如何解决错误：“警告：mysqli_fetch_array（）期望参数1是mysqli_result”？

Question

我在Windows 7上安装Xampp，并尝试从数据库表中选择数据。为此，我创建这样的代码：

<?php 
$con=mysqli_connect("localhost","test1","2jan1991","test1"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

$result = mysqli_query($con,"SELECT * FROM Persons"); 

while($row = mysqli_fetch_array($result)) 
    { 
    echo $row['FirstName'] . " " . $row['LastName']; 
    echo "<br>"; 
    } 

mysqli_close($con); 
?> 

但是当我尝试运行代码，它已经像错误：

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\t2\1.php on line 11

如何解决这个问题？

Answer 1

$result = mysqli_query($con, "SELECT * FROM Persons") or die("Error: " . mysqli_error($con)); 

可以告诉你原因。