php - 如果一个查询包含 Null 结果，如何从多个查询打印？

标签 php mysql null echo

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

第一个查询:

 $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 并在 client_id 的基础上我剩下的查询将工作。

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 query Bring client_id 那么query 2和query 4会被执行但是如果有没有 client_id 那么 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

有时我的查询 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。

最佳答案

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>

关于php - 如果一个查询包含 Null 结果，如何从多个查询打印？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32709282/

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