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Cron作业执行PHP文件

[英]Cron job execute php file

I set my cronjob to run a php file every one minute. 我设置我的cronjob每隔一分钟运行一个php文件。 The php file access to the database to update a value in a row. php文件访问数据库以连续更新一个值。 Here is the php code 这是PHP代码

<?php 

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
else {
mysql_query("UPDATE tanlename SET status = 2 WHERE created < (NOW() - 
INTERVAL 1 MINUTE)");
}
mysqli_close($conn);
?>

I want the code to set the status of a post to 2 after 1 minute of the its posting time. 我希望代码在其发布时间1分钟后将其状态设置为2。

The problem is in the else here: 问题出在else地方:

You need to change mysql_ to mysqli_ - not only is mysql_ deprecated but also because you're already using mysqli_ in $conn . 您需要更改mysql_mysqli_ -不仅mysql_过时但也因为你已经使用mysqli_$conn

Then you need to specify the connection in the mysqli_query . 然后,您需要在mysqli_query指定连接。

Update your old code from: 从以下位置更新旧代码:

else {
mysql_query("UPDATE tanlename SET status = 2 WHERE created < (NOW() - 
INTERVAL 1 MINUTE)");
}

to this: 对此:

else {
    mysqli_query($conn, "UPDATE tanlename SET status = 2 WHERE created < (NOW() - INTERVAL 1 MINUTE)");
}

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