為什么AJAX POST無法正常工作？

Question

誰能告訴我為什么我的AJAX POST不起作用...我添加了alert()並開始起作用... idk哪里錯了...這是我的代碼...

<script language="JavaScript" type="text/javascript">
    function initialCalendar() {
        var hr = new XMLHttpRequest();
        var url = "calendar.php";
        var currentTime = new Date();
        var month = currentTime.getMonth() + 1;
        var year = currentTime.getFullYear();
        showmonth = month;
        showyear = year;
        var vars = "showmonth="+showmonth+"&showyear="+showyear;
        hr.open("POST", url, true);
        hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        hr.onreadystatechange = function() {
            if(hr.readyState == 4 && hr.status == 200) {
                var return_data = hr.responseText;
                document.getElementById("showCalendar").innerHTML = return_data;
            }
        }
        hr.send(vars);
        document.getElementById("showCalendar").innerHTML = "processing..";
        alert(vars);
    }

    function next_month(){
        var nextmonth = showmonth + 1;
        if(nextmonth > 12) {
            nextmonth = 1;
            showyear = showyear + 1;
        }
        showmonth = nextmonth;
        var hr = new XMLHttpRequest();
        var url = "calendar.php";
        var vars = "showmonth="+showmonth+"&showyear="+showyear;
        hr.open("POST", url, true);
        hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        hr.onreadystatechange = function() {
            if(hr.readyState == 4 && hr.status == 200) {
                var return_data = hr.responseText;
                document.getElementById("showCalendar").innertHTML = return_data;
            }
        }
        hr.send(vars);
        document.getElementById("showCalendar").innertHTML = "processing...";
        alert(vars);
    }

    function last_month(){
        var lastmonth = showmonth - 1;
        if(lastmonth < 1) {
            lastmonth = 1;
            showyear = showyear - 1;
        }
        showmonth = lastmonth;
        var hr = new XMLHttpRequest();
        var url = "calendar.php";
        var vars = "showmonth="+showmonth+"&showyear="+showyear;
        hr.open("POST", url, true);
        hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        hr.onreadystatechange = function() {
            if(hr.readyState == 4 && hr.status == 200) {
                var return_data = hr.responseText;
                document.getElementById("showCalendar").innertHTML = return_data;
            }
        }
        hr.send(vars);
        document.getElementById("showCalendar").innertHTML = "processing...";
        alert(vars);
    }
</script>

而身體是

<body onLoad="initialCalendar();">
<div id="showCalendar"> </div>
</body>

和我以前和下一個的按鈕...這是從calendar.php

<?php
$showmonth = $_POST['showmonth'];
$showyear = $_POST['showyear'];
$showmonth = preg_replace('#[^0-9]#i', '', $showmonth);
$showyear = preg_replace('#[^0-9]#i', '', $showyear);

$day_count = cal_days_in_month(CAL_GREGORIAN, $showmonth, $showyear);
$pre_days = date('w', mktime(0,0,0, $showmonth, 1, $showyear));
$post_days = (6 - (date('w', mktime(0,0,0, $showmonth, $day_count, $showyear))));


echo '<div class="title_bar">';
echo '<div class="previous_month"><input name="myBtn" type="submit" value="Previous Month" onClick="javascript:last_month();"></div>';
echo '<div class="show_month">' . $showmonth . '/' . $showyear . '</div>';
echo '<div class="next_month"><input name="myBtn2" type="submit" value="Next Month" onClick="javascript:next_month();"></div>';
echo '</div>';

Answer 1

更改innertHTML ，到innerHTML在您的next_month()和last_month()函數。 我建議使用像Komodo Edit 7.1這樣的編輯器。

為什么AJAX POST無法正常工作？

問題描述

1 個解決方案

解決方案1
1 已采納 2013-09-29 00:25:32

為什么AJAX POST無法正常工作？

問題描述

1 個解決方案

解決方案1 1 已采納 2013-09-29 00:25:32

解決方案1
1 已采納 2013-09-29 00:25:32