-1
讓我解釋一下背景故事,老師可以設置一個學生在場或不在場。這些學生的價值被放在一個數據庫中,這個查詢選擇了班級代碼,並計算了某個課程中學生的比例,但是我不知道該怎麼做,文件必須轉換爲JSON,放入一個ChartJS條形圖,但由於某種原因,我似乎無法弄清楚這個代碼,每個類的每個百分比都必須計算,所以我不能使用IN或類似的東西來計算整個課程的存在而不是每堂課的課程(這就是klas.code ='$ klas'的意思)是否有人知道我如何得到1的結果,但仍然能夠分別計算每個班級的百分比?MySQL查詢中的Foreach循環
謝謝。
$klassen = array("WFHBOICT.V1E", "WFHBOICT.V1F");
foreach($klassen as $klas){
//query to get data from the table
$query = ("SELECT klas.code klas, ROUND(
(
SELECT Count(aanwezigheid)
FROM aanwezigheid
JOIN college ON aanwezigheid.Ccode = college.code
JOIN klas ON college.Kcode = klas.code
WHERE klas.code = '".$klas."' AND vak.code = 'WFHBOICT.M032.16' AND college.college = '8'
AND aanwezigheid = '1'
)
/
(
SELECT Count(aanwezigheid)
FROM aanwezigheid
JOIN college ON aanwezigheid.Ccode = college.code
JOIN klas ON college.Kcode = klas.code
WHERE klas.code = '".$klas."' AND vak.code = 'WFHBOICT.M032.16' AND college.college = '8'
)
* 100)
as percentage
FROM aanwezigheid
JOIN college ON aanwezigheid.Ccode = college.code
JOIN klas ON college.Kcode = klas.code
JOIN vak ON college.Vcode = vak.code
WHERE klas.code = '".$klas."' AND vak.code = 'WFHBOICT.M032.16' AND college.college = '8'
GROUP BY klas.code");
//execute query
$result = $mysqli->query($query);
//loop through the returned data
$data = array();
foreach ($result as $row) {
$data[] = $row;
}
print json_encode($data);
這是結果:
[{"klas":"WFHBOICT.V1F","percentage":"67"}]
它必須在短短1代替返回括號中這兩個類...
查看https://meta.stackoverflow.com/questions/333952/why-should-i-provide-an-mcve-for-what-seems-to-me-to-be-a-very-simple- SQL查詢 – Strawberry