我有这个数据库
表“ ADS”
ID | Visible
-----------------
15 | 1
16 | 1
17 | 1
表“ IMAGES_ADS”
ID | NAME | ID_ADS
------------------------------
100 | xlasd.jpg | 15
101 | dadsa.jpg | 15
102 | dsfsf.jpg | 16
103 | ghdfd.jpg | 17
104 | jkyhg.jpg | 17
105 | rerem.jpg | 17
现在,我想创建一个从这2个表中获取值的php页面,并按ID打印ADS列表,其中图像名称的一行与用逗号分隔的ADS ID相关。
我写了这段代码
<?php
//ENTER YOUR DATABASE CONNECTION INFO BELOW:
$servername = "localhost";
$username = "user";
$password = "xxxxxx";
$dbname = "dbname";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else {
echo 'CONNECT TO DB';
}
$sql = "SELECT ADS.ID, IMAGES_ADS.NAME FROM ADS INNER JOIN IMAGES_ADS ON ADS.ID=IMAGES_ADS.ID_ADS WHERE ADS.Visible=1 ORDER BY ID ASC";
$results = $conn->query($sql);
echo '<table>';
if($results->num_rows > 0) {
while($row = $results->fetch_assoc()) {
echo '<tr>
<td>'.$row["ID"].'</td>
<td>'.$row["NAME"].'</td>
</tr>';
}
} else {
echo '0 rows';
}
echo '</table>';
$conn->close();
?>
但是使用此代码,我得到以下结果:
15 xlasd.jpg
15 dadsa.jpg
16 dsfsf.jpg
17 ghdfd.jpg
17 jkyhg.jpg
17 rerem.jpg
但是我需要这样的结果:
15 xlasd.jpg, dadsa.jpg
16 dsfsf.jpg
17 ghdfd.jpg, jkyhg.jpg, rerem.jpg
我怎么能得到这个结果?
感谢@Vipin Jain,我在下面编写了解决方案:
<?php
//ENTER YOUR DATABASE CONNECTION INFO BELOW:
$servername = "localhost";
$username = "user";
$password = "xxxxxx";
$dbname = "dbname";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else {
echo 'CONNECT TO DB';
}
$sql = "SELECT ADS.ID, GROUP_CONCAT(IMAGES_ADS.NAME SEPARATOR', ') as b
FROM ADS INNER JOIN IMAGES_ADS
ON ADS.ID=IMAGES_ADS.ID_ADS
WHERE ADS.Visible=1
ORDER BY ID ASC";
$results = $conn->query($sql);
echo '<table>';
if($results->num_rows > 0) {
while($row = $results->fetch_assoc()) {
echo '<tr>
<td>'.$row["ID"].'</td>
<td>'.$row["b"].'</td>
</tr>';
}
} else {
echo '0 rows';
}
echo '</table>';
$conn->close();
?>
您应该使用GROUP_CONCAT
SELECT ADS.ID, GROUP_CONCAT(IMAGES_ADS.NAME)
FROM ADS
INNER JOIN IMAGES_ADS
ON ADS.ID=IMAGES_ADS.ID_ADS
WHERE ADS.Visible=1
GROUP BY ID
ORDER BY ID ASC;
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句