我使用jQuery datepicker在输入日期中选择一个日期。将日期转换为日期时间时出错
,即时通讯在日期选择器脚本中使用的格式为:dateFormat: 'DD, d MM, yy',
所以我在输入获取此日期:“金塔 - 费拉,1马尤2014”(葡萄牙语日期)。
但现在我需要将此日期转换为保存,如日期时间在MySQL中。
我只需要下面使用此代码,如果日期是在英语:
$date = DateTime::createFromFormat('l, j F, Y', $_POST['date']);
echo $date->format('Y-m-d');
但我的约会心不是英语,所以我需要做的转换和我试着用下面的“convertDate”的功能做这个。
但是,当我调用函数,并我通过输入日期值是这样的:convertDate($_POST['date']);
我收到一个错误“电话在该成员函数格式()非对象上行:$day= $date->format("l");
你看到了什么在这里因为温控功能对我来说似乎不错
function convertDate($myDate){
$date = DateTime::createFromFormat('Ymd', $myDate);
$day = $date->format("l");
$daynum = $date->format("j");
$month = $date->format("F");
$year = $date->format("Y");
switch($day)
{
case "Segunda-Feira": $day = "Monday"; break;
case "Terça-Feira": $day = "Tuesday"; break;
case "Quarta-Feira": $day = "Wednesday"; break;
case "Quinta-Feira": $day = "Thursday"; break;
case "Sexta-Feira": $day = "Friday"; break;
case "Sábado": $day = "Saturday"; break;
case "Domingo": $day = "Sunday"; break;
default: $day = "Unknown"; break;
}
switch($month)
{
case "Janeiro": $month = "January"; break;
case "Fevereiro": $month = "February"; break;
case "Março": $month = "March"; break;
case "Abril": $month = "April"; break;
case "Maio": $month = "May"; break;
case "Junho": $month = "June"; break;
case "Julho": $month = "July"; break;
case "Agosto": $month = "August"; break;
case "Setembro": $month = "September"; break;
case "Outubro": $month = "October"; break;
case "Novembro": $month = "November"; break;
case "Dezembro": $month = "December"; break;
default: $month = "Unknown"; break;
}
echo $daynum . ", " . $month . ", " . $year;
}
与S尝试?! tr_ireplace:
$english = array("Segunda-Feira","Terça-Feira","Quarta-Feira","Quinta-Feira","Sexta-Feira","Sábado","Domingo");
$portuguese = array("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday");
$result= str_ireplace ($english , $portuguese, $_POST['date']);
$date = DateTime::createFromFormat('l, j F, Y', $result);
echo $date->format('Y-m-d');
我echo $date->format('Y-m-d')
得到了同样的错误
做一个'var_dump($ myDate)'给我们看结果请。 –
我收到这个:“string(25)”Sexta-feira,2 Maio,2014“”,这是我在日期选择器中选择的日期! – OzzC
你回到你最后一个问题中的同样的问题。该日期格式不适用于'DateTime()'(并且您在createFromFormat中的格式不正确)。我建议使用正则表达式将此日期字符串分解为您需要的部分。从那里你的代码的其余部分应该工作。 –