在下面的代码中,我正在从MariaDB数据库中检索数据,然后应该使用echo "Username: " . $row['username'];
列出数据库中的每个相关条目。它似乎没有打印,是否有一些我做错了。
(我知道这是不安全的)
<?php
$connection = @mysqli_connect("localhost","root","","jet")
OR die('Could not connect' .
mysqli_connect_error());
if($connection->connect_error){
$sql="SELECT username FROM userinfo WHERE username LIKE '%" . $_POST['search'] . "%';";
$res=$con->query($sql);
while($row=$res->fetch_assoc()){
echo "Username: " . $row['username'];
}
}
mysqli_close($connection);
?>
<form action="search.php" method="post">
<input type="text" name="search" id="bug">
<input type="submit" name="submit" value="search" id="rat">
</form>
我是新的堆栈溢出,我的英语不是很好,如果这是在错误的地方发布,请警告我,我会删除。谢谢
错误就在这一行
$res=$con->query($sql);
因为你的连接变量是$connection
。 Check This
将此更改为
$res=$connection->query($sql);
它可能会奏效。
代码如下
<?php
$connection = @mysqli_connect("localhost","root","","jet");
if($connection->connect_error){//show error if not connection failed
die("Connection failed: " . $connection->connect_error);
}
$sql="SELECT username FROM userinfo WHERE username LIKE '%" . $_POST['search'] . "%';";
$res=$connection->query($sql);
while($row=$res->fetch_assoc()){
echo "Username: " . $row['username'];
}
mysqli_close($connection);
?>