如何在 PHP Mysql 中用选定的旧记录回显新记录？答案

【问题标题】：HOW to echo New record with selected old record in PHP Mysql?如何在 PHP Mysql 中用选定的旧记录回显新记录？
【发布时间】：2015-08-05 10:48:29
【问题描述】：

我是 PhP 的新手。我有一个问题

"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

现在我想打印结果查询的第一行并且我成功了。但现在我只想从第二行回显两列ltitle 和Stitle。这里我失败了。

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误的。你们能帮帮我吗？

【问题讨论】：

您是否在不添加html代码的情况下将结果打印在php中？我建议你先这样做。
@cedricliang 我发布了一张没有 html 的查询结果图片
尝试在你的sql中将DESC改为ASC？
@cedricliang 还没有工作
对，这就是我在之前评论中的意思。为了避免这种情况，你应该修改你的 sql.尝试“ORDER BY r.id DESC LIMIT 1,1”直接获取第二行。

标签： php mysql database echo

【解决方案1】：

用“ORDER BY r.id DESC LIMIT 1,1”修改sql直接得到第二行:)

【讨论】：

非常感谢。按照你的提示我做到了:)

【解决方案2】：

雅虎...！！！我已经做了。谢谢cedricliang的帮助

看看下面我更新的代码

<?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";


$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

【讨论】：