Question

为什么我的ajax函数没有使用get方法提交这两个值

<script language="javascript" type="text/javascript">
<!-- 
//Browser Support Code
function ajaxFunction( ){
    var ajaxRequest;  // The variable that makes Ajax possible!

    try{
        // Opera 8.0+, Firefox, Safari
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }
    // Create a function that will receive data sent from the server
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            document.getElementById("emailconfirm").innerHTML= ajaxRequest.responseText;
        }
    }

    ajaxRequest.open("GET", "Atest.php?<?php echo $PDFPassingArray ?>?Email="+ document.getElementById("email").value, true);
    ajaxRequest.send(null); 
}

我希望函数提交数组$ pdfpassingArray，这是一个html_build_query和表单中的Email elemment id。目前我只能反转$ pdfPassingarray值，而不是电子邮件值。所以我的问题，我做了这个部分吗

ajaxRequest.open("GET", "Atest.php?<?php echo $PDFPassingArray ?>?Email="+ document.getElementById("email").value, true);

我的表格

       <form name='myForm'>

    <input name="email"   id="email"type="text" />
    <input name="button" onclick="ajaxFunction()" type="button" />
</form>

Answer 1

您的网址字符串中有两个?，

它的格式应为url.php?a=1&b=2

Ajax提交多元化

1 个答案: