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Question

我有三个表（个人，群组和记录标签），并希望使用它们之间的信息来制作记录标签中的人员列表。示例：它应该找到＃的companyID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？

// 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

给予（希望）更好的视觉效果。

鲍勃（个人）
Lisa（个人）
疯狂（团体）

我在寻求帮助之前的测试给出了这个结果：

鲍勃（个人）
Lisa（个人）
疯狂（团体）
疯狂（团体）

那是因为我将个人和群体的输出放在他们自己的单独价值中。然后我回应$var1 . $var2我认为会出错。

更新：如果没有人能搞清楚，我只会重做我的数据库和表格。谢谢你们所有的帮助。

Answer 1

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，companyId 和 individual都有group，individual有groupId外国人键。

鉴于这种关系，你很可能希望按照这些方式写下你的陈述：

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

其他小调：
从我所听到的情况来看，最佳做法通常会给出表格单数名称，而不是复数名称 ....En和...Ko列的内容是什么？在我看来，你可能正在做一些国际化（英语和韩语？） - 如果是这样，你将要从原始表中提取这些列，并启动转换表。如果您以后需要添加对其他语言的支持，这将有助于您极大。

编辑：

虽然有多种方法可以处理转换表，但您可能希望每个表需要一个需要翻译数据的转换表。您可能还需要一个标准语言表供参考：

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

以recordLabel为例：

1）创建您的翻译表，其中包含需要翻译的所有列：

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：

- ＆GT;简单的语言连接

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

- ＆GT;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

3个表= 1个列表

1 个答案:

编辑：