Question

我有一个jquery日历，用于存储日期onselect，我想将所选日期发送到php文件，以便根据所选日期从数据库中获取结果..这是我的代码： -

enter code here<div id="calendar"></div>

日期：

my scripr:-<script> var x =''; $（ '＃日历'）。日期选择器（{

altField: '#datepicker_send',

    inline: true,

    firstDay: 1,

    showOtherMonths: true,

    altFormat: "dd/mm/yy",

    dateFormat: "dd/mm/yy",

    dayNamesMin: ['Sun', 'Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat'],

     onSelect: function(dateText){
     $('#event-date').text(dateText)
      x=dateText;          
     // alert(x);
     $.ajax({ type: "GET", 
     url: 'check_events.php', 
     data: { y : x}, 
     //dataType: 'json',
     success: function(data)
     { alert(data); }   /// tried to get the value of date
          } );
    }
    });
</script>

Check_events.php： -

enter code here<?php mysql_connect("localhost", "root", "") or die(mysql_error());

mysql_select_db("users") or die(mysql_error());

if(isset($_REQUEST['y']))

{

$y = date_format($_REQUEST['y'],'YYYY-MM-DD');

echo $y;

exit; //trying to print the date

//$uname = mysql_real_escape_string($x);

$sql_check = mysql_query("SELECT * FROM events WHERE date='$x'");

while($res=mysql_fetch_assoc($sql_check))

{

print_r($res);

}

}

?>

Answer 1

$.ajax({ type: "GET", 
 url: 'check_events.php', 
 data: { y : x}, 
 //dataType: 'json',
 success: function(data)
 { 
      alert(data); //Alert Data on success not result.
 }   
 });

Answer 2

只需从此处删除不必要的,：

success: function(data),

所以新代码应该是：

$.ajax({ type: "GET", 
  url: 'check_events.php', 
  data: { y : x}, 
  //dataType: 'json',
  success: function(data) { 
    alert(data); //Alert Data on success not result.
  }
});

通过Ajax on Select将Javascript变量传递给PHP文件的问题

2 个答案: