我在显示数据库表时遇到问题。我的数据库表(活动)具有以下值,ID(主键),类型,标题,描述和图片。我试图以表格格式检索和显示标题,描述和图片,但面临错误
"注意:未定义索引:标题,注意:未定义索引:描述 并注意:未定义索引:图片"
我无法解决它。 以下是我的PHP代码:
<div class="col-md-4 col-md-offset-1">
<div id="content">
<table><?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sentosa_resort";
$conn = new mysqli($servername, $username, $password,
$dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$query = "SELECT title,description,picture FROM activites";
$result = $conn->query($sql);;
// start a table tag in the HTML
while($row=mysqli_fetch_array($result)){
$title = $row['title'];
$description = $row['description'];
$picture = $row['picture'];
echo "<tr>";
echo "<td>" . $title . "</td>";
echo "<td>" . $description . "</td>";
echo "<td>" . $picture . "</td>";
echo "</tr>";
}
?></table>
</div>
</div>
答案 0 :(得分:0)
请检查您的代码,以获取第16行中的额外分号...
OR
您可以使用mysqli_fetch_row
<div class="col-md-4 col-md-offset-1">
<div id="content">
<table>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sentosa_resort";
$conn = new mysqli($servername, $username, $password,
$dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$query = "SELECT title,description,picture FROM activites";
$result = $conn->query($query);
// start a table tag in the HTML
while($row=mysqli_fetch_row($result)){
$title = $row['0'];
$description = $row['1'];
$picture = $row['2'];
echo "<tr>";
echo "<td>" . $title . "</td>";
echo "<td>" . $description . "</td>";
echo "<td>" . $picture . "</td>";
echo "</tr>";
}
?>
</table>
</div>
</div>
答案 1 :(得分:0)
您的代码中存在错误,即:
$query = "SELECT title,description,picture FROM activites";
$result = $conn->query($sql);; // here is error
请更改上面的错误行,如下所示:
$result = $conn->query($query);
解决方案:将$ sql更改为$ query,因为您的查询字符串存储在$ query变量而非$ sql中并删除了额外的分号
答案 2 :(得分:0)
我已更改您的代码,请检查
<div class="col-md-4 col-md-offset-1">
<div id="content">
<table><?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sentosa_resort";
$conn = mysqli_connect($servername, $username, $password,
$dbname);
$query = "SELECT title,description,picture FROM activites";
$result = mysqli_query($conn,$query);
while($row=mysqli_fetch_assoc($result)){
$title = $row['title'];
$description = $row['description'];
$picture = $row['picture'];
echo "<tr>";
echo "<td>" . $title . "</td>";
echo "<td>" . $description . "</td>";
echo "<td>" . $picture . "</td>";
echo "</tr>";
}
?></table>
</div>
</div>
答案 3 :(得分:-1)
您收到这些通知是因为您使用的是mysqli_fetch_array()
,而是需要使用mysqli_fetch_assoc()
。
如果您想继续使用mysqli_fetch_array()
,请将代码更改为:
$title = $row[0];
$description = $row[1];
$picture = $row[2];