我有一个网站与4 wordpress安装,(不需要为wp-MU) x4数据库,每个唯一的用户名和通过…
我将其称为wp1=mainsite wp2=childsite wp3=childsite wp4=childsite
在wp1中,有一个页面,与标签,这应该加载最近的帖子,从wp2,wp3,wp4安装,这个页面是使用一个自定义模板,以提供与php代码的页面,应该连接到每个数据库,并返回每个…
代码:
$db_wp2_user = 'db_user'
$db_wp2_pwd = 'db_pass';
$db_wp2_db = 'db_name';
$db_wp2_host = 'localhost';
$wpdb_wp2 = new wpdb($db_wp2_user, $db_wp2_pwd, $db_wp2_db, $db_wp2_host);
//query
$wp2_latestposts = $wpdb_wp2->get_results("SELECT ID, post_title FROM $wpdb_wp2->posts WHERE post_status = 'publish' AND post_type = 'post' ");
foreach ($wp2_latestposts as $latestpost) {
echo $latestpost->post_title."<br />";
}
$db_wp3_user = 'db_user'
$db_wp3_pwd = 'db_pass';
$db_wp3_db = 'db_name';
$db_wp3_host = 'localhost';
$wpdb_wp3 = new wpdb($db_wp3_user, $db_wp3_pwd, $db_wp3_db, $db_wp3_host);
//query
$wp3_latestposts = $wpdb_wp3->get_results("SELECT ID, post_title FROM $wpdb_wp3->posts WHERE post_status = 'publish' AND post_type = 'post' ");
foreach ($wp3_latestposts as $latestpost) {
echo $latestpost->post_title."<br />";
}
$db_wp4_user = 'db_user'
$db_wp4_pwd = 'db_pass';
$db_wp4_db = 'db_name';
$db_wp4_host = 'localhost';
$wpdb_wp4 = new wpdb($db_wp4_user, $db_wp4_pwd, $db_wp4_db, $db_wp4_host);
//query
$wp4_latestposts = $wpdb_wp4->get_results("SELECT ID, post_title FROM $wpdb_wp4->posts WHERE post_status = 'publish' AND post_type = 'post' ");
foreach ($wp4_latestposts as $latestpost) {
echo $latestpost->post_title."<br />";
}对每个数据库的每次调用都应该显示来自每个数据库的最新帖子。在那一页上。加载到每个选项卡中...
当代码运行时,页面返回所有样式和页脚,但信息应加载其空白,没有错误消息,
您能告诉我这是否是执行此操作的正确方法,或者如果数据存储在主wp安装之外的另一个数据库中,是否有更简单的方法来填充数组?
发布于 2010-02-18 23:54:56
在安装过程中,您可以添加一个前缀,这样您就可以在同一数据库中安装任意数量的WP安装程序。对于你正在尝试做的事情来说,这可能会更容易。
https://stackoverflow.com/questions/2289511
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