我想要一个json来知道在“classid”:1中,“studentsid”:[1,2,3,4]在类中没有apear,我试图做一个查询,给我带来“classid”和“studentsid”,但是它们都在同一个表中,我不知道仅仅更改查询部分是否能带来我需要的东西。所以我的table会像:
Table: School
ClassID | StudentsID
1 1
1 2
2 2
2 3
2 4
2 5
3 1
3 2
因此,正如您在“classid”中看到的:1“studentsid”:[1和2]丢失了。我需要的json是这样的:
{"Misses":[{"classID":1,"StudentsMissing":[1,2]},{"classID":2,"StudentsMissing":[2,3,4,5]}]}
我已经尝试过arraypush,array\u merge,但是由于我是php新手,所以我不能像这样使用json。
我只能得到这个json:
{"Misses":[{"classID":"1"}]}{"Students":[{"StudentsMissing":"1"}]}
我试过的代码是:
<?php
$list = array();
$bd = mysqli_connect("localhost","root","","web");
if ($bd) {
$qry = mysqli_query($bd, "SELECT ClassID FROM School ORDER BY ClassID");
$qry1 = mysqli_query($bd, "SELECT StudentID FROM School");
if (mysqli_num_rows($qry) > 0){
while ($row = mysqli_fetch_object ($qry)){
$list[] = $row;
}
$output = new StdClass();
$output->ClassID = $list;
echo json_encode($output);
}
$list = array();
if (mysqli_num_rows($qry1) > 0){
while ($row = mysqli_fetch_object ($qry1)){
$list[] = $row;
}
$output = new StdClass();
$output->Students = $list;
echo json_encode($output);
}
else {
echo "Erro";
}
}
else {
echo "Erro";
}
我有这样一个json:
{"Misses":[{"classID":"1"},{"classID":"2"},{"classID":"3"}]}{"Students":[{"StudentsMissing":"1"},{"StudentsMissing":"2"},{"StudentsMissing":"3"}]}
1条答案
按热度按时间ogsagwnx1#
只需一个查询,使用
GROUP_CONCAT
把同一个班的所有学生合并在一起。$list = array();
while ($row = mysqli_fetch_object($qry)) {
$list[] = array('ClassID' => $row['ClassID'], 'StudentsID' => explode(',', $row['StudentsID']);
}
echo json_encode($list);