php - MYSQL 查询并将输出显示为 html

Question

我想显示每个公司的总评级，并且总评级将放在特定类别

代码是这样的

if (isset($_POST['search'])){

    $key = $_POST['keyword'];

    $com = mysql_query("
SELECT * 
  FROM company
 WHERE id LIKE '%$key%' 
    OR name LIKE '%$key%' 
    OR address LIKE '%$key%' 
    OR city LIKE '%$key%' 
    OR province LIKE '%$key%' 
    OR region LIKE '%$key%' 
    OR country LIKE '%$key%' 
    OR description LIKE '%$key%'
 ");
    while($comshow = mysql_fetch_array($com)){
        $comshowid = $comshow['id'];

        echo '
    <form  method="POST">   
    <div class = "row">
<div class = "text-right">                      
    <b><input  name = "reviewid" style = "margin-right:50%;" class = "text-right" id = "id" type = "text" value = "'.$comshow['id'].'" readonly></b>                    
</div>
</div>
<div class = "row" style = "margin-left:1%;">
    <div class ="col-md-4">
        <img src = "'.$comshow['dp'].'" style = "width:80%;">
    </div>
    <div class = "col-md-6">

        <h2><i>'.$comshow['name'].'</i></h2>
        <hr />
        <p>'.$comshow['description'].'</p>
        <h4><b>Ratings</b></h4>
        <p>Recruitment:</p>
        <p>Tenure:</p>
        <p>Separation:</p>  
        <input class = "btn btn-info" type = "submit"  name ="review" value = "REVIEW">
    </div>                      
</div><br><br></form>

    ';

    }

这是评分表

这是公司表 这就是输出

Answer 1

@Barmar 是正确的。您需要将您的公司表与评级表连接起来。我还建议有一个名为 ratings_id 的列，它是 rating.id 的外键。然后，您可以在评级表中添加一个名为 rating_value 的列。

插入以下内容:

    ALTER TABLE ‘rating’ ADD ‘rating_value’ INT NULL;
    ALTER TABLE ‘company’ ADD ‘ratings_id’ INT NULL;

    SELECT c.id, c.name, c.address, c.city, c.province, 
    c.region, c.country, c.description, c.ratings_id, 
    r.id, r.rating_value 
    FROM company c 
    JOIN rating r
    ON c.ratings_is = r.id
    AND
    c.id LIKE '%$key%' 
    OR c.name LIKE '%$key%' 
    OR c.address LIKE '%$key%' 
    OR city LIKE '%$key%' 
    OR c.province LIKE '%$key%' 
    OR c.region LIKE '%$key%' 
    OR c.country LIKE '%$key%' 
    OR c.description LIKE '%$key%';

我强烈建议您研究一下 PDO 准备语句，正如许多人在面临 SQL 注入(inject)风险之前所说的那样。