Question

这是我的第一篇文章，请原谅我的任何错误

我是JQUERY和AJAX的新手，所以请非常清楚地说明你的答案

我做了很多搜索，但无济于事，我正在使用PHP MYSQLI和JQUERY构建公寓预订系统，除了签到日期之外，我在同一页面上有2个jquery datepickers，除了结账日期之外，使用的MYSQL表称为预订，它有这些列 -

id，first_name，email，check_in_date，check_out_date

让我们说Mysql表返回2行，第一行返回check_in_date（2017-05-02）和check_out_date（2017-05-10）

和第二次返回check_in_date（2017-06-05）和check_out_date（2017-06-20）

如何在两个日期选择器上禁用这两个范围内的所有日期这就是我提出的问题

这是HTML表单的一部分

<div class="form-group">
<label for="check_in_date">Check In Date</label>
<input type="text" class="form-control" id="datepicker" placeholder="Check in Date">
</div>
<div class="form-group">
<label for="check_out_date">Check Out Date</label>
<input type="text" class="form-control" id="datepicker1" placeholder="Check Out Date">
</div>

这是Javascript - 它的不完整

$(function() {

  function checkAvailability(){

}

$( "#datepicker" ).datepicker({
    dateFormat : 'yy-mm-dd',
    beforeShowDay: checkAvailability        
    }       
});

$( "#datepicker1" ).datepicker({
    dateFormat : 'yy-mm-dd',
    beforeShowDay: checkAvailability

    }       
});

  });

PHP文件availability.php如下

$result = mysqli_query("SELECT check_in_date, check_out_date FROM bookings");

while ($row = mysqli_fetch_assoc($result)){

 $check_in_dates[] =  $row['check_in_date'];

 $check_out_dates[] =  $row['check_out_date'];  
 }

先谢谢

Answer 1

我在这里想出来的是解决方案

    <?php
    //below function returns all the dates within a given range
     function date_range($first, $last, $step = '+1 day', $output_format = 'Y-m-d' ) {//if it was simple y example y-m-d then it will show 17 instead of 2017

       $dates = array();
       $current = strtotime($first);
       $last = strtotime($last);

while( $current <= $last ) {

    $dates[] = date($output_format, $current);
    $current = strtotime($step, $current);
}//end of while loop

    return $dates;//returns a array
   }//end of function
   //select all the  date ranges that exist in the database 
     $query = 'SELECT check_in, check_out FROM bookings';

 $result = mysqli_query($db, $query);

    $i = 0;
    while ($row=mysqli_fetch_assoc($result)){
    //output all the dates within the ranges in the database
    $range[$i] = date_range($row['check_in'], $row['check_out']);//creates a associative array with numerical index values



  $i++;     


  }

  $individual_dates = array();

 //converts the associative array  into a regular array
  foreach($range as $ranges){
     foreach ($ranges as $many_ranges){
       $individual_dates[] = $many_ranges;
     }
  }


 $json_array = json_encode($individual_dates);

 ?>

这里是javascript

<script>
            var dateToday = new Date(); //creating new date obj. used to disable dates before today used in minDate: dateToday
              $(function() {

                $( "#datepicker" ).datepicker({
                    minDate: dateToday,     
                        dateFormat : 'yy-mm-dd',
                    beforeShowDay: checkAvailability

                    });

                $( "#datepicker1" ).datepicker({
                        minDate: dateToday,
                        dateFormat : 'yy-mm-dd',
                    beforeShowDay: checkAvailability        
                    });

              })
              /******************THE json array returned by php is used here*****/
                  var $disabledDates = <?php echo $json_array; ?>

                function checkAvailability(mydate){
                var $return=true;
                var $returnclass ="available";
                $checkdate = $.datepicker.formatDate('yy-mm-dd', mydate);
                for(var i = 0; i < $disabledDates.length; i++)
                    {    
                       if($disabledDates[i] == $checkdate)
                          {
                        $return = false;
                        $returnclass= "unavailable";
                        }
                    }
                return [$return,$returnclass];
                }

              </script>

在PHP和Mysql的jquery datepickers中禁用给定范围内的日期

1 个答案: