Php mysql_close()要求参数1为资源,给定为空
导致错误的行是这一行Php mysql_close()要求参数1为资源,给定为空,php,parameters,null,Php,Parameters,Null,导致错误的行是这一行 mysql_close($con); 下面是整个代码 $con=mysqli_connect("localhost","root","pass","db_name"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT
mysql_close($con);
下面是整个代码
$con=mysqli_connect("localhost","root","pass","db_name");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM medicos");
while($row = mysqli_fetch_array($result)) {
echo '.....';
}
mysql_close($con);
行mysql\u close($con)代码>必须更改为mysqli\u close($con)代码>
您不能互换使用mysql
和mysqli
函数,例如:
mysqli\u connect
需要使用mysqli\u close
-同样地-
mysql\u connect
需要使用mysql\u close
您正在混合使用mysql和mysqli…您必须使用全部mysql或全部mysqli