php - 在一个数组中从 mysql 检索多个结果

Question

我面临的问题是我想将从 mysql 检索到的结果放入一个数组中，如下所示:{“日期”:“2014-09-28”，“2014-09-29”，“2014-09-30”}尽管有这个结果{“日期”:“2014-09-28”}{“日期”:“2014-09-29”}{“日期”:“2014-09-30”}它将用于禁用另一个页面中我的日期选择器上的这些日期这是我的源代码

    <?php
@session_start;
include '../includes/db_login.php';
$flat_id= $_GET['flat_id']; 

 $result = mysql_query("SELECT startDate,endDate FROM reservation WHERE flat_id='$flat_id'");
 $i=0;
 while($row = mysql_fetch_array($result))
    {


$s_dte=$row['startDate'];
$s_dte2=strtotime($s_dte);
$s=date("Y-m-d",$s_dte2);
$e_dte=$row['endDate'];
$diff = abs(strtotime($e_dte) - strtotime("-1 day",strtotime($s))); 



$yrs   = floor($diff / (365*60*60*24)); 
$mnth  = floor(($diff - $yrs * 365*60*60*24) / (30*60*60*24)); 
$days    = floor(($diff - $yrs * 365*60*60*24 - $mnth*30*60*60*24)/ (60*60*24));
$t=1;


while($t <= $days){

//echo $s."\n";
$confirmedSends = array ( "dates" => $s);
$date = strtotime("+1 day", strtotime($s));
$s=date("Y-m-d", $date);
$t++;
$jsonConfirmedSends =  json_encode($confirmedSends); 
    echo $jsonConfirmedSends;
}

}

?>

我现在得到了所需的输出，但我仍然无法禁用 jquery 上的日期，这是我用于 jquery 的源代码

jQuery(document).ready(function () {

                    $("#datepick").datepicker({
                        defaultDate: "d",
                        dateFormat: 'yy-mm-dd',
                        changeMonth: true,
                        numberOfMonths: 1,
                        beforeShowDay: checkAvailabilityStart,
                        onClose: function (selectedDate) {
                            $("#datepick2").datepicker("option", "minDate", selectedDate);
                        }
                    });



                    var dates = [];
                    var flat_id = $("#flat_id").val();



                    $.getJSON("ajax.php?flat_id=" + flat_id, function (data) {
         $.each(data, function(index, value) {
            dates.push(value.data); // i don't know what's (value.data) refer to !
        });
    });


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

                            {
                                $return = false;
                                $returnclass = "unavailable";
                            }
                        }
                        return [$return, $returnclass];
                    }



                });


            </script>

Answer 1

短一点

$dates = array('dates' =>array());
while ($row = mysqli_fetch_assoc($result)) {
    $date = strtotime($row['startDate']);
    $enddate = strtotime($row['endDate']);
    while ($date <= $enddate) {
        $dates['dates'][] = date('Y-m-d', $date);
        $date += 86400;
    }

}
echo json_encode($dates);