PHP POST选择显示

时间:2013-09-04 11:14:56

标签: php post

我有这个代码,我遇到的问题是,一旦我选择一个选项并保存它将不记得下次选择。但是帖子已经在数据库中注册了正确的值。是否有引起您注意的语法错误?谢谢你的帮助。

平台:PHP,SQLite

<tr>
<?php 
//location
?>
<td class='tdt' class='tdt'><?php te("Department");?>:</td>
<td>
<select id='locations' name='locations'>
<option value=''><?php te("Select");?></option>
<?php 
    foreach ($locations  as $key=>$location ) {
        $dbid=$location['id']; 
        $itype=$location['name'];
        $s="";
        if (($locations=="$dbid")) $s=" SELECTED "; 
            echo "    <option $s value='$dbid'>$itype</option>\n";
    }
?>
</select>

</td>
</tr>

<tr>

1 个答案:

答案 0 :(得分:0)

你可以试试这个(假设你准备了$locations数组)

<select id='locations' name='locations'>
    <option value=''><?php te("Select");?></option>
    <?php 
        foreach ($locations  as $key => $location ) {
            $s = $_POST['locations'] == $location['id'] ? 'selected' : '';
            echo "<option $s value='" . $location['id'] . "'>" . $location['name'] . "</option>";
        }
    ?>
</select>