Question

在我的php代码中，我在我的数据库中插入一些数据。一个是约会。我希望以27-08-2016的格式显示，但我回到01-01-1970

这是我的代码。

<?php 
include("./init.php");

if(isset($_POST['submit'])){

 $post_game = $_POST['game'];

 $time = strtotime($_POST['date']);

 $post_date = date("d-m-Y", strtotime($time));

if($post_game==''){

    echo "<script>alert('Please fill in all fields')</script>";
    exit();
    }
else {

    $insert_game = "insert into last_game (game,date) values ('$post_game','$post_date')";

    $run_posts = mysqli_query($con,$insert_game); 

        echo "<script>alert('Post Has been Published!')</script>";

        echo "<script>window.open('index.php?last_game_details','_self')   
   </script>";

     }

   }

?>

关键路线是：

 $post_date = date("d-m-Y", strtotime($time));

我的php版本是： 5.6.23

Answer 1

slabService

Answer 2

请在查询中使用Now（）来插入当前日期。 NOW（）返回当前日期和时间。

if(isset($_POST['submit'])){

 $post_game = $_POST['game'];

 $time = strtotime($_POST['date']);

 $post_date = date("d-m-Y", strtotime($time));

if($post_game==''){

    echo "<script>alert('Please fill in all fields')</script>";
    exit();
    }
else {

    $insert_game = "insert into last_game (game,date) values ('$post_game',NOW())";

    $run_posts = mysqli_query($con,$insert_game); 

        echo "<script>alert('Post Has been Published!')</script>";

        echo "<script>window.open('index.php?last_game_details','_self')   
   </script>";

     }

   }

?>

要检索所需的日期格式，请使用php

`echo date_format($date,"Y-m-d ");`

Answer 3

你的问题在这里：

 $time = strtotime($_POST['date']);    
 $post_date = date("d-m-Y", strtotime($time)); //why call strtotime again?

第二次拨打strtotime时，您正在为其添加时间戳（第一次通话的结果），但它需要一个日期字符串。因此，它无法解析输入并返回无效时间。只需将以上两行替换为：

 $time = strtotime($_POST['date']);    
 $post_date = date("d-m-Y", $time);

php日期格式返回01-01-1970

3 个答案: