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我有三个表（个人、组和唱片公司），并希望使用它们之间的信息来列出唱片公司内的人员。示例：它应该找到 companyID 为 # 的个人和组并列出它们。如何在不重复结果的情况下正确创建它？

// Connect to database
include "mysqli_connect.php";

// Set variables
$url_num = $_GET['company_id'];
$company_num = "";
$company_members = "";
$company_members2 = "";

//Check for artist id pagination
if(is_numeric($url_num)){
    $company = intval($url_num);
}else{
    $company = 1;
}

// Construct our join query
$sqli = "SELECT DISTINCT * FROM recordlabels
INNER JOIN individuals ON individuals.companyID=recordlabels.companyID
INNER JOIN groups ON groups.companyID=recordlabels.companyID
WHERE recordlabels.companyID = '{$company}'";

// Create results
$result = mysqli_query($link, $sqli);

//Check for albums
$totalmembers = mysqli_num_rows($result);

// Checking if query is successful
if($result){

// Print out the contents of each row into a table 
while($row = mysqli_fetch_array($result, MYSQLI_BOTH)){

// Assign value of column if not empty, otherwise "DamJuNoImage" (Thanks to Jonathan Sampson from Stack Overflow)
    $companyen = empty( $row["companyen"] ) 
        ? "Unknown" 
        : $row["companyen"];
    $companyko = empty( $row["companyko"] ) 
        ? "Unknown" 
        : $row["companyko"];
    $companyType = empty( $row["companyType"] ) 
        ? "Unknown" 
        : $row["companyType"];
    $founded = empty( $row["established"] ) 
        ? "Unknown" 
        : $row["established"];
    $founder = empty( $row["companyFounder"] ) 
        ? "Unknown" 
        : $row["companyFounder"];
    $information = empty( $row["information"] ) 
        ? "Unknown" 
        : $row["information"];
    $location = empty( $row["companyLocation"] ) 
        ? "Unknown" 
        : $row["companyLocation"];
    $homepage = empty( $row["companyPage"] ) 
        ? "#" 
        : $row["homepage"];
    $solopic = empty( $row["solopic"] ) 
        ? "DamjuNoImage" 
        : $row["solopic"];
    $soloen = empty( $row["soloen"] )
        ? "Unknown"
        : $row["soloen"];
    $solokn = empty( $row["solokn"] )
        ? "Unknown"
        : $row["solokn"];
    $grouppic = empty( $row["grouppic"] ) 
        ? "DamjuNoImage" 
        : $row["grouppic"];
    $groupen = empty( $row["groupen"] )
        ? "Unknown"
        : $row["groupen"];
    $groupkn = empty( $row["groupkn"] )
        ? "Unknown"
        : $row["groupkn"];
    $company_members .= '<li><a href="#">
      <div class="image"><img src="pathhere/' . $solopic . '"></div>
      <p class="datatitle2">' . $soloen . '</p>
      <p class="data-info2">' . $solokn . '</p>
      </a></li>';
    $company_members2 .= '<li><a href="#">
      <div class="image"><img src="pathhere' . $grouppic . '"></div>
      <p class="datatitle2">' . $groupen . '</p>
      <p class="data-info2">' . $groupkn . '</p>
      </a></li>';
    $listofmembers = $company_members . $company_members2; 
} // End of while statement
}else{
    echo "No people under $companyen";
} // End of If statement

提供（希望）更好的视觉效果。

  • 鲍勃（个人）
  • 丽莎（个人）
  • 狂热（团体）

我在寻求帮助之前的测试给出了这个结果：

  • 鲍勃（个人）
  • 丽莎（个人）
  • 狂热（团体）
  • 狂热（团体）

那是因为我把个人和群体的产出放在了各自独立的价值中。然后我附和$var1 . $var2了我觉得会出错的地方。

更新：如果没有人能弄清楚，我将重做我的数据库和表。谢谢你们所有的帮助。

4

1 回答 1

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问题很可能来自您的 SQL 代码。具体来说，您似乎有一些非规范化，在andcompanyId中都有，并且有一个外键。individual groupindividualgroupId

鉴于这种关系，您很可能希望按照以下方式编写您的陈述： 

SELECT companyEn, companyKo, companyType, established, companyFounder, 
       information, companyLocation, companyPage,
       soloPic, soloEn, soloKn, 
       groupPic, groupEn, groupKn
FROM recordLabels
JOIN groups 
ON groups.companyId = recordLabels.companyId
JOIN individuals
ON individuals.groupId = groups.groupId
WHERE recordLabels.companyId = $company

其他次要注意事项：
据我所知，最佳实践通常给表格命名单数，而不是复数。和列
是怎么回事？在我看来，您可能正在进行一些国际化（英语和韩语？） - 如果是这样，您将希望从原始表中提取这些列，并开始翻译表。如果您以后需要添加对其他语言的支持，这将对您有很大帮助。....En...Ko

编辑：

尽管有多种方法可以处理翻译表，但您可能希望每个需要翻译数据的表都有一个翻译表。您可能还需要一个标准语言表以供参考：

language
============
id  -- autoincrement - or possibly just use code
code  -- ISO 3-letter code, unique
shortCode  -- ISO 2-letter code, unique (I think?)
name  -- ISO standard name

举recordLabel个例子：

1）创建翻译表，所有需要翻译的列： 

recordLabelTranslation
=====================
companyId  -- fk to recordLabel.companyId, or whatever the primary key of that table is
languageId  -- fk to language.id
company  -- whatever 'companyEn' and 'companyKo' was (company name?  why translate?)

2) 删除所有已翻译的列 ( companyEn, companyKo)

3）（可选）对视图进行编码，以便您轻松参考。有两种口味：

-> 语言的简单连接

CREATE VIEW Record_Label_Language as 
SELECT recordLabel.companyId, recordLabel.information, -- all current columns...
       recordLabelTranslation.languageId, recordLabelTranslation.company
FROM recordLabel
JOIN recordLabelTranslation
ON recordLabelTranslation.companyId = recordLabel.companyid

-> 加入语言，默认为英语（或其他语言）

CREATE VIEW Record_Label_Language_Default as 
SELECT recordLabel.companyId, recordLabel.information, -- all current columns...
       COALESCE(recordLabelTranslation.languageId, language.languageId),
       COALESCE(recordLabelTranslation.company, dflt.company)
FROM recordLabel
JOIN recordLabelTranslation as dflt
ON dflt.companyId = recordLabel.companyid
AND dflt.languageId = [englishLanguageId]
CROSS JOIN language
LEFT JOIN recordLabelTranslation
ON recordLabelTranslation.companyId = recordLabel.companyId
AND recordLabelTranslation.languageId = language.id
于 2012-06-14T15:43:27.643 回答