Question

我试图从网址获取：http://redsox.tcs.auckland.ac.nz/BC/Open/Service.svc/booklist 然后将其解析为JSON，将JSON格式发送到表创建函数showTable（）。

html文件：

<!DOCTYPE html>
<html>
<head>
<title>Online Shop</title>
<link rel="stylesheet" type="text/css" href="stylesheet.css">
<script src="myscripts.js"></script>
</head>
<body>
    <h1 id="mainheader">Boutique</h1>
    <table id="menuTable" align="center">
        <tr>
            <td><button onclick="getResults()">Browse shop</button></td>
            <td><button onclick="joinShop()">join the shop</button></td>
            <td><button onclick="showComments()">Your comments</button></td>
        </tr>
    </table>
    <hr></hr>
    <div id="bodyDiv">
    </div>
</body>
</html>

相关的javascript：

function getResults(){  //gotta change it to take appended shit after ".svc/"
    var uri = "http://redsox.tcs.auckland.ac.nz/BC/Open/Service.svc/booklist";
    var xhr = new XMLHttpRequest();
    xhr.open('GET', uri, true);
    xhr.onload = function (){
    var resp = JSON.parse(xhr.responseText);
    showTable(resp.value);
    }
    xhr.send(null);
}

function showTable(item) {
    var tableContent = "<tr class='orderTitle'><td>book id</td><td>book title</td></tr>\n"//creating unique table heading column
    for(var i = 0; i < item.length; ++i) {
    var record = item[i];
    if (i & 1 == 1){
        tableContent += "<tr class='orderOdd'>";
    }
    else { //even row
        tableContent += "<tr class='orderEven'>";
    }
    tableContent += "<td>" + record.Id + "</td><td>" + record.Title + "</td></tr>\n"
    }
    document.getElementById("bodyDiv").innerHTML = tableContent;
}

Xhr没有生成，无法弄清楚我哪里出错了

0 个答案: