我是PhP的新手。我有一个查询
"SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as cname,t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle
FROM og_ratings r
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
WHERE c.id= 338
ORDER BY r.id DESC
LIMIT 2";
我的查询结果是
现在我要打印第一行结果查询,我成功了。但现在我想从第二行只回显两列ltitle
和Stitle
。我在这里失败了。
这是我的代码
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pacra1";
$conn = new mysqli($servername, $username, $password, $dbname);
//$id2 = $_GET['id'];
$sql= "SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as cname,t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle
FROM og_ratings r
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
WHERE c.id= 338
ORDER BY r.id DESC
LIMIT 1";
$result = $conn->query($sql);
//$array = array('1','2','3');
while ($row = $result->fetch_assoc()){
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<table border="1">
<tr>
<td> ID </td>
<td> <?php echo $row['client_id'] ?> </td>
</tr>
<tr>
<td>Name </td>
<td><?php echo $row['cname'] ?> </td>
</tr>
<tr>
<td>Rating Type </td>
<td><?php echo $row['ttitle'] ?> </td>
</tr>
<tr>
<td>Action </td>
<td><?php echo $row['atitle'] ?> </td>
</tr>
<tr>
<td>Outlook </td>
<td><?php echo $row['otitle'] ?></td>
</tr>
<tr>
<td rowspan="2">Long Term Rating </td>
<td>Current (<?php echo $row['ltitle'] ?>) <tr><td>Previous (<?php echo $row['ltitle'][0] ?>)</td> </tr></td>
</tr>
<tr>
<td rowspan="2">Short Term Rating </td>
<td>Current (<?php echo $row['stitle'] ?>) <tr><td>Previous (<?php echo $row['stitle'][0] ?>)</td> </tr></td>
</tr>
</table>
</body>
</html>
<?php
}?>
我的代码的结果是
在我的代码的Previos列中,我想打印我的db表的第二行数据。 你可以看到我的结果是错的。你能帮帮我吗?
答案 0 :(得分:2)
使用&#34; ORDER BY r.id DESC LIMIT 1,1&#34;修改sql;直接获得第二行:)
答案 1 :(得分:0)
雅虎... !!!我已经做了。感谢您cedricliang的帮助
只需查看下面的更新代码
即可<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pacra1";
$conn = new mysqli($servername, $username, $password, $dbname);
//$id2 = $_GET['id'];
$sql= "SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as cname,t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle
FROM og_ratings r
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
WHERE c.id= 338
ORDER BY r.id DESC
LIMIT 1";
$sql1= "SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as cname,t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle
FROM og_ratings r
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
WHERE c.id= 338
ORDER BY r.id DESC
LIMIT 1,1";
$result = $conn->query($sql);
$result1 = $conn->query($sql1);
//$array = array('1','2','3');
while ($row = $result->fetch_assoc()){
while ($row1 = $result1->fetch_assoc()){
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<table border="1">
<tr>
<td> ID </td>
<td> <?php echo $row['client_id'] ?> </td>
</tr>
<tr>
<td>Name </td>
<td><?php echo $row['cname'] ?> </td>
</tr>
<tr>
<td>Rating Type </td>
<td><?php echo $row['ttitle'] ?> </td>
</tr>
<tr>
<td>Action </td>
<td><?php echo $row['atitle'] ?> </td>
</tr>
<tr>
<td>Outlook </td>
<td><?php echo $row['otitle'] ?></td>
</tr>
<tr>
<td rowspan="2">Long Term Rating </td>
<td>Current (<?php echo $row['ltitle'] ?>) <tr><td>Previous (<?php echo $row1['ltitle'] ?>)</td> </tr></td>
</tr>
<tr>
<td rowspan="2">Short Term Rating </td>
<td>Current (<?php echo $row['stitle'] ?>) <tr><td>Previous (<?php echo $row1['stitle'] ?>)</td> </tr></td>
</tr>
</table>
</body>
</html>
<?php
}
}?>
我的代码的结果是