php MySQL 选择优先级

Question

$query = "SELECT * FROM posts WHERE language='$lang' AND (title LIKE '%$search%' OR author LIKE '%$search%' OR year LIKE '%$search%')";  

这正是它应该做的。但我想做的是将“标题”作为优先事项。但是现在看起来(每个搜索都在 html 的下拉列表中)它简单地显示它没有优先级。所以标题可以在最底部，作者在顶部。错误的顺序。我想以某种方式始终将标题放在顶部。

如何？

$output = '';  
      $lang = $_SESSION["lang"];
      $search = $_POST["query"];
      $query = "SELECT * FROM posts WHERE language='$lang' AND (title LIKE '%$search%' OR author LIKE '%$search%' OR year LIKE '%$search%')";  
      $result = mysqli_query($connect, $query);  
      $output = '<ul class="list-unstyled">';  
      if(mysqli_num_rows($result) > 0)  
      {  
           while($row = mysqli_fetch_array($result))  
           { 

                $output .= '<a href="'.$url.'/'.$lang.'/'.$row["url"].'/"><li>'.$row["book"].'</li></a>';  
           }  
      }  
      else  
      {  
           $output .= 'Not found.';  
      }  
      $output .= '</ul>';  
      echo $output;  

Answer 1

您可以拆分查询。

  $output = '';  
  $lang = $_SESSION["lang"];
  $search = $_POST["query"];



      $query2 = "SELECT * FROM posts WHERE language='$lang' AND title LIKE '%$search%'";
      $result2 = mysqli_query($connect, $query2);  
      $output = '<ul class="list-unstyled">';  
      if(mysqli_num_rows($result2) > 0)  
      {  
           while($row = mysqli_fetch_array($result2))  
           { 

                $output .= '<a href="'.$url.'/'.$lang.'/'.$row["url"].'/"><li>'.$row["book"].'</li></a>';  
           }  
      }  
      else  
      {  
           $output .= 'Not found.';  
      }  

      $query = "SELECT * FROM posts WHERE language='$lang' AND (author LIKE '%$search%' OR year LIKE '%$search%')";  
      $result = mysqli_query($connect, $query);  


      if(mysqli_num_rows($result) > 0)  
      {  
       while($row = mysqli_fetch_array($result))  
       { 

            $output .= '<a href="'.$url.'/'.$lang.'/'.$row["url"].'/"><li>'.$row["book"].'</li></a>';  
       }  
     }  
      else  
     {  
       $output .= 'Not found.';  
     }  


  $output .= '</ul>'; 
  echo $output;