WordPress的，通過它的名字或URL獲得帖子內容

Question

我在這裏看到，我可以使用帖子ID在WordPress中獲得帖子的內容。就像：

<?php $my_postid = 83;//This is page id or post id 
$content_post = get_post($my_postid); 
$content = $content_post->post_content; 
$content = apply_filters('the_content', $content); 
$content = str_replace(']]>', ']]&gt;', $content); 
echo $content;?> 

我想要的是同樣的東西，但得到其名稱的職位。

Answer 1

您可以用做

$content_post = get_posts(array('name' => 'yourpostname')); // i.e. hello-world 
if(count($content_post)) 
{ 
    $content = $content_post[0]->post_content; 
    // do whatever you want 
    echo $content; 
} 

更新：你也可以在你的functions.php添加此功能，可以從任何地方

function get_post_by_name($post_name, $post_type = 'post', $output = OBJECT) { 
    global $wpdb; 
    $post = $wpdb->get_var($wpdb->prepare("SELECT ID FROM $wpdb->posts WHERE post_name = %s AND post_type= %s", $post_name, $post_type)); 
    if ($post) return get_post($post, $output); 
    return null; 
} 

// call the function "get_post_by_name" 
$content_post = get_post_by_name('hello-world'); 
if($content_post) 
{ 
    $content = $content_post->post_content; 
    // do whatever you want 
    echo $content; 
} 

更新叫它：爲了得到一個職位由它的標題你可以使用

// 'Hello World!' is post title here 
$content_post = get_page_by_title('Hello World!', OBJECT, 'post'); 

，或者你可以用你的$item->item_title變量

$content_post = get_page_by_title($item->item_title, OBJECT, 'post'); 
if($content_post) 
{ 
    $content = $content_post->post_content; 
    // do whatever you want 
    echo $content; 
}