如果一个查询包含 Null 结果，如何从多个查询中打印？答案

【问题标题】：How to print from multiple queries if one query contain Null result?如果一个查询包含 Null 结果，如何从多个查询中打印？
【发布时间】：2015-09-22 05:49:37
【问题描述】：

我是 PHP 新手。我在浏览器上打印数据时遇到问题。我有五个疑问。我的四个查询是基于第一个查询的结果

第一次查询：

 $opinion_id = "SELECT `client_id` FROM `pacra_client_opinion_relations` WHERE `opinion_id` = 379";
$result = mysql_query($opinion_id) or die;
$row = mysql_fetch_assoc($result);
$client_id = $row['client_id'];

此查询获取client_id 并基于client_id 我剩余的查询将有效。

查询 2：

$q_opinion="SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as opinion, r.notification_date, t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle, pr.opinion_id, pc.id, pr.client_id as pr_client, pc.address, pc.liaison_one, city.id, pc.head_office_id, city.city, pc.title as cname
FROM og_ratings r 
    inner join
(
  select max(notification_date) notification_date,
    client_id
  from og_ratings
  group by client_id
  ) r2
  on r.notification_date = r2.notification_date
  and r.client_id = r2.client_id
LEFT JOIN og_companies c
ON r.client_id = c.id
LEFT JOIN og_rating_types t
ON r.rating_type_id = t.id
LEFT JOIN og_actions a
ON r.pacra_action = a.id
LEFT JOIN og_outlooks o
ON r.pacra_outlook = o.id
LEFT JOIN og_lterms l
ON r.pacra_lterm = l.id
LEFT JOIN og_sterms s
ON r.pacra_sterm = s.id
LEFT JOIN pacra_client_opinion_relations pr
ON pr.opinion_id = c.id
LEFT JOIN pacra_clients pc
ON pc.id = pr.client_id
LEFT JOIN city
ON city.id = pc.head_office_id
WHERE r.client_id  IN (SELECT opinion_id FROM pacra_client_opinion_relations WHERE client_id = $client_id)
";

查询 3：

$q_opinion1 = "SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as opinion, r.notification_date, t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle, pr.opinion_id, pc.id, pr.client_id as pr_client, pc.address, pc.liaison_one, city.id, pc.head_office_id, city.city, pc.title as cname
FROM og_ratings r 
    inner join
(
  select max(notification_date) notification_date,
    client_id
  from og_ratings
  group by client_id
  ) r2
  on r.notification_date = r2.notification_date
  and r.client_id = r2.client_id
LEFT JOIN og_companies c
ON r.client_id = c.id
LEFT JOIN og_rating_types t
ON r.rating_type_id = t.id
LEFT JOIN og_actions a
ON r.pacra_action = a.id
LEFT JOIN og_outlooks o
ON r.pacra_outlook = o.id
LEFT JOIN og_lterms l
ON r.pacra_lterm = l.id
LEFT JOIN og_sterms s
ON r.pacra_sterm = s.id
LEFT JOIN pacra_client_opinion_relations pr
ON pr.opinion_id = c.id
LEFT JOIN pacra_clients pc
ON pc.id = pr.client_id
LEFT JOIN city
ON city.id = pc.head_office_id
WHERE r.client_id  IN (SELECT client_id FROM og_ratings WHERE client_id = 379)";

查询 4：

$q_opinion2="SELECT
   r.client_id,c.id,t.id,a.id,o.id,c.name as opinion, r.notification_date, t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle, pr.opinion_id, pc.id, pr.client_id as pr_client, pc.address, pc.liaison_one, city.id, pc.head_office_id, city.city, pc.title as cname
FROM
  og_ratings r 


  INNER JOIN (
    SELECT client_id, max(notification_date) notification_2nd_date
    FROM og_ratings
    WHERE client_id IN (SELECT `opinion_id` FROM `pacra_client_opinion_relations` WHERE `client_id` = $client_id) AND
      (client_id, notification_date) NOT IN (
        SELECT client_id, max(notification_date)
        FROM og_ratings GROUP BY client_id
          ORDER BY  client_id DESC)
    GROUP BY client_id
      ORDER BY  client_id DESC
   ) r2
  ON r.notification_date = r2.notification_2nd_date
     AND r.client_id = r2.client_id
  LEFT JOIN og_companies c ON r.client_id = c.id
  LEFT JOIN og_rating_types t ON r.rating_type_id = t.id
  LEFT JOIN og_actions a ON r.pacra_action = a.id
  LEFT JOIN og_outlooks o ON r.pacra_outlook = o.id
  LEFT JOIN og_lterms l ON r.pacra_lterm = l.id
  LEFT JOIN og_sterms s ON r.pacra_sterm = s.id
  LEFT JOIN pacra_client_opinion_relations pr ON pr.opinion_id = c.id
  LEFT JOIN pacra_clients pc ON pc.id = pr.client_id
  LEFT JOIN city ON city.id = pc.head_office_id
WHERE
  r.client_id IN (
    SELECT opinion_id FROM pacra_client_opinion_relations
    WHERE client_id = $client_id
  )";

查询 5：

$q_opinion3="SELECT
   r.client_id,c.id,t.id,a.id,o.id,c.name as opinion, r.notification_date, t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle, pr.opinion_id, pc.id, pr.client_id as pr_client, pc.address, pc.liaison_one, city.id, pc.head_office_id, city.city, pc.title as cname
FROM
  og_ratings r 


  INNER JOIN (
    SELECT client_id, max(notification_date) notification_2nd_date
    FROM og_ratings
    WHERE client_id IN (SELECT client_id FROM og_ratings WHERE client_id = 379) AND
      (client_id, notification_date) NOT IN (
        SELECT client_id, max(notification_date)
        FROM og_ratings GROUP BY client_id
          ORDER BY  client_id DESC)
    GROUP BY client_id
      ORDER BY  client_id DESC
   ) r2
  ON r.notification_date = r2.notification_2nd_date
     AND r.client_id = r2.client_id
  LEFT JOIN og_companies c ON r.client_id = c.id
  LEFT JOIN og_rating_types t ON r.rating_type_id = t.id
  LEFT JOIN og_actions a ON r.pacra_action = a.id
  LEFT JOIN og_outlooks o ON r.pacra_outlook = o.id
  LEFT JOIN og_lterms l ON r.pacra_lterm = l.id
  LEFT JOIN og_sterms s ON r.pacra_sterm = s.id
  LEFT JOIN pacra_client_opinion_relations pr ON pr.opinion_id = c.id
  LEFT JOIN pacra_clients pc ON pc.id = pr.client_id
  LEFT JOIN city ON city.id = pc.head_office_id
WHERE
  r.client_id IN (
    SELECT client_id FROM og_ratings WHERE client_id = 379)
  )";

如果query 1查询带client_id则query 2和query 4将被执行，但如果没有client_id则query 3和query 5将被执行。

if ($client_id == NULL)
{
    $query = $q_opinion1;
    $query1 = $q_opinion3;
    }
    else{
$query = $q_opinion;
$query1 = $q_opinion2;
    }
  $result1 = mysql_query($query) or die;
  $result2 = mysql_query($query1) or die;

剩下的PHP代码是

$opinion = array();

while($row1 = mysql_fetch_assoc($result1))
{        
    $opinion[]= $row1['opinion'];
    $action[]= $row1['atitle'];
    $long_term[]= $row1['ltitle'];
    $outlook[]= $row1['otitle'];
    $rating_type[]= $row1['ttitle'];
    $short_term[]= $row1['stitle'];


}
while($row2 = mysql_fetch_assoc($result2))
{
    $p_long_term[]= $row2['ltitle'];
    $p_short_term[]= $row2['stitle'];
}
?>

我的 HTML 代码是

<table width="657">
        <tr>
            <td width="225"> <strong>Opinion</strong></td>
            <td width="62"> <strong>Action</strong></td>
            <td colspan="4"><strong>Ratings</strong></td>
            <td width="54"><strong>Outlook</strong></td>
            <td width="67"><strong>Rating Type</strong></td>
        </tr>
        <tr>
          <td width="225">&nbsp;</td>
          <td width="62">&nbsp;</td>
          <td colspan="2"><b>Long Term</b></td>
          <td colspan="2"><b>Short Term</b></td>
          <td width="54">&nbsp;</td>
          <td width="67">&nbsp;</td>
        </tr>
        <tr>
          <td width="225">&nbsp;</td>
          <td width="62">&nbsp;</td>
          <td width="52"><b>Current</b></td>
          <td width="45"><b>Previous</b></td>
          <td width="49"><b>Current</b></td>
          <td width="51"><b>Previous</b></td>
          <td width="54">&nbsp;</td>
          <td width="67">&nbsp;</td>
        </tr>
        <?php
        for ($i=0; $i<count($opinion); $i++) {
    //if ($opinion[$i] == "")continue;
        ?>


    <tr>
           <td><?php echo $opinion[$i]?></td>
          <td><?php echo $action[$i] ?></td>
          <td><?php echo $long_term[$i] ?></td>
          <td><?php echo $p_long_term[$i]?></td>
          <td><?php echo $short_term[$i] ?></td>
          <td><?php echo $p_short_term[$i] ?></td>
          <td><?php echo $outlook[$i] ?></td>
          <td><?php echo $rating_type[$i] ?></td>
        </tr>

        <?php
        }
?>
 </table>

现在的问题是

有时我的query 5 包含空结果。由于这个问题，我的query 3 数据没有打印出来。我希望如果我的任何查询包含 Null 结果，我的其余数据将打印在我的页面上。

【问题讨论】：

不要使用 MySQL_* 函数这个扩展在 PHP 5.5.0 中被弃用，在 PHP 7.0.0 中被移除。相反，应该使用 MySQLi 或 PDO_MySQL 扩展名。
@sunny 我可能错了，但很难猜出你真正的问题是什么？并且请删除不相关的代码以仅关注问题。
@sitilge 所有代码都是相关的。我的问题是我显示来自两个查询组合的数据。有时我使用查询 2 和 4 的组合显示数据，有时我使用查询 3 和 5 的组合显示数据。如果我的查询 5 有空记录，那么它不会显示查询 3 的结果。我想那如果 Query 5 为 Null，那么它将能够显示 Query 3 的结果
@sunny 你试过if (empty($result5)) {while($row3 = mysql_fetch_assoc($result3)){print_r($row3}} 吗？
@sitilge 如果我创建另一个循环，那么我如何使用 HTML 显示我的记录 while($row2 = mysql_fetch_assoc($result2)) { $p_long_term[]= $row2['ltitle']; $p_short_term[]= $row2['stitle']; } 我正在使用这个循环并获取记录 <td><?php echo $p_long_term[$i]?></td> <td><?php echo $p_short_term[$i] ?></td>

标签： php mysql null echo

【解决方案1】：

您似乎正在遍历意见数组并使用索引来选择 $p_long_term[] 和 $p_short_term[] 数组中的相应值。如果查询 5 失败，这些数组将为空。

<tr>
       <td><?php echo $opinion[$i]?></td>
      <td><?php echo $action[$i] ?></td>
      <td><?php echo $long_term[$i] ?></td>
      <td><?php echo $p_long_term[$i]?></td>**
      <td><?php echo $short_term[$i] ?></td>
      <td><?php echo $p_short_term[$i] ?></td>
      <td><?php echo $outlook[$i] ?></td>
      <td><?php echo $rating_type[$i] ?></td>
    </tr>

在 echo 之前检查 key 是否存在。

<td><?php if(array_key_exists ($i, $p_long_term))echo $p_long_term[$i]?></td>
<td><?php if(array_key_exists ($i, $p_short_term))echo $p_short_term[$i] ?></td>

【讨论】：

尝试评论 echo $p_long_term 和 $p_short_term 并检查问题是否仍然存在。并在发现错误时向后工作

【解决方案2】：

问题在于循环期间的逻辑。您正在回显 query5 数组中不存在的内容。您可以尝试先检查元素是否存在，然后再回显到输出流。

for ($i=0; $i<count($opinion); $i++) {
    echo '<tr>';
    echo isset($opinion[$i])?  '<td>'. $opinion[$i] .'</td>' : '';
    echo isset($action[$i])?  '<td>'. $action[$i] .'</td>' : '';
    echo isset($long_term[$i])?  '<td>'. $long_term[$i] .'</td>' : '';
    echo isset($p_long_term[$i])?  '<td>'. $p_long_term[$i] .'</td>' : '';
    echo isset($short_term[$i])?  '<td>'. $short_term[$i] .'</td>' : '';
    echo isset($p_short_term[$i])?  '<td>'. $p_short_term[$i] .'</td>' : '';
    echo isset($outlook[$i])?  '<td>'. $outlook[$i] .'</td>' : '';
    echo isset($rating_type[$i])?  '<td>'. $rating_type[$i] .'</td>' : '';
    echo '</tr>';
}

【讨论】：

@sitilge 它显示Deprecated: mysql_connect() 错误
@sunny 您正在使用已弃用的方法。我想知道为什么这些充满错误信息的噩梦般的网站仍然从谷歌搜索中弹出......不要使用 mysql_* ，最好切换到 PDO 或其他抽象层，如 Doctrine
尝试启用

【解决方案3】：

使用这个

$max = max(count($opinion),count($p_long_term));
for ($i=0; $i<$max; $i++) {
    echo '<tr>';
    echo isset($opinion[$i])?  '<td>'. $opinion[$i] .'</td>' : '';
    echo isset($action[$i])?  '<td>'. $action[$i] .'</td>' : '';
    echo isset($long_term[$i])?  '<td>'. $long_term[$i] .'</td>' : '';
    echo isset($p_long_term[$i])?  '<td>'. $p_long_term[$i] .'</td>' : '';
    echo isset($short_term[$i])?  '<td>'. $short_term[$i] .'</td>' : '';
    echo isset($p_short_term[$i])?  '<td>'. $p_short_term[$i] .'</td>' : '';
    echo isset($outlook[$i])?  '<td>'. $outlook[$i] .'</td>' : '';
    echo isset($rating_type[$i])?  '<td>'. $rating_type[$i] .'</td>' : '';
    echo '</tr>';
}

【讨论】：

这与@faiz 的答案相同，除了$max 的派生方式