我的php文件名为view2.php:
<?php
include('db.php');
$select=mysqli_query($conn,"SELECT * FROM reservations where reservation_status ='CLAIMED' ");
while($userrow = mysqli_fetch_array($select)){
$reservationid=$userrow['reservation_id'];
$rtype=$userrow['room_type'];
$rstart=$userrow['reservation_start_date'];
$cname = $userrow['customer_name'];
$rend=$userrow['reservation_end_date'];
//$username=$userrow['username'];
//$password=$userrow['password'];
?>
<script>
var str = $rstart;
var year1 = str.substring(0, 4);
var month1 = str.substring(6, 7);
var day1 = str.substring(9, 10);
</script>
<?php }
funtion js_output(){
return {
title : 'STANDARD',
start : (year1, month1, day1),
end : (year1, month1, day1),
backgroundColor: 'red',
borderColor : 'pink'
},
}
?>
它从db获取事件,并据说在我的js文件上打印出部分脚本。完整的日历脚本:
<script>
$(function () {
var s = document.getElementById("START");
var date = new Date()
var d = date.getDate(),
m = date.getMonth(),
y = date.getFullYear()
$('#calendar').fullCalendar({
header : {
left : 'prev,next today',
center: 'title',
right : 'month,agendaWeek,agendaDay'
},
events : [
<?php include ('view2.php')?>
],
})
})
</script>
现在,我不知道为什么它不作为脚本的一部分打印。它只是在页面上回应。然后我用这段代码替换它:
<script>
$(function () {
var s = document.getElementById("START");
var date = new Date()
var d = date.getDate(),
m = date.getMonth(),
y = date.getFullYear()
$('#calendar').fullCalendar({
header : {
left : 'prev,next today',
center: 'title',
right : 'month,agendaWeek,agendaDay'
},
events : [
<?php
header("Content-type: application/x-javascript");
?>
var i = '<?php include( 'view2.php'); echo addslashes( js_output()); ?>';
console.log( i );
],
})
})
</script>
然后显示错误:
解析错误:语法错误,意外&#39; js_output&#39; (T_STRING)in 第21行的C:\ xampp \ htdocs \ system \ admin pages \ Reservations \ view2.php
答案 0 :(得分:1)
第21行,将funtion
替换为function
答案 1 :(得分:0)
因为它在view2.php(第21行)中显示语法错误,你可以在view2.php中共享/检查你的代码吗?
在我看来,它也可能是那个php文件中的问题。