Question

我想动态填充复选框，我从中获取数据库中的值并存储在复选框中，但不显示值如何显示值。

这是我的代码：

<div>
    <?
    //echo $eventid=$_POST['events'];
    $count=count($_POST['events']);
    for($i=0; $i<$count; $i++){
        $select="select b.first_name,b.last_name from buyers b,registrations r where b.buyer_id=r.buyer_id and r.event_id='".$_POST['events'][$i]."' group by r.buyer_id";
        $res = $GLOBALS ['mysqli']->query ($select) or die ($GLOBALS ['mysqli']->error . __LINE__);
        if ($res->num_rows > 0)
        {           
            while($row = $res->fetch_assoc ())
            {
            ?>  
                <input type="checkbox" name="receptionts" checked="checked" value="<? echo $row['first_name'];$row['last_name']?>"/><br />              
            <?
            }
        }
    }
    ?>
</div>

Answer 1

复选框的值属性应为

value="<?php echo $row['first_name']." ".$row['last_name']?>"

试试这个

  <input type="checkbox" name="receptionts" checked="checked" value="<?php echo $row['first_name']." ".$row['last_name']?>"/><br />

而不是

<input type="checkbox" name="receptionts" checked="checked" value="<? echo $row['first_name'];$row['last_name']?>"/><br />

更新2：

    <input type="checkbox" name="receptionts" checked="checked" value="<?php echo $row['first_name']." ".$row['last_name']?>"/> 
 <?php echo $row['first_name']." ".$row['last_name']?> <br />

Answer 2

从评论开始（您要将名字设置为值）：

$select="select b.first_name,b.last_name from buyers b,registrations r where b.buyer_id=r.buyer_id and r.event_id='".$_POST['events'][$i]."' group by r.buyer_id";

更改

<input type="checkbox" name="receptionts" checked="checked" value="<? echo $row['first_name'];$row['last_name']?>"/><br />

到

<input type="checkbox" name="receptionts" checked="checked" value="<?php echo $row['first_name'];?>"/><br />

并且（因为您从两个表中选择），您应该在查询中使用AS，如：

$select="select b.first_name AS first_name,b.last_name AS last_name from buyers b,registrations r where b.buyer_id=r.buyer_id and r.event_id='".$_POST['events'][$i]."' group by r.buyer_id";

（可能没必要）

另请查看prepared statements以防止SQL注入。

Answer 3

试试这个：

<div>
    <?php /* <-- Don't use PHP short tags, it's a bad practice, they are deprecated */

        /* Making select statement inside a loop is a very bad idea. 
           Also, you should make sure to properly escape
           $_POST['events']), as inserting unescaped inputs into a query 
           is very unsafe   */

        $post = array_map(
            array($GLOBALS['mysqli'], "escape_string"), 
            $_POST['events']
        );

         /* We apply mysqli::escape_string to every element of 
            $_POST['events'] and save result as $post */

        $select = "SELECT `b`.`first_name`, `b`.`last_name` FROM `buyers` `b` ".
                  "JOIN `registrations`.`r` ON `r`.`buyer_id` = `b`.`buyer_id` ".
                  "WHERE `r`.`event_id` IN ('".implode("','", $post)."') ".
                  "GROUP BY `r`.`buyer_id`";
        $res = $GLOBALS ['mysqli']->query ($select) 
               or die ($GLOBALS ['mysqli']->error . __LINE__);

        while($row = $res->fetch_assoc()) : ?>
           <label>
               <input type="checkbox" name="receptionts[]" checked="checked" value="<?= $row['first_name']." ".$row['last_name']?>"/> 
               <?= $row['first_name']." ".$row['last_name'] ?>
           </label>
        <?php endwhile; ?>

</div>

Answer 4

像这样给予价值

    <input type="checkbox" name="receptionts" checked="checked" value="<?php echo $row['first_name'].' '.$row['last_name'];?>"/><br />

如何在php中动态填充复选框？

4 个答案: