Question

如何在php中将INT转换为日期，而不是在mysql表中找到这个日期？

让我说我想转换

$month = (int)$_GET['month'];
$month = 42014 ;//This means 04th month in 2014 year
$month = 04-2014; // I want this

而不是在mysql表中找到

$query=mysql_query("SELECT id FROM matches WHERE date=$month");// This need to select
                                                               // all data that is
                                                               // created in 04th month
                                                               // of 2014 year .



echo "Thanks :)";

Answer 1

不要像你那样传递$month参数。

发送并传递$month和$year并在以下情况下处理：

$month = (int)$_GET['month'];
$year = (int)$_GET['year'];
if($month < 10) { // add this check for months lesser then october (they containt 1 digit, which is wrong for t-sql)
    $month = "0".$month;
}
$query=mysql_query("SELECT id FROM matches WHERE DATE_FORMAT(datefield, '%m-%Y') = '$month-$year'");

source

如果您不能同时发送不同变量的月份和年份，请执行此操作，例如M Khalid Junaid建议：

$query=mysql_query("SELECT id FROM matches WHERE DATE_FORMAT(datefield,'%m%Y')='$month'");

Answer 2

使用 DateTime::createFromFormat

尝试以下操作

$month = $_GET['month'];
$date = DateTime::createFromFormat('mY',$month);
echo $date->format('m-Y');

Answer 3

试试这个

$input=42014;
$year = strlen(substr($input,-4));
$month = substr(0,(strlen($input)-strlen($year)));
$final = $month."-".$year

MySQL - 将INT转换为DATE格式。

3 个答案: