php - 如何回显数据库查询结果？

Question

我编写了一些代码来访问我的数据库并将结果回显到网页上。不幸的是，我的代码一定有问题。非常感谢您的帮助。

$conn = new mysqli("website.com","Ericjs","Password","i4706992_wp1");
if ($conn) {
echo "Connection established! <br>";
} else {
die("Connection failed. Reason: ".mysqli_connect_error());
}
echo "Eric, I'm here first <br>";
$sql="SELECT Facility Name, Address, City, State, Zip FROM 'wp_facilities' 
WHERE 'Zip' = '01040' LIMIT 50";
echo "Eric, I'm here second <br>";
$results=mysqli_query($conn,$sql);
echo "Eric, I'm here third <br>";
if (mysqli_num_rows($results)>0) {
while($row=mysqli_fetch_array($results))  {
echo "Let's get ready to roll! <br>"; 
echo $row[1]." ".$row[2]." ".$row[3]." ".$row[4]." ".$row[5];
echo "<br>";
}
}
mysqli_close($conn);

我得到的输出是上面 if 语句之前的回显。

Connection established!
Eric, I'm here first
Eric, I'm here second
Eric, I'm here third

我正在寻找显示为的结果:

"Ryan Rink" "524 Pleasant St" "Watertown" "Massachusetts" "01040"

Answer 1

我会发表评论，但没有足够的声誉。有两个问题。

1. 由于您的查询不正确，因此未返回任何结果。正如 Sally CJ 所指出的。
2. 数组索引从 0 开始，但您尝试从索引 1 开始打印结果。这将导致 undefined offset 错误。