Php 警告:mysqli_fetch_array()显示SQL代码时出错
我有一个小php显示我写的脚本问题。我无法在表中显示代码。我做错了什么?在某处也找不到答案 错误代码:警告:mysqli_fetch_array()要求参数1为mysqli_结果,布尔值在Php 警告:mysqli_fetch_array()显示SQL代码时出错,php,mysql,Php,Mysql,我有一个小php显示我写的脚本问题。我无法在表中显示代码。我做错了什么?在某处也找不到答案 错误代码:警告:mysqli_fetch_array()要求参数1为mysqli_结果,布尔值在 $con=mysqli_connect("localhost","database_connected","password"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: "
$con=mysqli_connect("localhost","database_connected","password");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM links");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['links'] . "</td>";
echo "<td>" . $row['url'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
$con=mysqli_connect(“本地主机”、“数据库已连接”、“密码”);
//检查连接
if(mysqli\u connect\u errno()){
echo“未能连接到MySQL:”.mysqli_connect_error();
}
$result=mysqli_查询($con,“从链接中选择*);
回声“
名字
姓氏
";
while($row=mysqli\u fetch\u数组($result)){
回声“;
回显“$row['links']”;
回显“$row['url']”;
回声“;
}
回声“;
mysqli_close($con);
我建议离开mysqli,改用mysqli;除了有一个更合理的模型外,它更容易安全地使用。下面是如何将上面的代码重写为PDO代码的快速示例:
<?php
$db = new PDO('mysql:host=localhost;dbname=SOME_DB_NAME_HERE;charset=utf8', 'database_connected', 'password');
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
try {
$rows = $db->query('SELECT * FROM links');
} catch(PDOException $ex) {
echo "An Error occured: ", $ex->getMessage(); //user friendly message
exit();
}
foreach($rows as $row) {
echo '<tr>';
foreach($row as $column=>$value) { echo '<td>', $value, '</td>'; }
echo '</tr>';
}
请查看最好在mysqli\u查询中提供数据库名称,尝试这样做