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我想通过从 mysql 表中获取值来创建一个下拉区域。此代码似乎不起作用；它打印出来的只是没有内容的下拉框，我怎样才能让它工作？或者有没有其他程序可以做到这一点？

<?
$connection = mysql_connect("localhost", "root", "");
mysql_select_db("test", $connection);
$query = "SELECT full_name FROM test";
$names = mysql_query($query);
function dropDown($content, $result)
 {
while($row = mysql_fetch_row($result))
    {
        $name = $row[0];
        $content .= "<option value='$name'>$name</option>";
    }
 }
$content=<<<EOT
           <html>
              <body>
                 <select name="name">
EOT;

dropDown($content, $names)

$content.=<<<EOT
                 </select>
              </body>   
            </html>          


EOT;

echo $content;
?>
4

3 回答 3

2

return字符串。PHP 不是 C 语言，您使用out参数只是因为它们有时很方便。

function dropDown($result, $fieldName)
{
    $content = '';
    while($row = mysql_fetch_row($result)) {
        $name = $row[0];
        $content .= "<option value='$name'>$name</option>";
    }
    return '<select name="'.$fieldName.'">'.$content.'</select>';
}

$content = '<html><body>';
$content .= dropDown($names, 'name');
于 2012-05-03T06:19:16.073 回答
0 
<?php echo $form->dropDownList($model,'courseprefer', array(

'(*value in database table*'=>'displaying value',
 )); ?>

您可以手动添加它们，只需确保数组中的第一个值在数据库表中.. :)

以免出错！

于 2012-05-03T07:46:22.587 回答
0 
    here database details      
  mysql_connect('hostname', 'username', 'password');
mysql_select_db('database-name');

     $sql = "SELECT username FROM userregistraton";
        $result = mysql_query($sql);

         echo "<select name='username'>";
  while ($row = mysql_fetch_array($result)) {
   echo "<option value='" . $row['username'] ."'>" . $row['username'] ."</option>";}
   echo "</select>";


      here username is the column of my table(userregistration)
     it works perfectly

or simply the code is 
        <?
         $sql = "SELECT * FROM userregistraton ";
             $result = mysql_query($sql); ?>


 <select name="username" id="username">
<option value="">Select user</option>
    <?php
        while ($row = mysql_fetch_array($result))
        { ?>
          <option value="<?php echo $row['uid']?>"         <?php    if($row['uid']==$_POST["username"]) echo "selected"; ?> >
   <?php echo $row['username'];?></option>
   <?php
         } ?>

于 2012-06-09T10:20:16.300 回答