Question

我的代码下面包含一个MySQL查询：

// These gets the values entered in the drop-down lists and stores the values as
// variables.
$event   = $_GET['cmbEvent'];
$yrGroup = $_GET['cmbYrGroup'];
$gender  = $_GET['cmbGender'];

/* The query $eygQuery, where the 'eyg' stands for event, year group and gender, 
grabs the stud_fname, stud_sname, result_studpoints, result_position, studevent_result 
where the requirements of the specific stud_gender, stud_yrgroup and event_id is met */
$eygQuery = "
    SELECT  stud_id,
            stud_fname, 
            stud_sname, 
            result_studpoints, 
            result_position, 
            studevent_result 
    FROM 
        students s
            INNER JOIN 
        result r on 
        s.stud_id = r.stud_id
    WHERE 
        stud_gender  = '$gender'
    AND stud_yrgroup = '$yrGroup'
    AND r.event_id   = '$event'  
";

$result = mysqli_query($conn, $eygQuery);

由于'SELECT stud_id'（这是主键和外键）而失败。当我删除stud_id时，查询看起来像这样：

        SELECT  stud_fname, 
                stud_sname, 
                result_studpoints, 
                result_position, 
                studevent_result 
        FROM 
            students s
                INNER JOIN 
            result r on 
            s.stud_id = r.stud_id
        WHERE 
            stud_gender  = '$gender'
        AND stud_yrgroup = '$yrGroup'
        AND r.event_id   = '$event'

它有效，但是如何修改查询以输出stud_id呢？

提前致谢。

Answer 1

stud_id在您的查询中含糊不清。使（表示）该字段更具体：

    SELECT  s.stud_id,
            stud_fname, 
            stud_sname, 
            result_studpoints, 
            result_position, 
            studevent_result 
    FROM 
        students s
            INNER JOIN 
        result r on 
        s.stud_id = r.stud_id
    WHERE 
        stud_gender  = '$gender'
    AND stud_yrgroup = '$yrGroup'
    AND r.event_id   = '$event'

编辑： 为了更方便的结果提取，我建议对某些选定的字段使用别名：

SELECT  s.stud_id AS student_id,
        ....

获取数据示例（在PHP中）：

... 
while ($data = mysqli_fetch_assoc($result)) { 
    $studId = $data['student_id']; 
    echo $studId; 
}
...

Answer 2

很可能您会收到此错误，因为列名称不明确，因此您需要指定要从中获取结果的表。尝试在列名称前添加表格的简称，例如：

SELECT  s.stud_id,
        stud_fname, 
        stud_sname, 
        result_studpoints, 
        result_position, 
        studevent_result 
FROM 
    students s
        INNER JOIN 
    result r on 
    s.stud_id = r.stud_id
WHERE 
    stud_gender  = '$gender'
AND stud_yrgroup = '$yrGroup'
AND r.event_id   = '$event'

Answer 3

stud_id是students和result表中的字段集。当您要求数据库检索该数据时，他不知道要选择什么值。它足以指定从哪个表中获取该数据：

SELECT  s.stud_id.
        stud_fname, 
        stud_sname, 
        result_studpoints, 
        result_position, 
        ìstudevent_result 
FROM 
        students s
INNER JOIN 
        result r on 
        s.stud_id = r.stud_id
WHERE 
    stud_gender  = '$gender'
AND stud_yrgroup = '$yrGroup'
AND r.event_id   = '$event'

Answer 4

stud_id含糊不清 - 它出现在students和result表格中。只需用表别名来限定它，你应该没问题：

SELECT   s.stud_fname, 
-- Here -^
         stud_sname, 
         result_studpoints, 
         result_position, 
         studevent_result 
 FROM 
     students s
 INNER JOIN 
     result r ON s.stud_id = r.stud_id
 WHERE 
     stud_gender  = '$gender'
 AND stud_yrgroup = '$yrGroup'
 AND r.event_id   = '$event'

选择某些数据

4 个答案: