Question

这将被生成

我已经能够产生什么

我只需要GUI部件的帮助。请不要投票，否则我将无法提出问题。

    <body>

<div class="container-fluid" style="margin-top:5%;">
    <h3><b>PRODUCTION</b></h3>
    <table id="StatusTable">
        <tr>
            <th>Phase</th>
            <th>Error Code</th>
            <th>Error Description</th>
            <th>Turnback</th>
            <th>Score</th>
            <th>Start Date</th>
            <th>Incorporation Date</th>
            <th>Completion Date</th>
        </tr>
        <tr>
            <th>Authoring</th>
            <td>---</td>
            <td>---</td>
            <td>---</td>
            <td>---</td>
            <?php
            session_start();
            $WO_ID = $_SESSION['SELECTED_WO_ID'];
            $sql="SELECT * FROM `associated_wos` WHERE `ID` = '$WO_ID'";
            //echo $sql;
            $result = mysqli_query($conn,$sql);
            while ($row = mysqli_fetch_array($result))
            { ?>
                <td><?php echo $row["StartDate"]; ?></td>
                <td>---</td>
                <td><?php echo $row["SFCK SubmissionDate"];?></td>
                <?php
            }?>
            ?>
        </tr>
        <tr>
            <?php
            $value = 'Peer Review';
            $sql="SELECT * FROM `wo_errorinfo` WHERE `WO_NO` = '$WO_ID' AND `Review_Type` = '$value'";
            $result = mysqli_query($conn,$sql);
            $num_rows = mysqli_num_rows($result);
            ?>
        <tr>
            <th rowspan = "<?php echo $num_rows + 1 ;?>">Peer Review</th>
            <?php
            while ($row = mysqli_fetch_array($result)) {?>
                <tr>
            <td><?php echo $row["Error_Code"];?></td>
            <td><?php echo $row["Error_Description"];?></td>
            <td><?php echo $row["Error_Quantity"];?></td>
        <?php
        }?>
            <?php
            $select = "SELECT * FROM `wo_reviewerqa` WHERE `WO_ID` = '$WO_ID' and `reviewType` = '$value'";
            $result1 = mysqli_query($conn,$select);
            $num_rows1 = mysqli_num_rows($result1);
            while($row1 = mysqli_fetch_array($result1))
            {?>
                <td rowspan = "<?php echo $num_rows + 1 ?>"><?php echo $row1["reviewScore"];?></td>
                <td rowspan = "<?php echo $num_rows + 1 ?>"><?php echo $row1["reviewStartDate"];?></td>
                <td rowspan = "<?php echo $num_rows + 1 ?>">---</td>
                <td rowspan = "<?php echo $num_rows + 1 ?>"><?php echo $row1["reviewCompleteDate"];?></td>
            <?php
            }
             ?>

     </tr>
    </tr>
    </table>
</div>
</body>

我应该进行哪些更改以获得所需的表。我能够生成该表的唯一方法是交换列。我应该使用JOIN命令吗？

Answer 1

您需要GUI，这是表格的HTML结构

 <table border="1" cellpadding="5">
  <tr>
    <td colspan="8"><center>PRODUCTION</center></td>
  </tr>
    <tr>
        <td>Phase</td>
        <td>Error Code</td>
        <td>Error Description</td>
        <td>Turnback</td>
        <td>Score</td>
        <td>Start Date</td>
        <td>Incorporation Date</td>
        <td>Completion Date</td>
    </tr>
    <tr>
        <td>Authoring</td>
        <td></td>
        <td></td>
        <td></td>
        <td></td>
        <td></td>
        <td></td>
        <td></td>
    </tr>
    <tr>
        <td rowspan="2">Peer Review</td>
        <td></td>
        <td></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
    </tr>
    <tr>
        <td></td>
        <td></td>
    </tr>
    <tr>
        <td rowspan="2">Internal Review</td>
        <td></td>
        <td></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
        <td rowspan="2"></td>
    </tr>
    <tr>
      <td></td>
      <td></td>
    </tr>
</table>

如何使用html生成此表？

1 个答案: