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我是一個初學者，我試圖做一個簡單的應用程序來顯示從MySQL使用JavaScript和jQuery，json對象的表。這不是給我任何錯誤，但我失去了如何繼續。請提供您的反饋！使用jquery和ajax顯示來自mysql的數據
這是我books.java
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.ArrayList;
public class Books {
private int id;
private String title;
private String author;
private float price;
private int qty;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getTitle() {
return title;
}
public void setTitle(String title) {
this.title = title;
}
public String getAuthor() {
return author;
}
public void setAuthor(String author) {
this.author = author;
}
public float getPrice() {
return price;
}
public void setPrice(float price) {
this.price = price;
}
public int getQty() {
return qty;
}
public void setQty(int qty) {
this.qty = qty;
}
public Books(int id, String title, String author, float price, int qty) {
super();
this.id = id;
this.title = title;
this.author = author;
this.price = price;
this.qty = qty;
}
public Books() {
}
public ArrayList <Books> getBooks(ArrayList<Books> bookList){
Connection conn =null;
ResultSet rset = null;
try {
Class.forName("com.mysql.jdbc.Driver");
conn = DriverManager.getConnection(
"jdbc:mysql://localhost:3306/ebookshop", "root", "root");
PreparedStatement pst = (PreparedStatement) conn
.prepareStatement("SELECT * from books");
rset = pst.executeQuery();
while (rset.next()){
Books books = new Books();
books.setId(rset.getInt("id"));
books.setAuthor(rset.getString("author"));
books.setTitle(rset.getString("title"));
books.setPrice(rset.getFloat("price"));
books.setQty(rset.getInt("qty"));
bookList.add(books);
}
} catch (SQLException e) {
System.out.println("Could not connect to DB" + e);
} catch (ClassNotFoundException classexpt){
System.out.println("Couldn't find the class" + classexpt);
}
return bookList;
}
}
Servlet的文件：
import java.io.IOException;
import java.util.ArrayList;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* Servlet implementation class BooksServlet
*/
@WebServlet("/BooksServlet")
public class BooksServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public BooksServlet() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("application/json");
Books books = new Books();
ArrayList <Books> bookList = new ArrayList <Books>();
bookList = books.getBooks(bookList);
request.setAttribute("bookList", bookList);
JSONObject obj=new JSONObject();
JSONArray arr = new JSONArray();
for(int i = 0 ; i< bookList.size() ; i++)
{
Books b = bookList.get(i);
obj.put("id", b.getId());
obj.put("title", b.getTitle());
obj.put("author", b.getAuthor());
obj.put("price", b.getPrice());
obj.put("qty", b.getQty());
arr.add(obj.toString());
obj = new JSONObject();
}
String address = "DisplayBook.jsp";
RequestDispatcher dispatcher =
request.getRequestDispatcher(address);
dispatcher.forward(request, response);
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
}
}
JSP文件：
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<script type="text/javascript" src="js/jquery-1.11.0.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
function showBook(){
$.ajax({
type:"GET",
url:"BooksServlet.java",
dataType: "json",
data: JSON.stringify({ bookList: bookdata }),
success:function(data){
$("#content").html(data);
}
});
}
showBook();
});
</script>
<h3 align="center">Manage Book Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="content" align="center"></div>
<table>
<tr>
<td>id</td>
</tr>
</table>
</body>
</html>
JSP文件是我迷路了。我想我會將數據正確地發送到列表中，但沒有收到。或者我發送錯誤的數據在servlet中？ 任何幫助，將不勝感激！
我添加了response.setContentType（「application/JSON」），那是好還是我該如何讓它接收JSON？ – zydexo 2014-10-20 10:24:21
將contentType更改爲JSON不會自動將您的內容更改爲JSON。首先熟悉Ajax和HTML，然後從JSON開始。 – PbxMan 2014-10-20 10:49:06