PDO选择并显示具有相同日期的数据数组

Question

我已经编写了这段代码。显示基于同一日期(天(的数据

$group = array();
$stmt = $conn->prepare("
SELECT *,date(date_added) as dateadded
FROM rr
ORDER BY dateadded DESC
");
$stmt->execute();
$stmt->setFetchMode(PDO::FETCH_ASSOC);
while ($res = $stmt->fetch()){
$rtype = array();
$rtype['type'] = "date";
$rtype['date'] = date_format (new DateTime($res['dateadded']), 'M jS, Y');
$rtype[] = $res;
$result[] = $rtype;
}

我从查询 sql 中得到的结果

array
(
[0] => Array
(
[type] => date
[date] => Dec 18th, 2019
[0] => Array
(
[id] => 1
[user_id] => 1
[status] => 1
[date_added] => 2019-12-18 13:44:30
[dateadded] => 2019-12-18
)
)
[1] => Array
(
[type] => date
[date] => Dec 20th, 2019
[0] => Array
(
[id] => 3
[user_id] => 2
[status] => 2
[date_added] => 2019-12-20 14:53:04
[dateadded] => 2019-12-20
)
)
[2] => Array
(
[type] => date
[date] => Dec 29th, 2019
[0] => Array
(
[id] => 5
[user_id] => 3
[status] => 0
[date_added] => 2019-12-29 00:39:21
[dateadded] => 2019-12-29
)
)
)

有些数据未在同一日期显示为结果。

我想展示的是这样的。

Array
(
[0] => Array
(
[type] => date
[date] => Dec 18th, 2019
)
[1] => Array
(
[id] => 2
[user_id] => 1
[status] => 1
[date_added] => 2019-12-18 13:44:30
)
[2] => Array
(
[type] => date
[date] => Dec 20th, 2019
)
[3] => Array
(
[id] => 1
[user_id] => 1
[status] => 0
[date_added] => 2019-12-20 14:53:04
)
[4] => Array
(
[id] => 3
[user_id] => 1
[status] => 0
[date_added] => 2019-12-20 14:53:04
)
)

结果变成了一个数组，就像按日期(天(分组一样。

如何获得此结果？ 谢谢你帮助我。

Answer 1

您需要将$group_date设置为$res['dateadded']的值，以便知道日期何时更改。没有必要使用substr()，因为dateadded只包含日期而没有时间。

你应该把$res推到$result上，而不是$rtype。

$group_date = "";
$group = array();
$stmt = $conn->prepare("
SELECT *,date(date_added) as dateadded
FROM rr
ORDER BY dateadded 
");
$stmt->execute();
$stmt->setFetchMode(PDO::FETCH_ASSOC);
while ($res = $stmt->fetch()){
$rtype = array();
if ($group_date !== $res["dateadded"]) {
$rtype['type'] = "date";
$rtype['date'] = date_format (new DateTime($res['dateadded']), 'M jS, Y');
$result[] = $rtype;
$group_date = $res["dateadded"];
}
$result[] = $res;
}