2017-05-23 26 views
-5

我想从表td调用一个javascript函数,并将此td值作为此javascript函数的参数传递。这是我的脚本和表格。我想在表中调用一个JavaScript函数td php

<table width="100%" class="track"> 
        <tr> 
         <th colspan="2">Load Information</th> 
        </tr>      
        <tr> 
         <td>LOAD DATE:</td> 
         <td id="ldate"><?php echo $datas[0]->DATE?> </td> 
        </tr> 
       </table> 
<script> 
     function formatDate(date_string) { 
      var date = new Date(date_string); 
      var monthNames = [ 
       "January", "February", "March", 
       "April", "May", "June", "July", 
       "August", "September", "October", 
       "November", "December" 
      ]; 
      var day = date.getDate(); 
      var monthIndex = date.getMonth(); 
      var year = date.getFullYear(); 

      if (day < 10) { 
       day = '0' + day; 
      } 
      return monthNames[monthIndex] + ' ' + day + ', ' + year ; 
     } 
    </script> 

我想显示像MARCH 03,2016这样的日期格式,但我的实际格式将显示为2016-03-03。

+0

通话formatDate功能。 – spankajd

+0

然后是什么问题? –

+0

你在哪里调用javascript函数'formatDate'? – lkdhruw

回答

1

使用strtotime()date()使用PHP:文件准备好后

$originalDate = "2010-03-21"; 
$newDate = date("d-m-Y", strtotime($originalDate)); 
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