我的代码有一个小问题,可以说我有一个像这样的json:
Select Max(x.price), x.month, x.year from(
SELECT SUM(price) as price, Year(paymensDate) as year, Month(paymentDate) as month FROM `payments`
where idGroup=27 group by Year(paymentDate), Month(paymentDate)
) as x
我想像这样整理数据:
John @john john.png
Mark @mark mark.png
但是每次数据出来都是这样的:
John Mark @john @mark john.png mark.png
这是我的密码:
[{"img":"john.png","name":"John","username":"@john"},
{"img":"mark.png","name":"mark","username":"@mark"}]
这是索引代码:
<?php
class search{
public function gettingvalues($search_value){
require_once('conx.php');
$dir = "usersimage/";
$sql = "SELECT name,img,username FROM users WHERE username like '$search_value%' || name like '$search_value%'";
$query = mysqli_query($conx,$sql);
if ($query) {
if (mysqli_num_rows($query) > 0) {
while ($row = mysqli_fetch_array($query)) {
$img = $row['img'];
$name = $row['name'];
$username = $row['username'];
$json = array('img' => $img, 'name' => $name, 'username' => $username);
$results[] = $json;
}
echo json_encode($results);
}
}
}
}
?>
实际上,这是我第一次使用json,我不知道出了什么问题,或者我错过了一些事情,希望得到一些答案。
答案 0 :(得分:0)
您需要按照显示顺序构建HTML。
{{1}}