Wordpress wpdb->导入数据的查询问题

Question

我下面的代码行应该使用（在这种情况下）预定义的值来更新 NULL 值字段。当我执行我的 wpdb 查询时，页面给出了 500 错误。

$wpdb->query( $wpdb->query( "UPDATE ziekbeter SET healthy= '1994-06-04' WHERE person = 5 AND sick IS NOT NULL AND healthy IS NULL") );

有人可以看看这行代码并可能告诉我有什么问题吗？

代码在单击按钮时被执行。

添加了我的 wp 表的屏幕截图。

个人 ID 和健康日期将是动态的，但现在我保持静态。

profile.php

<?php
$user_ID = get_current_user_id();
echo $user_ID;
global $wpdb;

if ($wpdb->get_results( "SELECT * FROM ziekbeter WHERE person = $user_ID AND healthy IS NULL"))
{
    $row = $wpdb->get_results( "SELECT * FROM ziekbeter WHERE person = $user_ID AND healthy IS NULL");
    {
        ?>
            <form action="<?php bloginfo('url'); ?>/wp-content/themes/stk/ziekbeter.php" method="post">
                    <input type="submit" name="submitBeter" value="Meld mij beter!">
            </form>
        <?php
    } 
}
elseif ($wpdb->get_results("SELECT healthy FROM ziekbeter WHERE person = $user_ID AND healthy IS NOT NULL"))
{
    $row = $wpdb->get_results( "SELECT * FROM ziekbeter WHERE person = $user_ID AND healthy IS NOT NULL");
    {
        ?>
            <form action="<?php bloginfo('url'); ?>/wp-content/themes/stk/ziekbeter.php" method="post">
                    <input type="submit" name="submitZiek" value="Meld mij ziek!">
            </form>
        <?php
    } 
}
else {
        ?>
            <form action="<?php bloginfo('url'); ?>/wp-content/themes/stk/ziekbeter.php" method="post">
                    <input type="submit" name="submitZiek" value="Meld mij ziek!">
            </form>
        <?php
}
?>

ziekbeter.php

if(isset($_POST['submitZiek']))
{
     /* This function will come after i got the submitBeter working */
}
elseif(isset($_POST['submitBeter']))
{
     $wpdb->query( $wpdb->query( "UPDATE ziekbeter SET healthy= '1994-06-04' WHERE person = 5 AND sick IS NOT NULL AND healthy IS NULL") );

     echo "submitBeter wordt uitgevoerd";
}

如果我用 echo 替换 wpdb-> 查询，代码将正确执行并运行 echo 没有任何问题。

Answer 1

尝试修改逻辑，例如：

profile.php

<?php
global $wpdb;
$user_ID = get_current_user_id();
echo $user_ID;

//First DB query
$row1 = $wpdb->get_results("SELECT * FROM ziekbeter WHERE person = $user_ID AND healthy IS NULL");
//Second DB query
$row2 = $wpdb->get_results("SELECT healthy FROM ziekbeter WHERE person = $user_ID AND healthy IS NOT NULL");

 // I.e. greater than zero draw the HTML form
 if (count($row1)>0) {

 ?>
  <form action="<?php esc_url(bloginfo('url')); ?>/wp-content/themes/stk/ziekbeter.php" method="post">
     <input type="submit" name="submitBeter" value="Meld mij beter!">
  </form>
 <?php
    } 

  // Second DB query draw different HTML form
  if (count($row2)>0)   {
  ?>
  <form action="<?php esc_url(bloginfo('url')); ?>/wp-content/themes/stk/ziekbeter.php" method="post">
     <input type="submit" name="submitZiek" value="Meld mij ziek!">
  </form>
<?php

}

// Draw third HTML form otherwise
if (!$row1 || !$row2) {
?>
<form action="<?php esc_url(bloginfo('url')); ?>/wp-content/themes/stk/ziekbeter.php" method="post">
   <input type="submit" name="submitZiek" value="Meld mij ziek!">
 </form>
<?php
}
?>

现在，另一个文件，不要忘记将$wpdb 变量全球化：

ziekbeter.php （已编辑）：

<?php
global $wpdb;
if(isset($_POST['submitZiek']))  {
     /* This function will come after i got the submitBeter working */
}




if(isset($_POST['submitBeter']))  {
     $result = $wpdb->query( "UPDATE ziekbeter SET healthy= '1994-06-04' WHERE person = 5 AND sick IS NOT NULL AND healthy IS NULL");
    /*Or, use the native WordPress function:
    $result = $wpdb->update( $table, $data, $where, $format = null, $where_format = null ); */     
     if ($result) echo "submitBeter wordt uitgevoerd";
}
?>

Answer 2

如果您尝试直接访问这些文件，即http://example.com/wp-content/themes/stk/profile.php 和http://example.com/wp-content/themes/stk/ziekbeter.php，那么您需要包含 WordPress：

<?php
require_once '../../../wp-load.php';

// rest of profile.php

<?php
require_once '../../../wp-load.php';

// rest of ziekbeter.php