如何检查PHP中DATE和DATETIME的每一行？

问题描述

我有三组 datetime 类型，两列 dtfrom 和 dtto 在mysql数据库中

 第1行：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 
  / pre> 
 
 
如果我在数据库中输入相同的日期，问题是输出无法显示重复日期 
 
 
 
它始终显示找不到数据！ 
 
 
 
有人可以帮助？ 
 <？php 
 $ connect = mysqli_connect（localhost，root，root数据库）; 
 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））
 {
回覆日期重复; 
} 
 mysqli_free_result（$ run）; 
} 
 else 
 {
 echoDate not found！; 
} 
} 
？> 
< form action =datetime.phpmethod =post> 
< table> 
< tr> 
< td>< i class =fa fa-unlock-alt>< / i> < / td> 
< td>用户ID：< / td> 
< td>< input type =textname =user_idsize =30>< / td> 
< / tr> 
< tr> 
< td>< i class =fa fa-unlock-alt>< / i> < / td> 
< td> Date from：< / td> 
< td>< input type =datename =dtfromsize =30>< / td> 
< / tr> 
< tr> 
< td>< i class =fa fa-unlock-alt>< / i> < / td> 
< td>日期到：< / td> 
< td>< input type =datename =dttosize =30>< / td> 
< / tr> 
< / table> 
 
< p>< input class =btnSuccesstype =submitname =Submitvalue =Submit> < / p> 
< / form> 
  
 
 
解决方案
尝试检查错误。使用echo mysqli_error（$ con）;而是回覆Date not found！; 
 
 
 
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。'和dtto <='$ dtto_user  
  
 
I have three sets of datetime type with two columns dtfrom and dtto in mysql database 
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
The problem is the output cannot show "Duplicate date" if I input the same date in database. 
It always show "Data not found !"
Can somebody help here? 
<?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>
 解决方案 
Try to check error . Use echo mysqli_error($con); instead  echo "Date not found !";
And Change your Query to
 $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."'";
                        
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