想要从数据库中获取用户信息,并希望在日历上显示

时间:2016-05-18 07:06:29

标签: javascript php jquery jquery-ui

我有这个PHP代码我得到了一个用户预订的完整日期,如12-15 17-18,对于第二个用户20-21,我在日历上标记了这些日期。

现在我想显示鼠标悬停的用户信息:

<?php 

    include 'dbconfig.php';


    $query_getaway = "SELECT * FROM getaway_table; ";
    $result_getaway = $db->query($query_getaway);
     $category = mysqli_real_escape_string($db, filter_input(INPUT_POST, 'category'))  .'<br>'; 
     $resort = mysqli_real_escape_string($db, filter_input(INPUT_POST, 'resorts'))  .'<br>'; 

    if( $category != 0 ){

    $query = "SELECT * FROM booking_request where room_id = '$resort' '<br>' ";
    $result = $db->query($query);
            if( $result -> num_rows > 0 ) {




                while( $row = $result-> fetch_assoc() )
                {
              $name = $row['getaway_name']." <br>";
               $start = $row['check_in_date']." <br>";
                $end = $row['check_out_date']." <br>";
                $guest_n[]= $row['guest_name'];
                $guest_number[]= $row['guest_phone'];


                $arr1 = range(strtotime($row['check_in_date']),strtotime($row['check_out_date']), "86400");


                array_walk_recursive($arr1, function(&$element) { $element = date("j-n-Y", $element); });

              foreach ($arr1  as $value)

              {
                  $arr2[]= $value;
              }
           $arrt= (array_merge($guest_n, $guest_number));
           //print_r($arrt);
           foreach($arrt as $val)
           {
               echo $arr_val[]=$val;
           }

    $arr[]= $value;

    }
    }
    else { echo "select" ;}

?>

然后在js部分我显示了不可用的日期,机器人在悬停时我无法向用户显示完整的信息:

<script type="text/javascript">
    // Add New Dates and Block Out Booked Days Below 
    /*var unavailableDates1 = ["20-5-2016", "21-5-2016", "22-5-2016", "23-5-2016", "5-6-2016", "7-6-2016", "8-6-2016",
                          "15-6-2016", "16-6-2016", "25-6-2016", "26-6-2016", "27-6-2016", "28-6-2016", "14-2-2016",
                          "15-7-2016", "16-7-2016", "17-7-2016", "18-7-2016", "19-7-2016","20-7-2016","21-7-2016",
                          ];   */
    var unavailableDates = <?php echo json_encode($arr2); ?>;


    document.write(unavailableDates);

    function unavailable(date) {

        dmy = date.getDate() + "-" + (date.getMonth() + 1 ) + "-" + date.getFullYear();

        if($.inArray(dmy, unavailableDates) == -1) {

            return [true, ""];

        }else {

this line is showing unavlaible on hover, i want from database user info            
            return [false, "","unavalible"];

                    <?php echo $val; ?>
        }
    }
    $(function() {
        $("#Datepicker1").datepicker({
            numberOfMonths:3,
            beforeShowDay: unavailable
        }); 
    });

</script>

0 个答案:

没有答案