我在Windows 7上安装Xampp,并尝试从数据库表中选择数据。为此,我创建了这样的代码:
<?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时,它有如下错误:
警告:mysqli_fetch_array()要求参数1为mysqli_result,第11行的C:\ xampp \ htdocs \ t2 \ 1.php中给出布尔值
如何解决这个问题?
答案 0 :(得分:5)
$result = mysqli_query($con, "SELECT * FROM Persons") or die("Error: " . mysqli_error($con));
可以告诉你原因。