PHP - 将日期转换为 UTC 日期

发布于 2024-12-02 03:08:26 字数 925 浏览 0 评论 0原文

我从 mysql 数据库中提取了一个格式为 YYYY-MM-DD 的日期。

现在我有一些 javascript 需要这样的日期：

data: [
            [Date.UTC(2010, 0, 1), 29.9], 
            [Date.UTC(2010, 2, 1), 71.5], 
            [Date.UTC(2010, 3, 1), 106.4]
        ]

所以我想我可以执行以下操作：

while($row = mysql_fetch_array($sql)){  
      $thedate = explode('-', $row['date']); //$row['date'] = YYYY-MM-DD      
      $str .= '[Date.UTC(' . $thedate[0].','.$thedate[1].','.$thedate[2].'), '.$row['answer'].'],';
      $i++;
    }

然后使用变量 data 中的 AJAX 请求在浏览器端简单地捕获此内容，给出：

data: [data]

发出警报时我注意到 22 日和 22 日的数据内容如下： 8 月 23 日：

[Date.UTC(2011, 08, 22), 55], [Date.UTC(2011, 08, 23), 65]

当我需要的是

[Date.UTC(2011, 7, 22), 55], [Date.UTC(2011, 7, 22), 65]

有人可以告诉我如何将从数据库关系中提取的日期转换为正确的格式....

YYYY

原文

I have a date pulled from my mysql database in the form YYYY-MM-DD.

Now I have some javascript that requires the dates like this:

data: [
            [Date.UTC(2010, 0, 1), 29.9], 
            [Date.UTC(2010, 2, 1), 71.5], 
            [Date.UTC(2010, 3, 1), 106.4]
        ]

So I thought I could do the following:

while($row = mysql_fetch_array($sql)){  
      $thedate = explode('-', $row['date']); //$row['date'] = YYYY-MM-DD      
      $str .= '[Date.UTC(' . $thedate[0].','.$thedate[1].','.$thedate[2].'), '.$row['answer'].'],';
      $i++;
    }

And then simply capture this in the browser side using an AJAX request in a variable data, giving:

data: [data]

When alerting the contents of data I notice I get something like the following for 22nd & 23rd August:

[Date.UTC(2011, 08, 22), 55], [Date.UTC(2011, 08, 23), 65]

When what I need is

[Date.UTC(2011, 7, 22), 55], [Date.UTC(2011, 7, 22), 65]

Can someone tell me how to convert convert the date pulled from the db relation into the correct format....

YYYY

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赢得她心 2024-12-09 03:08:26

那是什么格式？无论如何，使用 DateTime::createFromFormat 解析日期，之后您可以按照您喜欢的方式格式化它。

示例：

while($row = mysql_fetch_array($sql)){  
    $Date = DateTime::createFromFormat( 'Y-m-d', $row['date'] ); 
    $str .= '[Date.UTC(' . $Date->format('Y') . ', ';
    $str .= $Date->format('n') - 1 . ', ';
    $str .= $Date->format('j') . '), ' . $row['answer'] . '],';
}

顺便看一下 json_encode http ://php.net/manual/en/function.json-encode.php

Which format is that? Anyway, use DateTime::createFromFormat to parse the date, after that you can format it anyway you like.

Example:

while($row = mysql_fetch_array($sql)){  
    $Date = DateTime::createFromFormat( 'Y-m-d', $row['date'] ); 
    $str .= '[Date.UTC(' . $Date->format('Y') . ', ';
    $str .= $Date->format('n') - 1 . ', ';
    $str .= $Date->format('j') . '), ' . $row['answer'] . '],';
}

By the way, have a look at json_encode http://php.net/manual/en/function.json-encode.php

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居里长安 2024-12-09 03:08:26

while($row = mysql_fetch_array($sql)){ 
  $timestamp = strtotime($row['date']); 
  $str .= '[Date.UTC(' . date('Y', $timestamp). ', ';
    $str .= date('n', $timestamp) - 1 . ', ';
    $str .= date('j', $timestamp) . '), ' . $row['answer'] . '],';
}

关于 DateTime - 这是一个很棒的类，但仅适用于 pphp >= 5.2

while($row = mysql_fetch_array($sql)){ 
  $timestamp = strtotime($row['date']); 
  $str .= '[Date.UTC(' . date('Y', $timestamp). ', ';
    $str .= date('n', $timestamp) - 1 . ', ';
    $str .= date('j', $timestamp) . '), ' . $row['answer'] . '],';
}

about DateTime - this is a great class but avaliable only for pphp >= 5.2

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