[英]How to convert array to date format in PHP?
我在PHP中得到一個數組:
Array
(
[1] => 2019
[2] => 5
[3] => 7
[4] => 0
)
其中[1]始終為年份,[2]始終為月份,[3]始終為日期。
如何將該數組轉換為date("Ymd")
格式?
假設此數據輸入:
$data = [null, 2019, 5, 7, 0];
使用日期時間
$dt = new DateTime(sprintf( "%04d-%02d-%02d", $data[1], $data[2],
$data[3]));
echo $dt->format('Y-m-d') . "\n";
使用Sprintf
// use this if you really trust the data
$dt = sprintf( "%04d-%02d-%02d", $data[0], $data[1], $data[2]);
echo $dt . "\n";
使用碳
// Carbon is a fantastic Date and Time class -> https://carbon.nesbot.com/
$dt = \Carbon\Carbon::create($data[0], $data[1], $data[2], 0, 0, 0);
echo $dt->format('Y-m-d') . "\n";
您可以使用DateTime
$timeArray = [2019,5,7,0];
$dateTime = new DateTime(printf( "%d-%d-%d", $timeArray[0],$timeArray[1],$timeArray[2] ));
echo $dateTime->format('Y-m-d'); // output: 2019-05-07
像這樣做
$arr = array( '2019', '5', '7', '0' );
echo date('Y-m-d',strtotime("$arr[0]/$arr[1]/$arr[2]"));
盡管可以將這些值連接成一個字符串,然后讓PHP將字符串解析為Ymd
格式,但我個人認為mktime()
是更好的解決方案:
echo date("Y-m-d", mktime(0, 0, 0, $arr[2], $arr[3], $arr[1]));
// 2019-05-07
這消除了PHP意外地以錯誤的順序解釋日期和月份的風險。
您可以簡單地使用concat並將它們連接到字符串中:
$arr = array(
"1" => "2019",
"2" => "5",
"3" => "7",
"4" => "0",
);
$datetime_format = $arr["1"] . "-" . $arr["2"] . "-" . $arr["3"];
var_dump($datetime_format);
string(8) "2019-5-7"
如果您希望使用4-2-2格式,則可以這樣做:
$arr = array(
"1" => "2019",
"2" => "5",
"3" => "7",
"4" => "0",
);
$datetime_format = '';
foreach ($arr as $key => $value) {
if ($key == "4") {break;}
echo strlen($value);
if (strlen($value) >= 2) {
$datetime_format .= $value;
} elseif (strlen($value) == 2) {
$datetime_format .= $value;
} elseif (strlen($value) == 1) {
$datetime_format .= "0" . $value;
} else {
echo "Something is not right!";
}
if ($key <= "2") {$datetime_format .= '-';}
}
var_dump($datetime_format);
string(10) "2019-05-07"
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