Question

我正在尝试基于我从先前选择中获得的嵌套数组进行选择。

这是我第一次选择的地方：

$coursequery = " 
            SELECT
                courseID
            FROM enrollments 
            WHERE 
                studentID = '$userid' 
        "; 

        try 
        { 
            $stmt = $db->prepare($coursequery); 
            $result = $stmt->execute(); 
        } 
        catch(PDOException $ex) 
        {  
            die("Failed to run query: " . $ex->getMessage()); 
        } 

        $rows = $stmt->fetchAll(); 

            $_SESSION['studentcourses'] = $rows;

这将以下列格式获取所有courseID：

Array ( [0] => Array ( [courseID] => 6 ) [1] => Array ( [courseID] => 7 ) )

我希望能够访问这些ID以从其他表中选择信息。我开始尝试使用for循环来获取所有“课程信息”，具体取决于ID。

for($i = 0; $i < $count; $i++) {

    $coursequery = " 
            SELECT
                *
            FROM courses 
            WHERE courseID = '$studentcourses[$i]'

        "; 

        try 
        { 
            $stmt = $db->prepare($coursequery); 
            $result = $stmt->execute(); 
        } 
        catch(PDOException $ex) 
        {  
            die("Failed to run query: " . $ex->getMessage()); 
        } 

        $row = $stmt->fetchAll(); 

            $_SESSION['studentcourseinfo'] = $row;

    }

在完成此任务时，我们将非常感谢您的帮助！

Answer 1

您可以使用一个SQL查询来完成此操作，从而消除所有这些循环

SELECT
    *
FROM courses
     INNER JOIN enrollments ON 
         enrollments.courseID = courses.courseID 
         AND enrollments.studentID = '$userid'
WHERE 1

Answer 2

您可以在查询中使用相同的数组，可以使用implode函数

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{
//Do something once the key is pressed
static_cast<MainWindow*>(userData)->new_screenshot();
return noErr;
}

只要更改输出格式

即可

  $coursequery =  "SELECT  * FROM courses 
  WHERE courseID  IN (" . implode(",", $studentcourses) . ");";

其中6和7将是课程的ID

同样在你的代码中有重写var Array (6, 7)的错误，这些错误一次覆盖同一个变量

从嵌套数组中选择SQL

2 个答案: