Question

我正在尝试使用PHP和MySQL自动生成这样的列表。

主题＃1

资源＃3
资源＃5
资源＃12

主题＃2

资源＃7
资源＃4
资源＃1

等

以下是表格：

eresources

ERID
etitle

受试者

SID
stitle

subjectmap

SID
ERID

以下是按字母顺序成功提供所有主题的代码：

$subjectQuery = "SELECT * FROM subjects WHERE sid != 17 ORDER BY stitle ASC";
$subjectResult = $mysqli->query($subjectQuery);

while ($subjectArray = $subjectResult->fetch_assoc()) { 
print "<h5 class='subcategory'>" . $subjectArray['stitle'] . "</h5>";
print "<div class='inner'>  

// Need a list of resources that match each subject ID (sid) here!

</div>";
}

现在，这里的代码成功地为我提供了具有固定主题ID的所有电子资源：

$getERBySubjectQuery = " SELECT erid FROM subjectmap WHERE sid=11 ";
$getERBySubjectResult = $mysqli->query($getERBySubjectQuery);
if($getERBySubjectResult && $getERBySubjectResult->num_rows >= 1){
 while($getERBySubjectArray = $getERBySubjectResult->fetch_assoc() ){
        $query = " SELECT * FROM eresources WHERE erid = " .$getERBySubjectArray['erid']. " ORDER BY ertitle ASC ";
    $result = $mysqli->query($query);

    if($result && $result->num_rows >= 1){
       while($array = $result->fetch_assoc() ){
          print("<a href=\"" . $array['link'] . "\">" .   "<h5 class='subcategory'>" . $array['ertitle'] . "<div class='accessnote'>"   .  $array['access']    . "</div></h5></a>");

 }}}}

当然，我并不倾向于为每个主题ID（sid）复制相同的节。我希望自动生成主题列表，并自动生成具有该主题的资源列表。

基本上我需要将第一节的sid提供给每个主题的第二节。

Answer 1

如果没有执行n+1查询，n是主题数，则无法在MySQL中执行此操作。

你可以使用JOIN和一些聪明的排序。

SELECT
    subject.stitle, eresource.etitle
FROM subject
    LEFT JOIN subjectmap ON subject.id = subjectmap.sid
    LEFT JOIN eresource ON eresource.id = subjectmap.erid
ORDER BY subject.stitle, eresource.etitle

您将得到如下结果：

+------------+--------------+
| stitle     | etitle       |
+------------+--------------+
| Subject #1 | Resource #3  |
| Subject #1 | Resource #5  |
| Subject #1 | Resource #12 |
| Subject #2 | Resource #1  |
| Subject #2 | Resource #4  |
| Subject #2 | Resource #7  |
+------------+--------------+

Answer 2

通过使用WHERE子句，我认为Frits van Campen的查询可能更具人性化，特别是对于初学者：

SELECT subject.stitle, eresource.etitle
FROM subject,subjectmap,eresource 
WHERE subject.id = subjectmap.sid
  AND eresource.id = subjectmap.erid
ORDER BY subject.stitle, eresource.etitle

使用来自一个MySQL查询的结果插入到另一个查询中

2 个答案: