php和mysql中的链接
我从mysql数据库加载数据并将其显示到listview jqm中。这是我的密码:php和mysql中的链接,php,mysql,Php,Mysql,我从mysql数据库加载数据并将其显示到listview jqm中。这是我的密码: <?php include "connect.php"; $result = mysqli_query($con,"SELECT * FROM story"); id==$row['id']; echo "<ul data-role='listview' data-inline='true'>"; while($row
<?php
include "connect.php";
$result = mysqli_query($con,"SELECT * FROM story");
id==$row['id'];
echo "<ul data-role='listview' data-inline='true'>";
while($row = mysqli_fetch_array($result)) {
echo "<li>" ;
echo "<a href='view.php?id=".$rows['id']."' data-rel='external'>";
echo $row['name'];
echo "</a>";
echo "</li>";
}
echo "</ul>";
?>
此错误表明SQL查询无法将数据转换为$result,如果您尝试在屏幕上打印此结果,它将显示bool(false)作为输出。此参数在传递到mysqli_fetch_array()时不会为您提取数组,因为它需要一个真布尔值
首先尝试将数据库中的数据输入$result,打印它,然后继续
也许这会有帮助
$result=mysqli_query($con,“SELECT*FROM story,其中id=”.$id.”)
请记住始终使用错误检查方法,如die、mysqli\u error、mysqli\u connect\u error()、mysqli\u connect\u errno()等,以便更好地进行调试
您不会检查任何错误。你以为一切都很好
而不是:
$result = mysqli_query($con,"SELECT * FROM story WHERE id='".$id."'");
使用:
在connect.php
文件中使用:
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
不要忘记从全局变量中转义值。您的查询似乎返回了布尔值(false)。检查您的连接和查询是否正确。可能$id不正确
$result = mysqli_query($con,"SELECT * FROM story WHERE id='".$id."'") or die(mysqli_error($con));
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}