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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[]= $row2['stitle'];
}
?>
我的 HTML 代码是
<table width="657">
<tr>
<td width="225"> <strong>Opinion</strong></td>
<td width="62"> <strong>Action</strong></td>
<td colspan="4"><strong>Ratings</strong></td>
<td width="54"><strong>Outlook</strong></td>
<td width="67"><strong>Rating Type</strong></td>
</tr>
<tr>
<td width="225"> </td>
<td width="62"> </td>
<td colspan="2"><b>Long Term</b></td>
<td colspan="2"><b>Short Term</b></td>
<td width="54"> </td>
<td width="67"> </td>
</tr>
<tr>
<td width="225"> </td>
<td width="62"> </td>
<td width="52"><b>Current</b></td>
<td width="45"><b>Previous</b></td>
<td width="49"><b>Current</b></td>
<td width="51"><b>Previous</b></td>
<td width="54"> </td>
<td width="67"> </td>
</tr>
<?php
for ($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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