Question

我正在创建事件CMS而且我在使用mysql中的表中获取数据时遇到了一些困难，以便显示我想要的格式。此外，由于某种原因，数据是在所有字段中复制事件编号。

<?php
    require("includes/con.php");
?>

<?php
    $queryVENUE = "SELECT * FROM venue ORDER BY event_date"; //pulls all data from the venue table.
    $resultVENUE = $mysqli->query($queryVENUE);

    $eventsNameArray = array();
    while ($rowEvents = $resultVENUE->fetch_assoc()) {
        $eventsNameArray[] = $rowEvents['event_id'] . " " . $rowEvents['event_date'] . " " . $rowEvents['event_name'] . " " . $rowEvents['event_location'] . " " . $rowEvents['event_details'] . " " . "&pound" . $rowEvents['event_ticket_cost']   ;
    }

    foreach($eventsNameArray as $eventsOutput){
        echo "<p>{$eventsOutput}</p>";

        echo "<H3>Event Number: {$eventsOutput['event_id']}</H3>";
        echo "<p>Event Date: {$eventsOutput['event_date']}</p>";
        echo "<p>Event Name: {$eventsOutput['event_name']}</p>";
        echo "<p>Event Location: {$eventsOutput['event_location']}</p>";
        echo "<p>Event Details: {$eventsOutput['event_details']}</p>";
        echo "<p>Event Ticket Cost: {$eventsOutput['event_ticket_cost']}</p>";
    }

我目前得到的输出是这样的： 1 2013-11-21 Xscape Castleford在中心下午5点见面。 £5.00

活动编号：1

活动日期：1

活动名称：1

活动地点：1

活动详情：1

活动门票费用：£1

2 2013-11-24乘坐Rossendale骑车前往曼彻斯特的Rossendale，并在假期前练习一下！ £5.00

活动编号：2

活动日期：2

活动名称：2

活动地点：2

活动详情：2

活动门票费用：2

目的是显示以下内容： --row 1 -

活动编号：

活动日期：

活动地点：

活动详情：

- 第2行 -

活动编号：

活动日期：

活动地点：

活动详情：

此外，我想将每个部分的事件编号等放入单独的Div中，以便我可以用css设置样式。

提前感谢您的帮助。

Answer 1

您将$eventsNameArray[]设置为同心字符串，看起来您需要将其作为数组进行操作。像 -

这样的东西

while ($rowEvents = $resultVENUE->fetch_assoc()) {
       $eventsNameArray[] = array(
              'event_id'=> $rowEvents['event_id'],
              'event_date'=> $rowEvents['event_date'], 
              'event_name'=> $rowEvents['event_name'],
              'event_location'=> $rowEvents['event_location'],
              'event_details'=> $rowEvents['event_details'],
              'event_ticket_cost'=> "&pound".$rowEvents['event_ticket_cost']
       );
}

或@ chas688指出，更简单 -

while ($rowEvents = $resultVENUE->fetch_assoc()) {
       $eventsNameArray[] = $rowEvents;
}

Answer 2

尝试使用这样的数据群，你是否直接在数据库上检查了你的查询？

<?php
require("includes/con.php");
?>

    <?php
    $queryVENUE = "SELECT * FROM venue ORDER BY event_date"; //pulls all data from the venue table.
    $resultVENUE = $mysqli->query($queryVENUE);

    $eventsNameArray = array();
    while ($rowEvents = $resultVENUE->fetch_assoc()) {
    $eventsNameArray[] = $rowEvents['event_id'] . " " . $rowEvents['event_date'] . " " . $rowEvents['event_name'] . " " . $rowEvents['event_location'] . " " . $rowEvents['event_details'] . " " . "&pound" . $rowEvents['event_ticket_cost']   ;
    }
    ?>
    <table>
    <tr>
    <td>Event Number</td>
    <td>Event Date</td>
    <td>Event Name</td>
    <td>Event Location</td>
    <td>Event Details</td>
    <td>Event Ticket Cost</td>
    </tr>
    <?pho foreach($eventsNameArray as $eventsOutput){
    ?>
    <tr>


    <td><?php echo $eventsOutput['event_id']?></td>
    <td><?php echo $eventsOutput['event_date']?></td>
    <td><?php echo $eventsOutput['event_name']?></td>
    <td><?php echo $eventsOutput['event_location']?></td>
    <td><?php echo $eventsOutput['event_details']?></td>
    <td><?php echo $eventsOutput['event_ticket_cost']?></td>
    </tr>
    <?php }?>
    </table>

获取php数组来调用以将每一行单独显示到表中

2 个答案: