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有两个参数,返回对象为JSON如何在PHP中使用输入变量调用函数?
我试图让帖子ID和连接到阵列:
// Get Considered Posts, Training, Service, Intall
public function get_considered_posts() {
$item = new stdclass();
$item->msg = 'Empty';
$item->status = false;
$result = Array();
$post = $get_post_by_id(156); // Training
if ($post) {
$result[] = $post ;
}
$post = $get_post_by_id(164); // Service
if ($post) {
$result[] = $post ;
}
$post = $get_post_by_id(161); // Intall
if ($post) {
$result[] = $post ;
}
if (count($result) == 0) {
$item->msg = 'Empty';
$item->status = false;
} else {
$item->msg = 'Success';
$item->status = true;
}
$item->result = $result;
return $item;
}
在这里,我希望只使用5每一个岗位性质:
// Get Post By Id
private function get_post_by_id($pid) {
$post = get_post($pid = 0);
if ($post) {
if ($post->post_status == 'publish') {
$object = new stdclass();
$object->id = $post->ID;
$object->cid = $pid;
$object->title = $post->post_title;
$object->content = $post->post_content;
$object->image = ''.wp_get_attachment_url(get_post_thumbnail_id($post->ID));
return $object;
}
}
return $post;
}
但它显示500 Internal Server Error
当我检查error_log
它显示:
[17-May-2016 14:27:54 UTC] PHP Notice: Undefined variable: get_post_by_id in /home/codypars/public_html/app/wp-webservice/helper.php on line 62
[17可能 - 2016 14时27分54秒UTC] PHP致命错误:函数名必须是线/home/codypars/public_html/app/wp-webservice/helper.php字符串62
我该如何修复它?
解决:谢谢CD-jS和JustOnUnderMillions。要在PHP类中调用自己的函数,正确的语法是:
this->myFunctionName($Param1, $Param2, $Param3, ...);
哪里是'$这个 - >'?? – JustOnUnderMillions
是的,$ post = $ this-> get_post_by_id(156); , 谢谢 – AndroSco