MySQL live Query

Question

我不知道是否有人可以帮助我试图将一个实时搜索到我的网站，我已经找到了一个教程和外观的一面现在正在工作，有人可以告诉我如何编辑后续PHP只从一个表中提取信息。

表 - 电影 字段呼应对于每个结果 - 电影，描述，图像

在它被从2个表中提取信息的时刻成功一个显示的查询的另一个用于类别分频器提供的信息的内容，我需要的是删除类别方面并从单个表中提取信息。

道歉我的PHP知识非常有限，希望能够最好地描述这个问题。

<p id="searchresults"> 
<?php 
// PHP5 Implementation - uses MySQLi. 
// mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase'); 
$db = new mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase'); 

if(!$db) { 
    // Show error if we cannot connect. 
    echo 'ERROR: Could not connect to the database.'; 
} else { 

    // Is there a posted query string? 
    if(isset($_POST['queryString'])) { 
     $queryString = $db->real_escape_string($_POST['queryString']); 

     // Is the string length greater than 0? 
     if(strlen($queryString) >0) { 
      $query = $db->query("SELECT * FROM search s INNER JOIN categories c ON s.cat_id = c.cid WHERE name LIKE '%" . $queryString . "%' ORDER BY cat_id LIMIT 8"); 

      if($query) { 
       // While there are results loop through them - fetching an Object. 

       // Store the category id 
       $catid = 0; 
       while ($result = $query ->fetch_object()) { 
        if($result->cat_id != $catid) { // check if the category changed 
         echo '<span class="category">'.$result->cat_name.'</span>'; 
         $catid = $result->cat_id; 
        } 
        echo '<a href="'.$result->url.'">'; 
        echo '<img src="search_images/'.$result->img.'" alt="" />'; 

        $name = $result->name; 
        if(strlen($name) > 35) { 
         $name = substr($name, 0, 35) . "..."; 
        }      
        echo '<span class="searchheading">'.$name.'</span>'; 

        $description = $result->desc; 
        if(strlen($description) > 80) { 
         $description = substr($description, 0, 80) . "..."; 
        } 

        echo '<span>'.$description.'</span></a>'; 
       } 
       echo '<span class="seperator"><a href="http://www.marcofolio.net/sitemap.html" title="Sitemap">Nothing interesting here? Try the sitemap.</a></span><br class="break" />'; 
      } else { 
       echo 'ERROR: There was a problem with the query.'; 
      } 
     } else { 
      // Dont do anything. 
     } // There is a queryString. 
    } else { 
     echo 'There should be no direct access to this script!'; 
    } 
} 
?> 
</p> 

Answer 1

更换

SELECT * FROM search s INNER JOIN categories c ON s.cat_id = c.cid WHERE name LIKE '%" . $queryString . "%' ORDER BY cat_id LIMIT 8 

与

SELECT * FROM movies WHERE movie LIKE '%" . $queryString . "%' ORDER BY movie LIMIT 8 

将是一个良好的开端。

你能从那里继续吗？

另外 - 您在上面的描述中为字段名称写了'Movie' - 我使用了小写'M'，因此您可能需要更改它。

[编辑]在回复评论1：

让我们先从简单 - 与

if ($query) { 
    while ($result = $query ->fetch_object()) { // this line loops through all the results 
     echo '<img src="search_images/'.$result->image.' />'; // check capital I on 'image' 
     echo '<strong>'.$result->movie.'</strong>'; 
     echo $result->description.'<br />'; 
    } 
} 

基本上取代if ($query) { ... }，你只要把$result->myFieldName while循环的内部来获取数据了你想。 希望这应该是足够让你去=]

Answer 2

我想：

<p id="searchresults"> 
<?php 
// PHP5 Implementation - uses MySQLi. 
// mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase'); 
$db = new mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase'); 

if(!$db) { 
    // Show error if we cannot connect. 
    echo 'ERROR: Could not connect to the database.'; 
} else { 

    // Is there a posted query string? 
    if(isset($_POST['queryString'])) { 
     $queryString = $db->real_escape_string($_POST['queryString']); 

     // Is the string length greater than 0? 
     if(strlen($queryString) >0) { 
      $query = $db->query("SELECT * FROM movies WHERE Movie LIKE '%" . $queryString . "%' ORDER BY Movie LIMIT 8"); 

      if($query) { 
       // While there are results loop through them - fetching an Object. 


       while ($result = $query ->fetch_object()) { 
        echo '<a href="'.$result->url.'">'; 
        echo '<img src="search_images/'.$result->img.'" alt="" />'; 

        $name = $result->movie; 
        if(strlen($name) > 35) { 
         $name = substr($name, 0, 35) . "..."; 
        }      
        echo '<span class="searchheading">'.$name.'</span>'; 

        $description = $result->description; 
        if(strlen($description) > 80) { 
         $description = substr($description, 0, 80) . "..."; 
        } 

        echo '<span>'.$description.'</span></a>'; 
       } 
       echo '<span class="seperator"><a href="http://www.marcofolio.net/sitemap.html" title="Sitemap">Nothing interesting here? Try the sitemap.</a></span><br class="break" />'; 
      } else { 
       echo 'ERROR: There was a problem with the query.'; 
      } 
     } else { 
      // Dont do anything. 
     } // There is a queryString. 
    } else { 
     echo 'There should be no direct access to this script!'; 
    } 
} 
?> 
</p>