我有这个查询
SELECT json_object('image_name', image_name, 'iid', iid) FROM `posts_images` WHERE `pid` = '62'
导致此
我可以将这些结果转换为同一查询中的一行json数组吗?
答案 0 :(得分:0)
您可以执行以下操作:
$query = "SELECT json_object('image_name', image_name, 'iid', iid) FROM `posts_images` WHERE `pid` = '62'";
$result = mysqli_query($query);
$data = array();
while($row = mysqli_fetch_assoc($result)) {
$data[] = $row;
}
print json_encode($data);
答案 1 :(得分:0)
var o = {"0":"1","1":"2","2":"3","3":"4"};
var arr = $.map(o, function(el) { return el; })
console.log(arr)
试试这个
答案 2 :(得分:0)
我想通了
SELECT CONCAT(
'[',
GROUP_CONCAT(json_object('image_name', image_name, 'iid', iid)),
']'
)
FROM posts_images WHERE pid = '62'
[{" iid":99," image_name":" 6_1499203101812.jpg"},{" iid":98, " image_name":" 6_1499203101708.jpg"},{" iid":97," image_name": " 6_1499203101651.jpg"},{" iid":96," image_name": " 6_1499203101574.jpg"},{" iid":95," image_name": " 6_1499203101472.jpg"},{" iid":100," image_name": " 6_1499203101886.jpg"},{" iid":101," image_name": " 6_1499203101960.jpg"},{" iid":102," image_name": " 6_1499203102020.jpg"}]