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php - 单击单选按钮时如何从 mysql 获取结果并使用 jquery 显示?

转载 作者:搜寻专家 更新时间:2023-10-31 22:14:06 25 4
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我想制定一个巴士座位计划。我有使用 javascript 函数的座位表。我从数据库中查询了两个名为 Bus_1Bus_2 的单选按钮。当我单击其中一个单选按钮时,我想获得可用座位以显示在座位表上。问题是我无法编写如何承载 radio 值以及如何在座位表上显示数据库结果。请帮助我。

<SCRIPT type="text/javascript">
$(function () {
var settings = { rowCssPrefix: 'row-', colCssPrefix: 'col-', seatWidth: 35, seatHeight: 35, seatCss: 'seat', selectedSeatCss: 'selectedSeat', selectingSeatCss: 'selectingSeat' };
var init = function (reservedSeat) {
var str = [], seatNo, className;
var shaSeat = [1,5,9,13,17,21,25,29,33,37,41,'@',2,6,10,14,18,22,26,30,34,38,42,'@','$','$','$','$','$','$','$','$','$','$',43,'@',3,7,11,15,19,23,27,31,35,39,44,'@',4,8,12,16,20,24,28,32,36,40,45];
var spr=0;
var spc=0;

for (i = 0; i<shaSeat.length; i++) {
if(shaSeat[i]=='@') {
spr++;
spc=0;
}
else if(shaSeat[i]=='$') {
spc++;
}
else {
seatNo = shaSeat[i];
className = settings.seatCss + ' ' + settings.rowCssPrefix + spr.toString() + ' ' + settings.colCssPrefix + spc.toString();
if ($.isArray(reservedSeat) && $.inArray(seatNo, reservedSeat) != -1) { className += ' ' + settings.selectedSeatCss; }
str.push('<li class="' + className + '"' +'style="top:' + (spr * settings.seatHeight).toString() + 'px;left:' + (spc * settings.seatWidth).toString() + 'px">' +'<a title="' + seatNo + '">' + seatNo + '</a>' +'</li>');
spc++;
}
}
$('#place').html(str.join(''));
}; //case I: Show from starting //init();

//Case II: If already booked
var bookedSeats = [2,3,4,5]; //**I don't know how to get query result in this array.This is problem for me **
init(bookedSeats);

$('.' + settings.seatCss).click(function () {
// ---- kmh-----
var label = $('#busprice');
var sprice = label.attr('pi');

//---- kmh ----
// var sprice= $("form.ss pri");
if ($(this).hasClass(settings.selectedSeatCss)){ alert('This seat is already reserved'); }
else {
$(this).toggleClass(settings.selectingSeatCss);
//--- sha ---
var str = [], item;
$.each($('#place li.' + settings.selectingSeatCss + ' a'), function (index, value) { item = $(this).attr('title'); str.push(item); });
var selSeat = document.getElementById("selectedseat");
selSeat.value = str.join(',');
//var amount = document.getElementById("price");
// amount.value = sprice*str.length;
document.getElementById('price').innerHTML = sprice*str.length;
return true;
}
});

$('#btnShow').click(function () {
var str = [];
$.each($('#place li.' + settings.selectedSeatCss + ' a, #place li.'+ settings.selectingSeatCss + ' a'), function (index, value) {
str.push($(this).attr('title'));
});
alert(str.join(','));
})

$('#btnShowNew').click(function () { // selected seat
var str = [], item;
$.each($('#place li.' + settings.selectingSeatCss + ' a'), function (index, value) { item = $(this).attr('title'); str.push(item); });
alert(str.join(','));
})
});
</SCRIPT>

最佳答案

您可以使用 onclick 告诉 AJAX 获取您的信息,然后使用 jQuery 处理这些信息。

<input type="radio" name="radio" onclick="ajaxFunction()" />

function ajaxFunction()
{
$.ajax({
type: "POST",
url: "you_script_page.php",
data: "post_data=posted",
success: function(data) {
//YOUR JQUERY HERE
}
});
}

如果您不传递任何变量,则不需要数据。

关于php - 单击单选按钮时如何从 mysql 获取结果并使用 jquery 显示?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8883731/

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