如何在PHP Mysql中用選定的舊記錄回顯新記錄？

Question

我是PhP新手。 我有一個查詢

"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";

我的查詢結果是

現在我想打印結果查詢的第一行，然后我成功。 但是現在我只想從第二行回ltitle和Stitle兩列。 在這里我失敗了。

這是我的代碼

<?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
}?>

我的代碼的結果是

在我的代碼的Previos列中，我要打印數據庫表的第二行數據。 您可以看到我的結果是錯誤的。 你們能幫我嗎？

Answer 1

用“ ORDER BY r.id DESC LIMIT 1,1”修改sql以直接獲取第二行:)

Answer 2

雅虎... !!! 我已經做了。 謝謝cedricliang對我的幫助

只需看下面我更新的代碼

<?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
    }
}?>

我的代碼的結果是

如何在PHP Mysql中用選定的舊記錄回顯新記錄？

問題描述

2 個解決方案

解決方案1
2 2015-08-05 12:26:37

解決方案2
0 2015-08-05 12:21:58

如何在PHP Mysql中用選定的舊記錄回顯新記錄？

問題描述

2 個解決方案

解決方案1 2 2015-08-05 12:26:37

解決方案2 0 2015-08-05 12:21:58

解決方案1
2 2015-08-05 12:26:37

解決方案2
0 2015-08-05 12:21:58