如何在 PHP MySQL 中检查 DATE 和 DATETIME 的每一行？答案

【问题标题】：How to check each rows for DATE and DATETIME in PHP MySQL?如何在 PHP MySQL 中检查 DATE 和 DATETIME 的每一行？
【发布时间】：2016-11-12 05:49:06
【问题描述】：

我在mysql数据库中有三组datetime类型，两列dtfrom和dtto

row 1 : dtfrom 2016-11-11 , dtto 2016-11-11
row 2 : dtfrom 2016-11-12 , dtto 2016-11-12
row 3 : dtfrom 2016-11-13 , dtto 2016-11-13

问题是如果我在数据库中输入相同的日期，输出无法显示“重复日期”。

它总是显示“数据未找到！”

有人可以帮忙吗？

<?php
    $connect      = mysqli_connect("localhost", "root", "root", "database");
    global $connect;   

    if(isset($_POST['Submit']))
    {
        $user_id        = $_POST['user_id'];
        $dtfrom         = $_POST['dtfrom'];
        $dtfrom_user    = strtotime($dtfrom);
        $dtto           = $_POST['dtto'];
        $dtto_user      = strtotime($dtto);

        $sql            = "SELECT * FROM table WHERE user_id='{$user_id}'  AND dtfrom >= '{$dtfrom_user}' AND dtto <= '{$dtto_user}'";
        $run            = mysqli_query($connect, $sql); 
        if($run && mysqli_num_rows($run) > 0 )
        {
            while($result = mysqli_fetch_assoc($run))
            {
                echo "Date duplicate";
            }
        mysqli_free_result($run);
        }
        else
        {
            echo "Date not found !";
        }
    }
?>
<form action="datetime.php" method="post">  
    <table> 
        <tr>
            <td><i class="fa fa-unlock-alt"></i> </td>
            <td>User ID : </td>
            <td><input type ="text" name="user_id" size="30"></td>
        </tr>
        <tr>
            <td><i class="fa fa-unlock-alt"></i> </td>
            <td>Date from : </td>
            <td><input type ="date" name="dtfrom" size="30"></td>
        </tr>
        <tr>
            <td><i class="fa fa-unlock-alt"></i> </td>
            <td>Date to : </td>
            <td><input type ="date" name="dtto" size="30"></td>
        </tr>
    </table>    

    <p><input class="btnSuccess" type ="submit" name="Submit" value="Submit"> </p>              
</form>

【问题讨论】：

标签： php mysql datetime

【解决方案1】：

尝试检查错误。使用 echo mysqli_error($con);而是回显“找不到日期！”；

并将您的查询更改为

 $user_id        = $_POST['user_id'];
 $dtfrom         = $_POST['dtfrom'];
 $dtfrom_user    = strtotime($dtfrom);
 $dtto           = $_POST['dtto'];
 $dtto_user      = strtotime($dtto);


 $dtfrom_user  = date("Y-m-d",$dtfrom_user );
 $dtto_user  = date("Y-m-d",$dtto_user);

 $sql = "SELECT * FROM table WHERE user_id='".$user_id."'  AND dtfrom >= '".$dtfrom_user."' AND dtto <= '".$dtto_user."'";

【讨论】：