我是javascript和ajax的新手。
自从我尝试通过查找某个教程来构建一个简单的预订系统已经过了一周,并且它并不顺利。我尝试修改,复制粘贴和谷歌搜索它,但我仍然不明白。我希望我能在这里找到更好的线索。
请帮助,这是我的代码。
index.php中的
<section>
<div id="stage">Bon Jovi Stage</div>
<div id="festival-a-left" class="festivalA seat-number"> Festival A Left
<?php
while($fesAL = mysql_fetch_assoc($festaleft)){
$fesAL_id = $fesAL['available_id'];
$fesAL_number = $fesAL['seat_number'];
$fesAL_seat = $fesAL['available_seat'];
$fesAL_price = $fesAL['seat_price'];
if($fesAL_seat == 0){
$display_fesAL .= '<div class="full"><div class="display-seat"> 0 Seats Available </div></div>';
}
else{
$display_fesAL .= '<div class="available"><div id="lock_'.$fesAL_id.'" class="display-seat" onClick="showSeats(this.id)" >'.$fesAL_seat.' Seats Available </div></div>';
} }
echo $display_fesAL;
?>
</div>
</section>
<section id="returnData">
return data
</section>
resProcessing.php中的
<?php
if(isset($_POST['getSeatButtons'])){
//initialize display to null
$spots = "";
//clean incoming POST to only allow numbers 0-9
//this is table id using to build seat button
$tid = preg_replace('#(^0-9)#', '', $_POST['getSeatButtons']);
//if post is empty after cleaning, exit script
if($tid == ""){
exit();
}
//query to fetch from available
$sql = "SELECT * FROM `available` WHERE `available_id` = '$tid'";
$query = mysql_query($sql) or die(mysql_error());
//verify result
$quickcheck = mysql_num_rows($query);
if($quickcheck !== 0){
while($row = mysql_fetch_array($query, MYSQL_ASSOC)){
$id = $row['available_id'];
$seat_number = $row['seat_number'];
$available_seat = $row['available_seat'];
$seat_price = $row['seat_price'];
$setButton = "";
//button that display output
$spots .= $seat_number." now has ".$available_seat." seat(s) available - ".$seat_price." per seat<br/>";
$spots .= "click the how many seat do you want to buy, we will held for 3 minutes once you click.<br/>";
//construct buttons for each seats
for($k=0; $k<$available_seat; $k++){
$k2 = $k+1;
//give each button the id
$setButton.= "<button id='tbid_".$id."_".$k2."' onClick='reserveSeats(this.available_id)'>" .$k2. "</button>";
}
//adding output buttons
$spots .= $setButton."<br/>";
}
}
else{
$spots = "sorry, there are no more available seats to buy";
}
//return output and exit
echo $spots;
exit();
}
?>
in reservation.js
//start initial state of common vars
var processStage = "open";
var reservationID = "open";
//end initial state of common vars
//start get buttons for available seats
function showSeats(lock){
//if they have an unfinished reservation, cancel reservation
if(processStage == "closed"){
cancelReserve(reservationID);
}
var a = lock.split("_");
document.getElementById("returnData").innerHTML = "Double Checking Availability ... ";
//developphp.com/view.php?tid=1185
var hr = new XMLHttpRequest();
var url = "resProcessing.php";
hr.open("POST", url, true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlendecoded");
hr.onreadystatechange = function(){
if(hr.readyState == 4 && hr.status == 200){
var return_data = hr.responseText;
document.getElementById("returnData").innerHTML = return_data;
}
}
hr.send("getSeatButtons="+a[1]);
}
//end get button for available seats
我的问题是,为什么我无法在section id="returnData"
中归还任何内容。应该如何,它应该返回按钮,其中包含我想要预订的座位数量?以及为什么section id =&#34; returnData&#34;甚至没有显示我在该部分输入的文字,我所说的是
<section id="returnData">
点击座位
后,这种情况正在消失return data
</section>
我尝试了什么:
这是chatlist.php
<?php
include 'core/init.php';
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
var xmlhttp = new XMLHttpRequest();
$(document).ready(function(e){
$.ajaxSetup({cache:false});
setInterval(function(){
$('#chatList').load('resProcessing.php?getSeatButtons=value')},2000);
});
</script>
<div id="chatList"></div>
当我在网络中检查时,它显示了这个东西。
请帮忙。