开始和结束之间的每个日期的 JSON MySQL 添加到 JSON 编码答案

【问题标题】：JSON MySQL for each date between start and end add to JSON encode开始和结束之间的每个日期的 JSON MySQL 添加到 JSON 编码
【发布时间】：2012-09-18 15:48:29
【问题描述】：

我在 MySQL 中有一个表，其中包含 startdate 和 enddate。

1 行的示例是（忽略整个命名）：

eg 
startdate -> 20121224
endate -> 20121226
title -> name
etc

我当前的脚本效果很好，并为每一行添加到 JSON，但是..... 我想....为每个 mysql 行获取 startdate 和 enddate，对于包含在 startdate 和 enddate 中的每个日期添加到 JSON 编码。

因此而不是上面返回 1 行的例子：

[{
    "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"
}]

它将为每个日期的一行返回 3 个 JSON：

[{
    "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));
?>

【问题讨论】：

标签： php arrays associative-array json

【解决方案1】：

将此添加到您的 sql 选择查询中：

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

然后做这样的事情：

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;
    }
}

你可能需要调整它，

【讨论】：