Question

下面的代码不按预期的方式工作

$t = date('Y-d-m H:i:s',time());
$query = "INSERT INTO time VALUES('$t')";
if(mysql_query($query))
echo "Date and Time are added";

我创建了一个名为time的表，其中只有一列now和DATETIME为数据类型。即使 PHP 脚本通过打印"Date and Time are added"在浏览器中成功执行。

数据库没有按照预期的方式更新。它不是当前时间，而是初始化为默认值。

建议纠正问题的方法。

Answer 1

您可以使用NOW（）

$query = "INSERT INTO `time` (`now`) VALUES (NOW())";

Answer 2

尝试使用此格式

$t = date('Y-m-d H:i:s');
$query = "INSERT INTO time (now) VALUES('$t')";
if(mysql_query($query))
echo "Date and Time are added";

OR

$query = "INSERT INTO time (now) VALUES(NOW())";

Answer 3

<?php

/* YOU NEED TO ESTABLISH FIRST A CONNECTION */

$con=mysqli_connect("Host","Username","Password","Database"); /* REPLACE THE NECESSARY HOST, USERNAME, PASSWORD, AND DATABASE */

if(mysqli_connect_errno()){

echo "Error".mysqli_connect_error();
}

$t = date('Y-d-m H:i:s');
$query = mysqli_query($con,"INSERT INTO time (now) VALUES ('$t')");
echo "Date and Time are added";

?>

Answer 4

从

更改$ t的值

$t = date('Y-d-m H:i:s',time());

到

$t = date('Y-m-d H:i:s');

更好的方法是编写插件查询，如下所示：

   $query = "INSERT INTO `time` VALUES ('".date('Y-m-d H:i:s')."')";

将时间插入数据库

4 个答案: