PHP中基于整数值的日或日格式

这是：-

现在，根据英语语法，我想输出+1天而不是+1天

请帮助我学习任何速记三值逻辑

谢谢你，

---

Sukumar

您可以执行以下操作，存储格式化的日期，然后使用三元运算符：

<?php

function ConvertintoDays($seconds) {
    $zeroSecond = new DateTime("@0");
    $givenSeconds = new DateTime("@$seconds");
    $day = $zeroSecond->diff($givenSeconds)->format('%a');
    return ($day == 1) ? '+'.$day.' day' : '+'.$day.' days';
}

echo ConvertintoDays(86400);

---

您可以使用三元运算符并在return语句中手动添加
+

<?php

    function ConvertintoDays($seconds)
    {
        $zeroSecond = new DateTime("@0");
        $givenSeconds = new DateTime("@$seconds");
        $day = $zeroSecond->diff($givenSeconds)->format('%a');
        return '+' . (($day == 1) ? $day . ' day' : $day . ' days');
    }

    echo ConvertintoDays(86400);
?>



示例：
格式（“%R%一天）
<?php

function ConvertintoDays($seconds) {
    $zeroSecond = new DateTime("@0");
    $givenSeconds = new DateTime("@$seconds");
    $day = $zeroSecond->diff($givenSeconds)->format('%a');
    return ($day == 1) ? '+'.$day.' day' : '+'.$day.' days';
}

echo ConvertintoDays(86400);

<?php

    function ConvertintoDays($seconds)
    {
        $zeroSecond = new DateTime("@0");
        $givenSeconds = new DateTime("@$seconds");
        $day = $zeroSecond->diff($givenSeconds)->format('%a');
        return '+' . (($day == 1) ? $day . ' day' : $day . ' days');
    }

    echo ConvertintoDays(86400);
?>