如何在php中修改json_encode函数文件？
如何在php中修改json_encode函数文件？
提问于 2018-08-28 15:58:38
代码：
<script>
$(document).ready(function(){
var events = <?php echo $results; ?>;
$("#calen").datepicker({
beforeShowDay: function(date) {
var result = [true, '', null];
var matching = $.grep(events, function(event) {
return event.Date.valueOf() === date.valueOf();
});
if (matching.length) {
result = [true, 'highlight', null];
}
return result;
},
onSelect: function(dateText) {
var date,
selectedDate = new Date(dateText),
i = 0,
event = null;
while (i < events.length && !event) {
date = events[i].Date;
if (selectedDate.valueOf() === date.valueOf()) {
event = events[i];
}
i++;
}
if (event) {
alert(event.Title);
}
}
});
});
</script>
<?php
$sqls = mysqli_query($con,"select pack_program, pack_duration, start_date, end_date from package_detail where pack_name = '".$row['pack_name']."'");
while($fetch = mysqli_fetch_array($sqls))
{
$data[] = array(
'Title' => $fetch['pack_program'],
'Duration' => $fetch['pack_duration'],
'Date' => "new Date(".$fetch['start_date'].")"
);
}
$results = str_replace('</', '<\/', json_encode($data, JSON_UNESCAPED_UNICODE | JSON_UNESCAPED_SLASHES));
?>预期输出：
var events = [{Title:"Tour Program : Delhi / Chandigargh Arrival - Shimla ( 02 Nights ) – Manali ( 03 nights ) – Chandigarh ( 01 night ) Delhi / Chandigargh Departure Total : 06 Nights & 07 Days Tour",Duration:"7 days 6 Nights",Date:new Date("08/28/2018")}];意外输出：
var events = [{"Title":"Tour Program : Delhi / Chandigargh Arrival - Shimla ( 02 Nights ) – Manali ( 03 nights ) – Chandigarh ( 01 night ) Delhi / Chandigargh Departure Total : 06 Nights & 07 Days Tour","Duration":"7 days 6 Nights","Date":"new Date(08/28/2018)"}];在下面的代码中，我使用jQuery日期选择器创建了一个事件日历，如上所述，它可以很好地用于expected output。但它不适用于上面提到的unexpected output。那么，如何在提供expected output的$data[]文件中进行修改呢？我该怎么做呢？请帮帮我。
谢谢
回答 2
Stack Overflow用户
发布于 2018-08-28 16:14:08
您需要的是JSON.parse()
var events = <?php echo $results; ?>;
var obj = JSON.parse(events);了解有关-> here的更多信息
Stack Overflow用户
发布于 2018-08-28 16:21:53
作为预期输出，new Date("08/28/2018")将成为Date对象，但在非预期输出中，它是一个字符串。
让您尝试更改代码，如下所示：
$data[] = array(
'Title' => $fetch['pack_program'],
'Duration' => $fetch['pack_duration'],
'Date' => date('Y-m-d', $fetch['start_date'])
);输出结果将是：
var events = [{"Title":"Tour Program : Delhi / Chandigargh Arrival - Shimla ( 02 Nights ) – Manali ( 03 nights ) – Chandigarh ( 01 night ) Delhi / Chandigargh Departure Total : 06 Nights & 07 Days Tour","Duration":"7 days 6 Nights","Date":"2018-08-28"}];并且它将与datepicker一起工作
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52052855
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