[英]Getting a resource not found response while developing a RESTful web service
我正在eclipse中使用Jersey框架開發一個示例RESTful Web服務。 它只需連接到mysql服務器就可以執行查詢並返回值。 當我編譯並運行(JVM)時,我得到了所需的輸出。
現在,我正在使用Tomcat v8.0服務器來處理請求,並希望XML格式的查詢結果相同。 但我總是得到404。
這是我的代碼
mysqlconn.java
package com.just_test_db;
import java.sql.*;
import javax.ws.rs.*;
import javax.ws.rs.core.MediaType;
@Path("/mysqlconn")
class mysqlconn {
@GET
@Path("/Customer")
@Produces(MediaType.APPLICATION_XML)
public void getdata() {
try {
Connection con = Connect_db.getConnection();
mysql_query(con);
mysql_connection_close(con);
}
catch (Exception e) {
e.printStackTrace();
}
}
public Customer mysql_query(Connection con) throws Exception {
Statement st = con.createStatement();
ResultSet rs = st.executeQuery("select Customer_id, Customer_email, Customer_phone from Customers");
Customer c = new Customer(rs.getInt(1),rs.getString(2),rs.getString(3));
return c;
}
public void mysql_connection_close(Connection con) throws Exception {
con.close();
}
}
這是Customer.java
package com.just_test_db;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
@XmlRootElement(name="Customer")
class Customer {
int CustomerID;
String Customer_email;
String Customer_phone;
public Customer() {}
public Customer(int c, String e, String p) {
this.CustomerID = c;
this.Customer_email = e;
this.Customer_phone = p;
}
public int getCustomerID() {
return CustomerID;
}
@XmlElement
public void setCustomerID(int customerID) {
this.CustomerID = customerID;
}
public String getCustomer_email() {
return Customer_email;
}
@XmlElement
public void setCustomer_email(String customer_email) {
this.Customer_email = customer_email;
}
public String getCustomer_phone() {
return Customer_phone;
}
@XmlElement
public void setCustomer_phone(String customer_phone) {
this.Customer_phone = customer_phone;
}
}
這是我的Connect_db.java:
package com.just_test_db;
import java.sql.Connection;
import java.sql.DriverManager;
class Connect_db {
static Connection con = null;
static String dbhost = "jdbc:mysql://localhost:3306/Test";
static String dbuser = "user";
static String dbpass = "pass";
public static Connection getConnection() {
if (con != null) return con;
return getConnection(dbhost,dbuser,dbpass);
}
public static Connection getConnection(String dbhost, String dbuser, String dbpass) {
try {
Class.forName("com.mysql.jdbc.Driver");
con = DriverManager.getConnection(dbhost,dbuser,dbpass);
return con;
}
catch (Exception c) {
System.out.println("Could not connect to the Database");
c.printStackTrace();
}
return con;
}
}
這是我的web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
id="WebApp_ID" version="3.0">
<display-name>Test_db</display-name>
<servlet>
<servlet-name>Jersey RESTful Application</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.just_test_db</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Jersey RESTful Application</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
我期望輸出是這樣的:
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Customer>
<CustomerID>1</Customerid>
<CustomerEmail>Bhanu</CustomerEmail>
<CustomerPhone>9999999999</CustomerPhone>
</Customer>
我使用的網址是:
http://localhost:8080/Test_db/rest/mysqlconn/Customer
嘗試
http://localhost:8080/<WebAppNameInTomcat>/rest/mysqlconn/Customer
Test_db是一個顯示名稱,可能不是此處使用的名稱(Test_db)。 Tomcat使用WebApp名稱作為上下文路徑
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