【发布时间】:2016-10-20 12:00:58
【问题描述】:
这是我的第一个 json
[
{
"future_sell": "2300.00",
"to_currency": "DKK",
"creation_datetime": "2016-10-20 12:23:29",
"title": "Total selling currency",
"title_two": "Date : 2016-10-20",
"description": "DKK = 2300.00",
"start": "2016-10-20",
"end": "2016-10-20"
},
{
"future_sell": "536.66",
"to_currency": "USD",
"creation_datetime": "2016-10-20 15:27:36",
"title": "Total selling currency",
"title_two": "Date : 2016-10-20",
"description": "USD = 536.66",
"start": "2016-10-20",
"end": "2016-10-20"
},
{
"future_sell": "600.00",
"to_currency": "USD",
"creation_datetime": "2016-10-21 13:51:28",
"title": "Total selling currency",
"title_two": "Date : 2016-10-21",
"description": "USD = 600.00",
"start": "2016-10-21",
"end": "2016-10-21"
}
]
我想从上面制作下面的json
[
{
"creation_datetime": "2016-10-20 12:23:29",
"title": "Total selling currency",
"title_two": "Date : 2016-10-20",
"description": "DKK = 2300.00,USD = 536.66",
"start": "2016-10-20",
"end": "2016-10-20"
},
{
"creation_datetime": "2016-10-21 13:51:28",
"title": "Total selling currency",
"title_two": "Date : 2016-10-21",
"description": "USD = 600.00",
"start": "2016-10-21",
"end": "2016-10-21"
}
]
表示如果 creation_datetime 字段中的相同日期和 Description 字段中的 to_currency 则要合并
我的数组在下面是第一个数组 json 对象
foreach ($future_sell_data as $key => $value)
{
$future_sell_data[$key]['title'] = "Total selling currency";
$future_sell_data[$key]['title_two'] = "Date : " .date("Y-m-d",strtotime($value['creation_datetime']));
$future_sell_data[$key]['description'] = $value['to_currency']." = ".$value['future_sell'];
$future_sell_data[$key]['start'] = date("Y-m-d",strtotime($value['creation_datetime']));
$future_sell_data[$key]['end'] = date("Y-m-d", strtotime($value['creation_datetime']));
}
谁能帮助我如何制作第二个 json 对象?
【问题讨论】:
-
在您的示例第二个 json 中,creation_datetime 不一样......
-
抱歉只是日期而不是时间
-
你能帮帮我吗?
-
为什么人们被否决了?我不明白?我的问题很真实,我想要解决方案。这就是我发布问题的原因....
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