode，然后选择业务\u名称如果使用2个查询，则可以先选择业务\u ID，然后在php中使用explode，然后选择业务名称如果使用2个查询，则可以首先选择业务ID，然后在php中使用explode，然后选择业务名称