Question

我有这个代码，它将从循环输入中获取日期，使用map函数使其成为数组。

var target_completion_date = $('input#target_completion_date').map(function() {
       return $(this).val();
       }).get();

这个JavaScript的输出在字符串数组中是这样的：

一分之二十七/ 2014,25 / 1 / 2014,29 /二千〇一十四分之一

问题是，如何编写更新MySQL或转换（27/1 / 2014,25 / 1 / 2014,29 / 1/2014）到（2014-01-27,2014-01-25,2014） -01-25）保存并更新到数据库之前：

$target_completion_date = $_POST['target_completion_date']; 



> for ($i = 0; $i <5;)  {

> $sql1 = "UPDATE AT_tna_assessment SET identify_needs = '$identify_needs[$i]', activity_required = '$part_c_activity[$i]',
> priority = '$part_c_priority[$i]', **target_completion_date = ?** , remarks = '$part_c_remark[$i]'
> WHERE id ='$part_c_id[$i]'";

> mysql_query($sql1) or die ("Cant Connect");           
$i++;       
}

我使用target_completion_date = DATE(STR_TO_DATE('$target_completion_date[$i]','%m-%d-%Y'));但没有工作

我已经得到了我的回答。谢谢所有

我已通过此代码解决了这个问题。谢谢：）

  for ($i = 0; $i <5;) {

    $tb = explode("/", ($target_completion_date[$i]));   
    date_duration_from[$i] = $tb[2] . "-" . $tb[1] . "-" . $tb[0].",";


    $sql1 = "UPDATE AT_tna_assessment SET identify_needs = '$identify_needs[$i]',   activity_required = '$part_c_activity[$i]', priority = '$part_c_priority[$i]',  target_completion_date= '$date_duration_from[$i]' , remarks = '$part_c_remark[$i]' WHERE id
    ='$part_c_id[$i]'";


     mysql_query($sql1) or die ("Cant Connect");
     $i++;   

    }

Answer 1

有类似......

<?PHP
$date_array = explode(",", "27/1/2014,25/1/2014,29/1/2014");
foreach($date_array as $date) {
    $timestamp = strtotime($date);
    $mysql_date[] = date("Y-m-d", $timestamp);
}
echo(implode(",", $mysql_date));

Answer 2

list($day, $month, $year) = explode('/', '27/1/2014');
echo sprintf('%4d-%02d-%02d', $year, $month, $day);

Answer 3

试试这段代码。

$date_array = explode(",", "27/1/2014,25/1/2014,29/1/2014");

echo 'Unformat dates: '.implode(",", $date_array).PHP_EOL;

foreach($date_array as $date) {
    list($day,$month,$year) = explode('/', $date);
    $date_fmt[] = date('Y-m-d', strtotime($year.'-'.$month.'-'.$day));
}

echo 'Format dates: '.implode(",", $date_fmt).PHP_EOL;

输出：

Unformat dates: 27/1/2014,25/1/2014,29/1/2014
Format dates: 2014-01-27,2014-01-25,2014-01-29

PHP：如何在更新mysql之前将日期的数组字符串转换为mysql日期（27/1/2014）到（2014-01-27）

3 个答案: