dongtanxi5676756 2018-03-16 05:48
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如果日期相同则只显示一次

if the date is same with database date just display one time only.Example:database date 16/2 then all 16/2 must display in a box, but other date also need to display. who help me to solve this problem.

   <h1>Payment Record</h1>
<?php

        $user_check=$_SESSION['login_user'];
        $query = "SELECT * FROM `confirm_order` where customer = '$user_check'";
        $result=mysqli_query($mysqli,$query);
        while($row=mysqli_fetch_array($result))
        {   

            ?>

            <table style="width:100%;border:1px solid black;margin-bottom:20px;">

            <tr>
                <td>Product Name: <?php echo $row['product_name'];?></td>
            </tr>
            <tr>
                <td>Quantity: <?php echo $row['quantity']?></td>
            </tr>
            <tr>
                <td>Receiver Name: <?php echo $row['receiver_name']?></td>
            </tr>
            <tr>
                <td>Receiver Address: <?php echo $row['receiver_address']?></td>
            </tr>
            <tr>
                <td>Receiver Contact: <?php echo $row['receiver_contact']?></td>
            </tr>
            <tr>
                <td style="float:right">Date: <?php echo $row['date']?></td>
            </tr>
            </table>

        <?php   


        }       
?>

Example

id | product | date |
---|---------|------|
1  | book    | 16/2 |
2  | pencil  | 16/2 |
3  | shoe    | 18/2 | 

Result

-------
book   |
pencil |
-------
shoe   |
-------
  • 写回答

1条回答 默认 最新

  • doutucui0133 2018-03-16 07:20
    关注

    Here is the complete solution:

        $user_check=$_SESSION['login_user'];
    $query = "SELECT * FROM `confirm_order` where customer = '$user_check'";
    $result=mysqli_query($mysqli,$query);
    $current_date=NULL;
    $previous_date=NULL;
    $table_open="<table border='1'>";
    $table_close="</table><hr>";
    
    $output=NULL;
    while($row=mysqli_fetch_array($result))
    {   
        $current_date=$row['date']; 
    
        if ($current_date==$previous_date) {            
            // SAME DATE
            $output.= "<tr><td>SAME DATE</td><td>".$current_date."</td></tr>";
            } else {
            // NEW DATE
            if ($previous_date<>NULL) {
                $output.= $table_close;
            }
            $output.= $table_open."<tr><td>NEW DATE</td><td>".$current_date."</td></tr>";
        }   
        $previous_date=$row['date'];
    }    
    
    echo $output;
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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