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javascript - Web 服务器代码无法正常工作

转载 作者：行者123 更新时间：2023-11-28 23:34:02 25 4
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我正在尝试编译一些代码，并通过虚拟机让它在我创建的这个 Web 服务索引程序中正常工作。 

package com.cs330;
import javax.ws.rs.*;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;

@Path("ws2")
public class IngredientServices 
{
 @Path("/ingredients")
 @GET
 @Produces("text/plain")
 public String getIngredients() throws SQLException, ClassNotFoundException {

 String connectStr="jdbc:mysql://localhost:3306/fooddb";
 //database username

 String username="root";
 //database password

 String password="csci330pass";
 /* The driver is the Java class used for accessing
  * a particular database. You must download this from
  * the database vendor.
  */

 String driver="com.mysql.jdbc.Driver";
 Class.forName(driver);
 //Creates a connection object for your database

 Connection con = DriverManager.getConnection(connectStr, username, password);
 /* Creates a statement object to be executed on
  * the attached database.
  */

 Statement stmt = con.createStatement();
 /* Executes a database query and returns the results
  * as a ResultSet object.
  */

 ResultSet rs = stmt.executeQuery("SELECT id, name, category FROM ingredient");
 /* This snippet shows how to parse a ResultSet object.
  * Basically, you loop through the object sort of like
  * a linkedlist, and use the getX methods to get data
  * from the current row. Each time you call rs.next()
  * it advances to the next row returned.
  * The result variable is just used to compile all the
  * data into one string.
  */

  String result = "";
  while (rs.next()) 
  {
   int theId = rs.getInt("id");
   String theName = rs.getString("name");
   String theCategory = rs.getString("category");
   result += "id: "+theId+ " , name: "+theName + "("+theCategory+")" + "\n" + "\n";
   }
    return result;
  }//END METHOD

 @Path("/ingredients/{id}")
 @GET
 @Produces("text/plain")
 public String getIngredientById(@PathParam("id") String theId) 
 throws SQLException, ClassNotFoundException {
 int intId = 0;
 try 
 {
  intId = Integer.parseInt(theId);
 }
 catch (NumberFormatException FAIL) 
 {
  intId = 1;
 }//Obtaining an ingredient from the database

 String connectStr="jdbc:mysql://localhost:3306/fooddb";
 String username="root";
 String password="csci330pass";
 String driver="com.mysql.jdbc.Driver";
 Class.forName(driver);
 Connection con = DriverManager.getConnection(connectStr, username, password);
 Statement stmt = con.createStatement();
 ResultSet rs = stmt.executeQuery("SELECT id, name, category FROM ingredient 
 WHERE id=" +intId);

 String result = "";
 while (rs.next()) 
 {
  int theId2 = rs.getInt("id");
  String theName2 = rs.getString("name");
  String theCategory = rs.getString("category");
  result += "id: "+theId2+ " , name: "+theName2 + "("+theCategory+")" + "\n" + "\n";
 }
  return result;
}//END METHOD

@Path("/ingredients/name")
@GET
@Produces("text/plain")
public String getIngredientByName(@QueryParam("name") String theName) 
throws SQLException, ClassNotFoundException
{
 //Obtaining an ingredient from the database
 String connectStr="jdbc:mysql://localhost:3306/fooddb";
 String username="root";
 String password="csci330pass";
 String driver="com.mysql.jdbc.Driver";
 Class.forName(driver);
 Connection con = DriverManager.getConnection(connectStr, username, password);
 Statement stmt = con.createStatement();
 ResultSet rs = stmt.executeQuery("SELECT id, name, category FROM ingredient WHERE 
 name='" + theName + "'");

 String result = "";
 while (rs.next()) 
 {
  int theId3 = rs.getInt("id");
  String theName3 = rs.getString("name");
  String theCategory = rs.getString("category");
  result += "id: "+theId3+ " , name: "+theName3 + "("+theCategory+")" + "\n" + "\n";
 }
  return result;
 }//END METHOD
}//END CODE

现在，前两种方法，即检索所有内容和通过 ID 检索项目，都可以正常工作，但通过 NAME 代码检索则不能。当我在我的虚拟机上的 cmd 上运行它并且在 Tomcat 8 上没有显示任何错误时它编译正确，唯一正确给我结果的代码是前两种方法。由于某种原因，第三种方法一直吐出第一个结果，并且只吐出第一个结果。

我还附上了 index.html 文件代码，以向您展示上面的代码的作用...

<html>
<head>
<title>Shakur (S-3) Burton's Web Services</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js">
</script>
<script>
$(document).ready( function() {
 alert("running script");
 $("#btnAll").click(function() {
 alert("clicked");
 $.ajax( {
 url:"http://localhost:8080/webserv1/resources/ws2/ingredients/",
 type: "GET",
 dataType: "text",
 success: function(result) {
 alert("success");
 $("#p_retrieveAll").html(result); },
 error:function(xhr) {
alert("error");
$("#p_retrieveAll").html("Error:"+xhr.status + " " + xhr.statusText);}
} );
});
$("#btnOneId").click(function() {
alert("clicked");
var inputId=document.getElementById("t_ingredId").value;
var theUrl = "http://localhost:8080/webserv1/resources/ws2/ingredients/"+inputId;
$.ajax( {
url: theUrl,
type: "GET",
dataType: "text",
success: function(result) {
alert("success");
$("#p_retrieveOneId").html(result); },
error:function(xhr) {
alert("error");
$("#p_retrieveOneId").html("Error:"+xhr.status+" "+xhr.statusText);}
} );
});
$("#btnOneName").click(function() {
alert("clicked");
var inputName=document.getElementByName("t_ingredName").value;
var theUrl: "http://localhost:8080/webserv1/resources/ws2/ingredients/ingredient?name="+inputName;
$.ajax( {
url: theUrl,
type: "GET",
dataType: "text",
success: function(result) {
alert("success");
$("#p_retrieveOneName").html(result); },
error:function(xhr) {
alert("error");
$("#p_retrieveOneName").html("Error:"+xhr.status+" "+xhr.statusText);}
} );
});
});
</script>
</head>
<body>
<h3>Testing Web Services</h3>
<div id="retrieveAll">
<button id="btnAll">Click to Retrieve All</button>
<p id="p_retrieveAll">Ingredients List Goes here</p>
</div>
<div id="retrieveOneId">
<input type="text" id="t_ingredId" value="type id here" />
<button id="btnOneId">Click to Retrieve by Id</button>
<p id="p_retrieveOneId">Ingredient By Id Goes here</p>
</div>
<div id="retrieveOneName">
<input type="text" id="t_ingredName" value="type name here"/>
<button id="btnOneName">Click to Retrieve by Name</button>
<p id="p_retrieveOneName">Ingredient By Name Goes here</p>
</div>
</body>
</html>

关于为什么我的 IngredientServices javascript 中的 GET by NAME 方法无法正常工作，这里是否可以提供任何建议？我错过了什么吗？

编辑 - 2014 年 11 月 4 日 - 16:05...

我认为这个问题可能出在数据库程序的这一部分中...与其通过按 ID 查找所述元素来按名称搜索成分，我应该在给定参数内按名称搜索它。希望这可以解决我遇到的问题...

顺便说一句，这是我之前修改的代码:var inputName=document.getElementByName("t_ingredName").value;

最佳答案

当我将您的代码添加到 Firefox 并单击名为 Firebug 的加载项时，它向我显示以下错误:

SyntaxError: missing ; before statement
var theUrl: "http://localhost:8080/webserv1/resources/ws2/ingredients/

因此它应该是 var theUrl= "http://localhost:8080/webserv1/resources/ws2/ingredients/ingredient?name="+inputName;

您尝试过调试吗？

此外，不要使用警报，而是使用 console.log("your message here"); - 它会显示在 Firebug 的控制台中。

关于javascript - Web 服务器代码无法正常工作，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26692738/

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