Question

我试图在两个日期之间进行比较，但不幸的是，通过strtotime将其转换为UNIX格式，这不起作用。我试图将日期与另一个日期进行比较。

但是这种格式有效：

if(strtotime("22-04-17") < strtotime("25-05-17")){
    echo 'Date One is smaller than date two';
}

但很多时候它都失败了。我在网上看过很多例子，但我找不到任何好的东西！

if(strtotime("22-04-17") < strtotime("04-05-17")){ //passing still the 
    // bigger on but not working
    echo 'Date One is smaller than date two';
}

Answer 1

来自manual（特别注意我输入的部分粗体）：

“m / d / y或dmy格式的日期通过查看各个组件之间的分隔符来消除歧义：如果分隔符是斜杠（/），则假定为美国m / d / y;而如果分隔符是短划线（ - ）或点（。），然后假定为欧洲dmy格式。但是，如果年份以两位数格式给出，而分隔符是短划线（ - ，日期string被解析为ymd。“

所以这就是你正在做的事情，PHP正在解释你的字符串的日期在评论中：

// Is 17 April 2022 earlier than 17 May 2025? Yes.
if(strtotime("22-04-17") < strtotime("25-05-17")){
    echo 'Date One is smaller than date two';
}

// Is 17 April 2022 earlier than 17 May 2004? No.
if(strtotime("22-04-17") < strtotime("04-05-17")){ //passing still the 
    // bigger on but not working
    echo 'Date One is smaller than date two';
}

我希望这会让您解决问题。

正如手册中所述，如果您想避免含糊不清，请使用DateTime::createFromFormat / date_create_from_format：

 $date = date_create_from_format('d-m-y', '04-05-17'); // 4 May 2017

Answer 2

试试这个

$date1 = date('d-m-y',strtotime("22-04-17"));
$date2 = date('d-m-y',strtotime("04-05-17"));;

if((int)strtotime($date1) < (int)strtotime($date2)){ //passing still the  
    echo 'Date One is smaller than date two';
}

您的年份格式17导致strtotime功能

中的问题

Answer 3

您的比较无效，因为strtotime("22-04-17")实际上会导致此日期的时间戳：17th April 2022;

执行以下操作，您将看到我的意思。以下代码将输出＆＃39; 2022-May-17`

$date = "22-05-17";
echo date ("Y-M-d ",strtotime($date))."<br>";

Answer 4

我知道这不是你想要的，但是你试过这样做来显示日期是否更大/更小？

// Dates
$Date1 = strtotime("22-04-17 GMT");
$Date2 = strtotime("04-05-17 GMT");

// Diff between dates
$Diff = (int)floor(($Date1 - $Date2) / (60*60*24));

if ($Diff < 0) {
   echo "The Diff is negative";
}

然后另一种方式就像这个答案：LINK

$date1 = strtotime("22-04-17 GMT");
$date2 = strtotime("04-05-17 GMT");

if((int)strtotime($date1) < (int)strtotime($date2)){ //passing still the  
    echo 'Date One is smaller than date two';
}

Answer 5

我写了一个函数，它接受你的日期字符串并正确返回timestmp。请注意，我遵循PHP的2位数年份惯例，i.e. 00-69 are mapped to 2000-2069 and 70-99 to 1970-1999

/* functon takes dateString of format dd-dd-yy and return proper timestamp */
function getTimestamp($dateString)
{
    $res = preg_match('/^([0-9]{2})\-([0-9]{2})-([0-9]{2})/', $dateString, $matches);
    if(!$res)
        return 0;
    //00-69 are mapped to 2000-2069 and 70-99 to 1970-1999
    if($matches[3]>=70 &&  $matches[3]<=99 )
        $year = "19".$matches[3];
    else
        $year = "20".$matches[3];
    $formatted_dat_string = $year."-".$matches[2]."-".$matches[1];
    return strtotime($formatted_dat_string);
}
getTimestamp("22-04-99");

因此，您现在可以使用此函数代替strtotime进行比较。

Answer 6

最后，我得到了解决方案。 dateTime对处理这种情况没有任何好处。您应该转而使用//First you need to pass the original format then you will need to pass new //Format to get this working properly. Hope this will help you guy's $myDateTimestart = DateTime::createFromFormat('d-m-Y', $dateString); $startDate = $myDateTimestart->format('m-d-Y'); //with Simply that you can format your date properly对象。

                //* Linear Layout *
            LinearLayout linearLayout = new LinearLayout(this);
            linearLayout.setTag("linearLayout" + Integer.toString(id));
            linearLayout.setOrientation(LinearLayout.VERTICAL);
            linearLayout.setBackgroundResource(R.drawable.background_border_1);
            linearLayout.setGravity(Gravity.CENTER_VERTICAL);

            LinearLayout.LayoutParams layoutParams = new LinearLayout.LayoutParams(300, 70);
            layoutParams.setMargins(20, 10, 20, 10);

            linearLayout.setWeightSum(10);
            linearLayout.setLayoutParams(layoutParams);

            Toast.makeText(getBaseContext(), Integer.toString((int)linearLayout.getWeightSum())+ ", " +  Integer.toString( linearLayout.getHeight()) + ", " + Integer.toString( linearLayout.getWidth()), Toast.LENGTH_LONG).show();

            //* Edit Text *
            EditText editText = new EditText(this);
            editText.setTag("editText" + Integer.toString(id));
            layoutParams.width = LinearLayout.LayoutParams.WRAP_CONTENT;
            layoutParams.height = 50;
            layoutParams.setMargins(10, 0, 10, 0);
            layoutParams.weight = 2;
            editText.setEms(10);
            editText.setInputType(InputType.TYPE_CLASS_NUMBER);
            editText.setLayoutParams(layoutParams);
            editText.setGravity(Gravity.CENTER);

            //* Text View *
            TextView textView = new TextView(this);
            textView.setTag("textView" + Integer.toString(id));
            layoutParams.height = 50;
            layoutParams.setMargins(10, 0, 10, 0);
            layoutParams.weight = 4;
            textView.setLayoutParams(layoutParams);
            textView.setGravity(Gravity.CENTER);
            textView.setText("Credit Hours");

            //* Spinner *
            Spinner spinner = new Spinner(this);
            spinner.setTag("spinner" + Integer.toString(id));
            layoutParams.width = 50;
            layoutParams.height = LinearLayout.LayoutParams.MATCH_PARENT;
            spinner.setGravity(Gravity.FILL_VERTICAL | Gravity.CENTER | Gravity.CENTER_VERTICAL);
            layoutParams.setMargins(10, 0, 10, 0);
            layoutParams.weight = 4;
            spinner.setLayoutParams(layoutParams);

            linearLayout.addView(editText);
            linearLayout.addView(textView);
            linearLayout.addView(spinner);

            llSubjectsContainer = (LinearLayout) findViewById(R.id.llSubjectsContainer);
            llSubjectsContainer.addView(linearLayout);

我只是把这件事搞砸了，这真的很糟糕

为什么strtotime不在php上工作

6 个答案: