Question

我正在关注newboston ajax教程，但遇到了错误。我已经给出了错误的快照。我正在使用wamp服务器并在chrome浏览器上运行它。

任何人都可以帮助我吗？

index.html

<!DOCTYPE html>
            <html>
             <head>
                    <meta content="text/html;charset=utf-8" http-equiv="Content-Type">
                    <meta content="utf=8" http-equiv="encoding">
                    <script type="text/javascript" src="food.js"></script>
            </head>
                <body onload="process()">
                    <h2>....Food Shop...</h2>
                    <input type="text" id="userinput" placeholder="enter the food u want" value="banana">
                    <div id="underinput" />
                </body>
            </html>

food.js

var xmlHttp= createXmlHttpRequestObject();

function createXmlHttpRequestObject(){

    var xmlHttp;

    if(window.ActiveXObjects){
        try{
            xmlHttp= new ActiveXObject("Microsoft.XMLHTTP");

        }
        catch(e)
        {
            xmlHttp = false;
        }
    }
    else
    {
        try{
                xmlHttp= new XMLHttpRequest();
        }catch(e)
        {
                xmlHttp=false;
        }
    }

    if(!xmlHttp)
    {
    alert("cant create the object");
    }
    else{
         return xmlHttp;
    }
}


function process(){
    if(xmlHttp.readyState == 4 || xmlHttp.readyState == 0){
        food=encodeURIComponent(document.getElementById("userinput").value);
        xmlHttp.open("GET","foodstore.php?food=" + food,true);
        xmlHttp.onreadystatechange= handleServerResponse;
        xmlHttp.send(null);


    }else{
        setTimeout('process()',1000);


    }
}
function handleServerResponse(){
    if(xmlHttp.readyState == 4){
        if(xmlHttp.status == 200){


             message = xmlHttp.responseXML.documentElement.firstChild.data;
            document.getElementById("underinput").innerHTML= '<span style="color:blue">' + message + '</span>';
            setTimeout('process()',1000);   
        }else{
            alert('something wrong...');
        }
    }
}

foodstore.php

<?php
    error_reporting(E_ALL);
    int_set('display_errors',true)
    header('Content-Type: text/xml');
    echo '<?xml version="1.0" encoding="UTF=8" standalone="yes" ?>';
    echo '<response>';
        $food = $_GET['food'];
        $foodArray= array('banana','apple','mango');
        if(in_array($food,$foodArray)){
            echo 'we do have'.$food;
        }
        else if($food ==''){
            echo 'enter the food';
        }
        else{
            echo 'sorry we do not have'.$food;
            }
    echo '</response>';
?>

Answer 1

我认为，错误在于生成XML数据的PHP代码。当我尝试使用该代码时，当我尝试使用javascript解析xml时出现错误。我得到的错误是：

此页面包含以下错误：第34行第1行的错误：字符串未关闭期待＆＃34;或者＆＃39;下面是页面的渲染到第一个错误。

当您使用PHP生成xml时，为什么不使用属于核心PHP的正确方法 - 即DOMDocument？

以下是您的代码的修改版本，应该返回您正在努力的响应。

<?php
    if( $_SERVER['REQUEST_METHOD']=='GET' && !empty( $_GET['food'] ) ){

        /* Filter the input data */
        $food = filter_input( INPUT_GET, 'food', FILTER_SANITIZE_STRING );

        /* an array of foodstuffs */
        $foods = array('banana','apple','mango','orange','plum','peach','cherry','kiwi','grape','lemon');

        /* Construct the response message based upon existance of food */
        $message = empty( $food ) ? 'enter the food' : ( in_array( $food, $foods ) ? 'We do have '.$food : 'Sorry, we do not have '.$food );


        /* using DOMDocument to output a correctly formed XML document */
        $dom=new DOMDocument('1.0', 'utf-8');
        $dom->appendChild( $dom->createElement( 'response', $message ) );
        $xml = $dom->saveXML();
        $dom=null;


        /* send the response */
        header('Content-Type: application/xml');
        exit( $xml );
    }
?>
<!doctype html>
<html>
    <head>
        <title>Ajax and XML</title>
        <script>
            /* lightweight utility function for making ajax requests */
            function ajax(url,method,params,callback){
                var xhr=new XMLHttpRequest();
                xhr.onload=function(){
                    callback.call( this, this.response )
                };
                xhr.onerror=function(e){
                    console.log(e)
                };
                var payload=[];
                for( var n in params )payload.push( n + '=' + params[ n ] );
                var p;

                switch( method.toLowerCase() ){
                    case 'post': p=payload.join('&'); break;
                    case 'get': url+='?'+payload.join('&'); p=null; break;
                }
                xhr.open( method.toUpperCase(), url, true );
                xhr.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
                xhr.send( p );
            }

            /* utility function to get a parser to process XML data */
            function get_parser(str_xml) {
                try {
                    if( window.XMLSerializer && typeof( DOMParser )!='undefined' ) {
                        var parser = new DOMParser();
                        return parser.parseFromString(str_xml, 'application/xml');

                    } else if( typeof( ActiveXObject ) != 'undefined' ) {
                        var parser = CreateMSXMLDocumentObject();
                            parser.async='false';
                            parser.load( str_xml );
                        return parser;

                    } else {
                        var url = 'data:text/xml;charset=utf-8,' + encodeURIComponent( str_xml );
                        var request = new XMLHttpRequest();
                            request.open( 'GET', url, false );
                            request.send( null );
                        return request.responseXML;

                    }
                } catch( err ) {
                    console.warn( err );
                }
            }

            function process(){
                /*
                    Change to suit your environment ~ for this demo I use same page ( location.href )
                */
                var food = document.querySelector('input[name="food"]').value;
                if( food && food!='' ){
                    //ajax.call( this, 'foodstore.php', 'get', { food:food }, handleServerResponse );

                    ajax.call( this, location.href, 'get', { food:food }, handleServerResponse );
                }
            }

            /* callback function to process xml response data */
            function handleServerResponse(response){
                if( response ){

                    /* parse xml response */
                    var xml=get_parser( response );

                    /* get the node which is of interest - only one in this case */
                    var message = xml.documentElement.textContent;

                    /* output message to browser */
                    document.getElementById("underinput").innerHTML= '<span style="color:blue">' + message + '</span>';
                }
            }


            function bindEvents(e){
                document.getElementById('bttn').onclick=process;
            }
            document.addEventListener('DOMContentLoaded', bindEvents,{ passive:true, capture:false } );
        </script>
    </head>
    <body>
        <form name='geronimo'>
            <label for='food'>Enter Food<input type='text' name='food' placeholder='eg: apple' size=60 /></label>
            <input type='button' id='bttn' value='Check' />
            <div id='underinput'></div>
        </form>
    </body>
</html>

未捕获的TypeError：无法读取null message = xmlHttp.responseXML.documentElement.firstChild.data的属性'documentElement';

1 个答案: