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我試圖編譯一些代碼，並使其在通過虛擬機創建的此Web服務索引程序中正常工作。Web服務器代碼無法正常工作
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
現在，前兩種方法，這是檢索所有內容，並檢索由ID都正常工作的項目，它是按名稱代碼，無法檢索。當我在我的虛擬機上的cmd上運行它時正確編譯，並且在Tomcat 8上沒有顯示任何錯誤，但正確給我結果的唯一代碼是前兩種方法。出於某種原因，第三種方法不斷吐出第一個結果，並且只是第一個結果。
我還附上index.html文件代碼向你展示什麼上面的代碼與...
<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>
是否有可以在這裏提出來的任何建議的工作，爲什麼按名稱獲取方法在我的IngredientServices中javascript無法正常工作？我錯過了什麼嗎？
編輯 - 2014年11月4日 - 16:05 ...
我想，這個問題可能是數據庫程序的這一部分...而是通過尋找說按名稱搜索的一個組成成分的元素的ID，我應該在給定的參數內搜索NAME。希望這可以解決我遇到的問題...
順便說一句，這是我修改過的代碼：var inputName = document.getElementByName（「t_ingredName」）.value;
當你直接對數據庫執行sql命令時，你會得到多個結果嗎？ – 2014-11-01 20:36:08
如果在對數據庫執行sql時只獲得一個結果，那意味着數據庫中只有一個匹配條目，並且代碼按預期工作。 – 2014-11-02 18:00:45
經過一段時間才能更好地瞭解事情。我按照指示採納了你的建議。它似乎可能是index.html中的某些東西導致我頭痛的問題。 – user2891351 2014-11-04 21:03:09