Php 开始和结束之间的每个日期的JSON MySQL添加到JSON编码

我在MySQL中有一个表，它有一个

startdate

和

enddate

一行的示例如下（忽略整个命名）：

我目前使用的脚本非常好，每行都添加了JSON，但是。。。。。 我想。。。。对于每个mysql行，获取startdate和enddate，对于startdate和enddate中包含的每个日期，添加到JSON编码

而不是上述返回一行的示例：

[{
    "user": "378",
    "date": "UNIX_TIMESTAMP(startdate)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(enddate)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}]

[{
    "user": "378",
    "date": "UNIX_TIMESTAMP(1st date)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(1st date)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}, {
    "user": "378",
    "date": "UNIX_TIMESTAMP(2nd date)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(2nd date)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}, {
    "user": "378",
    "date": "UNIX_TIMESTAMP(3rd date)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(3rd date)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}, ]

---

它将为每个日期的一行返回3个JSON:

[{
    "user": "378",
    "date": "UNIX_TIMESTAMP(startdate)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(enddate)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}]

[{
    "user": "378",
    "date": "UNIX_TIMESTAMP(1st date)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(1st date)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}, {
    "user": "378",
    "date": "UNIX_TIMESTAMP(2nd date)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(2nd date)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}, {
    "user": "378",
    "date": "UNIX_TIMESTAMP(3rd date)",
    "title": "name",
    "description": "6 Days",
    "url": "UNIX_TIMESTAMP(3rd date)",
    "bmanager": "manager name",
    "academic_year": "20120801",
    "division": "division name",
    "manager_id": "3"
}, ]

---

JSON

<?php
    header('Content-type: text/json');

    $lmanager = $_GET["lmanager"];

    $sqldata = mysql_query('
    SELECT
    requests.user,
    UNIX_TIMESTAMP(requests.start_date) AS date,
    requests.employee AS title, 
    requests.days AS description,
    UNIX_TIMESTAMP(requests.end_date) AS url,
    business.line_manager AS bmanager,
    requests.academic_year,
    business.academic_year,
    business.division,
    line_managers.userid AS manager_id
    FROM requests 
    INNER JOIN holiday_entitlement_business_manual AS business ON requests.user=business.userid AND requests.academic_year=business.academic_year
    INNER JOIN line_managers ON business.line_manager=line_managers.name
    WHERE requests.approved = 1 AND line_managers.userid = '.$lmanager.'
    ');

    $posts = array();

    while($row = mysql_fetch_assoc($sqldata))
    {
        $row['date'] = $row['date'].'000';
        $row['url'] = $row['url'].'000';
        $row['description'] = $row['description'].' Days';
        $posts[] = $row;
    }

    mysql_free_result($sqldata);
    die(json_encode($posts));
?>

---

将此添加到sql select查询：

select datediff(requests.end_date, requests.start_date) as numdays

然后像这样做：

while($row = mysql_fetch_assoc($sqldata))
{
    for($i = 0; $i < $row['numdays']; $i++){
        $row['date'] = $row['date']+$i*24*60*60; // add 24 hours in seconds to the start_date stamp
        $posts[] = $row;
    }
}

while（$row=mysql\u fetch\u assoc（$sqldata））
{
对于（$i=0；$i<$row['numdays']；$i++）{
$row['date']=$row['date']+$i*24*60*60；//在开始日期戳中以秒为单位添加24小时
$posts[]=$row；
}
}

你可能需要调整一下