gpt4 book ai didi

javascript - JQuery DatePicker 通过 PHP Json 不可用日期

转载 作者:行者123 更新时间:2023-12-03 07:20:17 25 4
gpt4 key购买 nike

我正在尝试从 MySQL 数据库检索日期,该数据库将用于动态禁用日期选择器 UI 中的日期。我已经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"}
]

我尝试将 getJSON 发送到 javascript 页面,在该页面中将检索该 JSON 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();
?>

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>

最佳答案

好的,我对你的 JavaScript 做了一些修改,这是我想出的代码:

$(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,更新后的 jsFiddle 在这里: https://jsfiddle.net/gregborbonus/v1dwqq5r/2/

你的 php 可能会吐出一些不同的东西,如果这对你不起作用,请链接到你的 php 脚本,我可以测试正确的 header 并测试 ajax 调用本身,但我怀疑这甚至是一个问题。

关于javascript - JQuery DatePicker 通过 PHP Json 不可用日期,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36237211/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com