php - 如何从 PHP 中的数据库中检索复选框值

Question

我的问题是如何在 PHP 中从数据库中检索复选框值。以下代码无效。

<td>Hobby</td>
<?php
    $result = mysqli_query($con,"SELECT hobby FROM simple WHERE id = id"); 
    while($row = mysqli_fetch_array($result))
    {
        $focus = explode(",", $row['hobby']);
?>    
<td>
    <input type="checkbox"  name="hobby[]"  value="Cricket"  size="17"   <?php if(in_array("Cricket",$hobby)) { ?> checked="checked" <?php } ?> >Cricket
    <input type="checkbox" name="hobby[]"  value="Music" size="17" <?php if(in_array("Music",$hobby)) { ?> checked="checked" <?php } ?> >Music
    <input type="checkbox" name="hobby[]"  value="Reading"   size="17" <?php if(in_array("Reading",$hobby)) { ?> checked="checked" <?php } ?> >Reading
    <input type="checkbox" name="hobby[]"  value="Study" size="17" <?php if(in_array("Study",$hobby)) { ?> checked="checked" <?php } ?> >Study</td>
<?php
   }
?>
</tr>

Answer 1

改变声明，

$focus=explode(",",$row['hobby']);

收件人:

$hobby=explode(",",$row['hobby']);

此外，您可以使用三元运算符而不是 if else 来最小化代码:

<td>
  <input type="checkbox"  name="hobby[]"  value="Cricket"  size="17" <?php echo (in_array("Cricket",$hobby)) ? 'checked="checked" : '';?>>Cricket
  <input type="checkbox" name="hobby[]"  value="Music" size="17" <?php echo (in_array("Music",$hobby)) ? 'checked="checked" : '';?>>Music
  <input type="checkbox" name="hobby[]"  value="Reading"   size="17" <?php echo (in_array("Reading",$hobby)) ? 'checked="checked" : '';?>>Reading
</td>

此外，@mapek 建议，您的 SQL 将获取所有记录。

$result = mysqli_query($con,"SELECT hobby FROM simple WHERE id = id");

因为 id 总是等于 id。

你应该把它改成:

$result = mysqli_query($con,"SELECT hobby FROM simple WHERE id = '$id'");

使用数组最小化此代码的另一个版本:

<td>
<?php
$hobbies = array('Cricket', 'Music', 'Reading');
if (! empty($hobbies)) {
  foreach ($hobbies as $myHobby) {
    $checked = (in_array($myHobby, $hobby)) ? 'checked="checked"' : '';
?>
<input type="checkbox" name="hobby[]" value="<?php echo $myHobby;?>" size="17" <?php echo $checked;?>><?php echo $myHobby;?>
<?php
    }
}
?>
</td>