[英]PHP MYSQL insert doesn't work and receive no error
I have this PHP code , and i would insert a data in a table. 我有这个PHP代码,我会在表中插入数据。 I don't know why , it doesn't work. 我不知道为什么,它不起作用。 I receive no error from the query. 我没有从查询中收到任何错误。 If i print the $update_miner
nothing is showed. 如果我打印$update_miner
不会显示任何内容。 This is the code: 这是代码:
<?php
session_start();
include("header.php");
$building['rock_miner'] =3;
$current_time = time();
$update_miner = mysqli_query($connection,"INSERT INTO `updates_queue` (`user_id`,`content`, `start_at`,`end_at`,`finished`,`old_level`) VALUES ('".$_SESSION['uid']."', 'rock_update', '".$current_time."','".$current_time."',0 ,'".$building['rock_miner']."'");
include("footer.php");
?>
This is the table: 这是表:
you are missing one parenthesis at last 您最后缺少一个括号
INSERT INTO `updates_queue` (`user_id`,`content`, `start_at`,`end_at`,`finished`,`old_level`)
VALUES ('".$_SESSION['uid']."', 'rock_update', '".$current_time."',
'".$current_time."',0 ,'".$building['rock_miner']."')")
^
Not too sure but wouldn't use int to store a time. 不太确定,但是不会使用int来存储时间。 Int only goes up to a certain value and then will fail-or just store the highest value it can. Int只会上升到某个值,然后会失败-或仅存储它可以的最高值。 I think that might depend on your mysql version though. 我认为这可能取决于您的mysql版本。
The comment before is a good start and make sure you have connected to the database as well. 之前的注释是一个好的开始,并确保您也已连接到数据库。
So your $connection
isn't false. 因此,您的$connection
不是假的。
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