Question

我正在尝试从servlet将JSON数据发送到我的JSP页面。一旦我运行JSP，整个JSP就会被JSON内容覆盖。为什么会这样？这是我的JSP页面的一部分。

    <%@ page import="java.util.ArrayList" %>
    <head>
    <script>
    function populateTypeOfWork(workType) {
        var xhttp = new XMLHttpRequest();
        xhttp.onreadystatechange = function () {
            if (this.readyState == 4 && this.status == 200) {
                var work = JSON.parse(this.responseText);
                workType.options.length = 0;
                for (i = 0; i < work.length; i++) {
                    createOption(workType, workDesc[i], workDesc[i]);
                }
            }
            xhttp.open("GET", "ServletRequest", true);
            xhttp.send();
        }
    }
    function createOption(dropdown, text, value) {
        var x = document.createElement('option');
        x.value = value;
        x.text = text;
        dropdown.options.add(x);
    }
</script>
</head>
<body>
    <label>Work Category **</label>
  <select id="workCategory" 
   onchange="populateTypeOfWork(document.getElementById('workTypeDesc1'))" 
   size=1
    style="width:95%;" required>
  <option value="" selected>Select A Work Category</option>
   <% ArrayList workCategory = (ArrayList) 
 request.getAttribute("workCategory");
    for (int i = 0; i < workCategory.size(); i++) {
%>
<option value="<%=workCategory.get(i)%>"><%=workCategory.get(i)%>
</option>
<% } %>
</select>&nbsp;<br>

<label>Type Of Work **</label>
<select id="workTypeDesc1" size=1 style="width:95%;" required>
<option value="" selected>Type of Work</option>
</select>
</body>
</html>

的Servlet

    String json = new Gson().toJson(arraylistvar);
    res.setContentType("application/json");
    PrintWriter pw = res.getWriter();
    pw.flush();
    pw.close();
    req.setAttribute("viewURL","/page1.jsp");

为什么stackoverflow要求我添加更多细节？因为它主要是代码.. ??我为什么要不必要地填写这篇文章？

Answer 1

我可以通过当前代码看到两个问题：（1）您正在引用一个不存在的变量workDesc;看来你应该引用名为work的变量，并且（b）发送AJAX请求的代码在onreadystatechange（）事件处理程序中，当它应该在设置事件处理程序之后实际执行时。以下是修订后的代码段：

xhttp.onreadystatechange = function () {
    if (this.readyState == 4 && this.status == 200) {
        var work = JSON.parse(this.responseText);
        workType.options.length = 0;
        for (i = 0; i < work.length; i++) {
            createOption(workType, work[i], work[i]);
        }
    }
}
xhttp.open("GET", "ServletRequest", true);
xhttp.send();

Answer 2

这可能是由于您的网页发出请求的原因。你应该使用AJAX调用;确保您没有在JavaScript中提交普通表单或更改页面的URL。

jsp页面被从Servlet发送的JSON数据覆盖

2 个答案: