检查函数返回值而不再执行它

Question

如何获得函数返回值而不再执行它？

我用这一点，但这个执行功能再次

我的功能：

function article() 
{ 
    if($_GET['action'] == "article" && !empty($_GET['id'])) 
    { 
     $id = intval($_GET['id']); 
     $article = array(); 
     $selectArticle = mysql_query("SELECT * FROM articles WHERE id='$id'"); 
     $rowArticle = mysql_fetch_array($selectArticle); 

     $id = $rowArticle['id']; 
     $title = stripcslashes($rowArticle['title']); 
     $category = stripcslashes($rowArticle['category']); 
     $image = stripcslashes($rowArticle['image']); 
     $description = stripcslashes($rowArticle['description']); 
     $full_description = stripcslashes($rowArticle['full_description']); 
     $keywords = stripcslashes($rowArticle['keywords']); 
     $url = "/article/" . $rowArticle['id'] . "/" . str_replace(" ","-",stripcslashes($rowArticle['title'])); 
     $article = array('id' => $id, 'title' => $title, 'category' => $category, 'image' => $image, 'description' => $description, 'full_description' => $full_description, 'keywords' => $keywords, 'url' => $url); 
     mysql_query("UPDATE articles SET visits=visits+1 WHERE id='$id'"); 
    } 
    return $article; 
} 

如何检查

if (article() != null) 
{ 
    $article = article(); 
    return $article['title']; 
} 

Answer 1

这应该做你想做的

if (null !== ($article = article())) { 
    return $article['title']; 
} 

这是“同时”分配和比较。首先，这部分评估为：($article = article())。它产生一个null值或一个array存储在$article。
其结果（null或array）然后通过if结构进行评估：if (null !== $article)并恢复正常流程。

Answer 2

修改检查一下。

$article = article(); 
return $article != null ? $article['title'] : null; 

Answer 3

像这样

$article_var = article(); 

if ($article_var!=null) 
{ 
     //do stuff 
     //return $article['title'] // etc 
}