我希望能够根据下拉菜单中的选择修改MySQL查询。这将更新页面上显示的后续表格。
我当前的代码如下（注意我正在使用PHP code snippets (Insert PHP) 插件）但是，这没有预期的结果，因为没有生成表。
如有任何帮助，我们将不胜感激。
<form method="POST">
<select name="post_ID">
<option value="21">21</option>
<option value="22">22</option>
</select>
</form>
[insert_php]
global $wpdb;
echo $_POST[\'post_ID\'];
$myQuery = $wpdb->get_results(\'SELECT * FROM \' . $wpdb->prefix.\'posts\' . \'WHERE ID = $_POST[\'post_ID\']\');
if($myQuery){
echo \'<table>\';
echo "<tr>";
echo "<th>Header 1</th>";
echo "<th>Header 2</th>";
echo "<th>Header 3</th>";
echo "<th>Header 4</th>";
echo "</tr>";
foreach ( $myQuery as $result )
{
echo \'<tr><td>\'.$result->post_title.\'</td><td>\'.$result->post_status.\'</td><td>\'.$result->post_date_gmt.\'</td><td>\'.$result->ID.\'</td></tr>\';
}
echo \'</table>\';
}
[/insert_php]
最合适的回答,由SO网友:Atlas_Gondal 整理而成
您不需要编写自己的查询，只需使用wordpress函数即可。
例如，通过id获取post；使用get_post
<form method="POST">
<select name="post_ID">
<option value="21">21</option>
<option value="22">22</option>
</select>
<input type="submit" value="Get Selected Value" />
</form>
<?php
$post_id = $_POST[\'post_ID\'];
if ( is_int( $post_id ) && !empty ($post_id ) ){
$post = get_post( $post_id );
if($post){
echo \'<table>\';
echo "<tr>";
echo "<th>Header 1</th>";
echo "<th>Header 2</th>";
echo "<th>Header 3</th>";
echo "<th>Header 4</th>";
echo "</tr>";
//foreach loop is removed because it will return only single object
echo \'<tr><td>\'.$post->post_title.\'</td><td>\'.$post->post_status.\'</td><td>\'.$post->post_date_gmt.\'</td><td>\'.$post->ID.\'</td></tr>\';
echo \'</table>\';
}
}
顺便说一句，如果您需要检索多篇帖子，请尝试
get_posts
