Question

我正在尝试从MySQL数据库中检索日期，该数据库将用于动态禁用datepicker UI中的日期。我从数据库中检索了日期并将其编码为JSON。这是echo JSON的输出：

[
 {"dates":"21-03-2016"},
 {"dates":"31-03-2016"},
 {"dates":"31-03-2016"},
 {"dates":"30-03-2016"}
 ]

我已经尝试将jSON添加到javascript页面，在那里它将被检索并用于消除日期。但是，它不起作用，因为日期选择器UI甚至不再出现。

有什么建议吗？谢谢。

checkDates.php

<?php
$servername = "localhost";
$username = "user";
$password = "user";
$dbname = "ebooking";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
 } 

$sql = "select booking_date from booking";
$result = $conn->query($sql);

$checkDates = array();

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $checkDate['dates'] = $row['booking_date'];
       
        $checkDates[] = $checkDate;
    }
} else {
    echo "0 results";
}
echo json_encode($checkDates);
 $conn->close();
 ?>

的index.jsp

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
    <title>Insert title here</title>
    <link rel="stylesheet"
          href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css">
    <script src="//code.jquery.com/jquery-1.10.2.js"></script>
    <script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script>
    <link rel="stylesheet" href="/resources/demos/style.css">

    <script>
        $(function() {
            $( "#datepicker" ).datepicker({
                dateFormat: 'dd-mm-yy',
                beforeShowDay: checkAvailability

            });
        })
        
        $.getJSON('checkDates.php?dld='+ id, function(json){dates=json;});

        function checkAvailability(mydate){
            var myBadDates = dates;

            var $return=true;
            var $returnclass ="available";
            $checkdate = $.datepicker.formatDate('dd-mm-yy', mydate);

            // start loop
            for(var x in myBadDates)
            {
                $myBadDates = new Array( myBadDates[x]['start']);

                for(var i = 0; i < $myBadDates.length; i++)
                    if($myBadDates[i] == $checkdate)
                    {
                        $return = false;
                        $returnclass= "unavailable";
                    }
            }
            //end loop

            return [$return,$returnclass];
        }
    </script>
</head>
<body>
Date:
<input type="text" id="datepicker">
</body>
</html>

Answer 1

好的，我已经改变了你的javascript，这就是我提出的代码：

$(function() {
  //ajax call better placed here.
  id="my ID"; //Define id, as it's not defined in the original post.
  /* Commented out just for JsFiddle, uncomment this for live version.
  $.getJSON('checkDates.php?dld=' + id, function(json) {
    dates = json;


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

    });
  });
  */
  //For JsFiddle ONLY remove this section of code for live version.
  dates = [{
    "dates": "21-03-2016"
  }, {
    "dates": "31-03-2016"
  }, {
    "dates": "31-03-2016"
  }, {
    "dates": "30-03-2016"
  }];
  $("#datepicker").datepicker({
    dateFormat: 'dd-mm-yy',
    beforeShowDay: checkAvailability

  });
  //End for JsFiddle
});




function checkAvailability(mydate) {


  var myBadDates = dates;

  var $return = true;
  var $returnclass = "available";
  $checkdate = $.datepicker.formatDate('dd-mm-yy', mydate);


  // start loop


  for (var x in myBadDates) {

    if (myBadDates[x].dates == $checkdate) {
      $return = false;
      $returnclass = "unavailable";
    }



  } //end loop



  return [$return, $returnclass];
}

请参阅jsFiddle： https://jsfiddle.net/gregborbonus/v1dwqq5r/1/

如果ajax失败，那么它就会崩溃。

我已经编辑了代码，更新的jsFiddle在这里： https://jsfiddle.net/gregborbonus/v1dwqq5r/2/

你的php可能会吐出不同的东西，如果这对你不起作用，链接到你的php脚本，我可以测试正确的标题并测试ajax调用本身，但是我怀疑这是一个问题。

Answer 2

checkdates.php

<?php
//Set document mime type... this is important since you want to return json not text!
header('Content-Type: application/json');

$servername = "localhost";
$username = "user";
$password = "user";
$dbname = "ebooking";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
 } 

$sql = "select booking_date from booking";
$result = $conn->query($sql);

$checkDates = array();

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $checkDates[] = $row['booking_date'];
    }
} else {
    echo "0 results";
}
echo json_encode($checkDates);
$conn->close();
?>

的index.jsp

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
    <title>Insert title here</title>
    <link rel="stylesheet"
          href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css">
    <script src="//code.jquery.com/jquery-1.10.2.js"></script>
    <script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script>
    <link rel="stylesheet" href="/resources/demos/style.css">


</head>
<body>
Date:
<input type="text" id="datepicker">
</body>
<script>
    var badDates;
    $.getJSON('checkDates.php, function(json){badDates=json;});

    $(function() {
        $( "#datepicker" ).datepicker({
            dateFormat: 'dd-mm-yy',
            beforeShowDay: function(date) {
                if($.inArray($.datepicker.formatDate('dd-mm-yy', date ), badDates) > -1)
                {
                    return [false,"Unavailable","Booked"];
                }
                else
                {
                    return [true,"Available","Not Booked"];
                }
            }
        });
    })

</script>
</html>

试试这个。我将内容类型添加到php文件中。我也改变了一下javascript。

在php文件中编辑平面数组，谢谢@Greg_Borbonus

JQuery DatePicker不可用日期通过PHP Json

2 个答案: