尝试使用 jquery 和 php 在日期选择器中“突出显示”我的数据库中的日期。我到处寻找可以帮助我的东西......
这是我目前正在使用的代码。 请注意，它有一个 onSelect 状态，可以单击日期并从数据库获取事件并将其打印到 div，这有效，现在我想让日期也突出显示事件..
$(document).ready(function() {
var SelectedDates = {};
$.getScript('eventdates.php', function(eventsdate) {
eventsdate;
});
$("#kalender").datepicker({
firstDay: 1,
dayNamesMin: ['Sö', 'Må', 'Ti', 'On', 'To', 'Fr', 'Lö'],
monthNames: ['Januari', 'Februari', 'Mars', 'April', 'Maj', 'Juni', 'Juli', 'Augusti', 'September', 'Oktober', 'November', 'December'],
dateFormat: "yy-mm-dd",
beforeShowDay: function(date) {
var Highlight = SelectedDates[date];
if (Highlight) {
return [true, "Highlighted", Highlight];
}
else {
return [true, '', ''];
}
},
onSelect: function(dateText, inst) {
var myDate = new Date(dateText);
var newFormat = $.datepicker.formatDate('yy-mm-dd', myDate);
$.ajax({
url: 'events.php',
type: 'POST',
dataType: 'json',
data: {selectedData: newFormat},
success: function(result) {
if (result.date === newFormat) {
$('#event').html(result['date'] + " " + result['event']);
}
else {
$('#event').html(result['none']);
}
}
});
}
});
});
事件日期.php
<?php
include_once ('connect.php');
$sql = mysql_query("SELECT * FROM `events`");
while ($row = mysql_fetch_array($sql)) {
echo "SelectedDates[new Date('{$row['date']}')] = new Date('{$row['date']}')<br/>";
}
?>
事件.php
<?php
include_once ('connect.php');
if (isset($_POST['selectedData'])) {
$sql2 = mysql_query("SELECT * FROM `events` WHERE `date`='{$_POST['selectedData']}'");
$row2 = mysql_fetch_array($sql2);
if ($row2['date'] == $_POST['selectedData']) {
$sql = mysql_query("SELECT * FROM `events` WHERE `date`='{$_POST['selectedData']}'");
while ($row = mysql_fetch_array($sql)) {
$data = array(
"date" => $row['date'],
"event" => utf8_encode($row['event'])
);
echo json_encode($data);
}
} else {
$data = array(
"none" => "Inget event denna dag!"
);
echo json_encode($data);
}
}
?>
最佳答案
我解决了这个问题，并想与其他遇到此类问题的人分享我的解决方案!
jQuery 脚本
<script>
$(document).ready(function() {
$("#kalender").datepicker({
firstDay: 1,
dayNamesMin: ['Sö', 'Må', 'Ti', 'On', 'To', 'Fr', 'Lö'],
monthNames: ['Januari', 'Februari', 'Mars', 'April', 'Maj', 'Juni', 'Juli', 'Augusti', 'September', 'Oktober', 'November', 'December'],
dateFormat: "yy-mm-dd",
onSelect: function(dateText, inst) {
var myDate = new Date(dateText);
var newFormat = $.datepicker.formatDate('yy-mm-dd', myDate);
$.ajax({
url: 'events.php',
type: 'POST',
dataType: 'json',
data: {selectedData: newFormat},
success: function(result) {
if (result.date === newFormat) {
$('#event').html(result['date'] + "<br />\n" + result['event']);
}
else {
$('#event').html(result['none']);
}
}
});
}, beforeShowDay: function(date) {
var yy = date.getFullYear(), mm = date.getMonth() + 1, dd = date.getDate();
if (dd < 10) {
var dt = yy + "-" + mm + "-0" + dd;
} else {
var dt = yy + "-" + mm + "-" + dd;
}
<?php include_once ('eventdates.php') ?>
return [true, ''];
}
});
});
</script>
事件.php
<?php
include_once ('connect.php');
if (isset($_POST['selectedData'])) {
$sql2 = mysql_query("SELECT * FROM `events` WHERE `date`='{$_POST['selectedData']}'");
$row2 = mysql_fetch_array($sql2);
if ($row2['date'] == $_POST['selectedData']) {
$sql = mysql_query("SELECT * FROM `events` WHERE `date`='{$_POST['selectedData']}'");
while ($row = mysql_fetch_array($sql)) {
$data = array(
"date" => $row['date'],
"event" => utf8_encode($row['event'])
);
echo json_encode($data);
}
} else {
$data = array(
"none" => "Inget event denna dag!"
);
echo json_encode($data);
}
}
?>
事件日期.php
<?php
include_once ('connect.php');
$sql = mysql_query("SELECT * FROM `events`");
while ($row = mysql_fetch_array($sql)) {
echo "var selDate" . $row['event_ID'] . " = '" . $row['date'] . "';";
echo "if (dt === selDate" . $row['event_ID'] . ") {";
echo "return [true, 'Highlighted'];";
echo "}";
}
?>
希望这对其他人有帮助!
关于php - 从 mysql 数据库中突出显示日期选择器中的事件，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13542733/