我从我写的一些代码中收到一个空指针异常,我看不出异常的原因.这是我的代码:
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.Scanner;
public class SearchingFilesMain {
public static void main(String[] arg) {
int checker4 = 0;
String checker3 = "";
String checker2 = "";
String checker1 = "";
String checker = "";
try {
Scanner scan = new Scanner(new BufferedReader(new FileReader(
"C:\\Users\\User\\Desktop\\asciiTracks.txt")));
while (checker != null) {
String pattern = "Array Start";
checker = scan.findWithinHorizon(pattern, 0);
if(checker.equals("Array Start")){
String pattern2 = "Array Size";
checker3 = scan.findWithinHorizon(pattern2, 300);
System.out.println(checker3);
if(checker3.equals("Array Size")){
checker4 = Integer.parseInt(scan.findInLine("(10000|\\d{1,4})")); …
Run Code Online (Sandbox Code Playgroud) 我有一些代码.在主要功能中,我在BST中推六个元素.当我看调试器时,我看到变量size = 5,但变量root = null.为什么变量root不会改变.
package Search;
public class BST<Key extends Comparable<Key>, Val> {
private class Node{
Key key;
Val val;
Node left;
Node right;
Node prev;
Node(Key k, Val v){
key = k;
val = v;
}
}
public void push(Key k, Val v){
push(root,k,v);
}
private void push(Node x, Key k, Val v){
if(x == null){
x = new Node(k,v);
size++;
return;
}
int cmp = x.key.compareTo(k);
if(cmp > 0)
push(x.left,k,v);
else if(cmp < 0)
push(x.right,k,v);
else
x.val …
Run Code Online (Sandbox Code Playgroud) 大家好,我想知道如何在 node.js 中创建一个带有函数的对象。你能给我一些关于如何做到这一点的例子,或者你能给我提供任何链接吗?我希望你能帮助我解决我的问题,因为我是 node.js 的新手。
我有更新到数据库表的方法,但是当我调用它时,我有一个异常"语法不正确'('."
这是方法
internal Boolean update(int customerID,int followingID, string fullName, string idNumber, string address, string tel, string mobile1, string mobile2, string email, string customerComment, DateTime timeStamp)
{
string sqlStatment = "update customers set (followingID, fullName,idNumber,address,tel,mobile1,mobile2,email,customerComment,timeStamp) = (@followingID, @fullName,@idNumber,@address,@tel,@mobile1,@mobile2,@email,@customerComment,@timeStamp) where customerID=@customerID";
SqlConnection con = new SqlConnection();
con.ConnectionString = connection;
SqlCommand cmd = new SqlCommand(sqlStatment, con);
cmd.Parameters.AddWithValue("@customerID", customerID);
cmd.Parameters.AddWithValue("@followingID", followingID);
cmd.Parameters.AddWithValue("@fullName", fullName);
cmd.Parameters.AddWithValue("@idNumber", idNumber);
cmd.Parameters.AddWithValue("@address", address);
cmd.Parameters.AddWithValue("@tel", tel);
cmd.Parameters.AddWithValue("@mobile1", mobile1);
cmd.Parameters.AddWithValue("@mobile2", mobile2);
cmd.Parameters.AddWithValue("@email", email);
cmd.Parameters.AddWithValue("@customerComment", customerComment);
cmd.Parameters.AddWithValue("@timeStamp", timeStamp);
bool success = false;
try …
Run Code Online (Sandbox Code Playgroud) 即使使用指针后,数字也不会在主函数中交换.我知道有一些原因却无法找到它是什么?
#include <stdio.h>
void swap(int*, int*);
int main()
{
int *ptr, *ptr2;
int num1 = 90;
int num2 = 900;
ptr = &num1;
ptr2 = &num2;
printf("Before swapping the values : %d : %d \n", *ptr, *ptr2);
swap(&num1,&num2);
printf("After calling the swap function : %d : %d \n", *ptr, *ptr2);
return 0;
}
void swap(int *ptr, int *ptr2)
{
int *temp;
temp = ptr;
ptr = ptr2;
ptr2 = temp;
printf("In the swap function : %d : %d\n", *ptr, *ptr2); …
Run Code Online (Sandbox Code Playgroud) 我写了下面的代码.已成功导入证书.谁能告诉我获得例外的原因是什么?我试图连接一个IP https:// xxxx但获得异常.我尝试了两件事
1)通过证书2)下载证书并使用keytool将其添加到java信任库.对我来说,没有人工作正常.
package com.dell;
import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import java.security.GeneralSecurityException;
import javax.net.ssl.HttpsURLConnection;
import javax.net.ssl.SSLContext;
import javax.net.ssl.TrustManager;
import javax.net.ssl.X509TrustManager;
public class HttpURLConnectionExample {
private final String USER_AGENT = "Mozilla/5.0";
public static void main(String[] args) throws Exception {
HttpURLConnectionExample http = new HttpURLConnectionExample();
System.out.println("Testing 1 - Send Http GET request");
http.sendGet();
System.out.println("\nTesting 2 - Send Http POST request");
http.sendPost();
}
// HTTP GET request
private void sendGet() throws Exception {
URL …
Run Code Online (Sandbox Code Playgroud) 我有一个人(员工)搜索应该按姓名,姓氏,出生日期和一些其他参数返回人员列表.搜索适用于每个参数免除日期参数,如出生日期.我的控制器代码是这样的:
$aWorker = new Worker();
$searchWorkerForm = $this->createFormBuilder($aWorker)
->add('omang', 'text', array('required' => false))
->add('workerName', 'text', array('required' => false))
->add('workerSurname', 'text', array('required' => false))
->add('birthDay', 'date',
array('required' => false, 'years' => range(1950, 2020)))
->add('dateOfEmployment', 'date', array('required' => false))
->add('search', 'submit')
->getForm();
//Handle Request
if ($request->getMethod() == 'POST')
$searchWorkerForm->handleRequest($request);
$aWorkerList = $this->getDoctrine()
->getRepository('OsdRetireBundle:Worker')
->findByPersonDetails($aWorker->getOmang(),
$aWorker->getWorkerName(),
$aWorker->getWorkerSurname(),
$aWorker->getBirthDay(),
$aWorker->getDateOfEmployment());
//...
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findByPersonDetails函数是这样的:
public function findByPersonDetails($omang, $WorkerName,
$workerSurname, $birthDay, $dateOfEmployment){
$qqq = $this->getEntityManager()
->createQuery('SELECT w FROM OsdRetireBundle:Worker w
WHERE w.omang LIKE :omang
AND w.workerName LIKE …
Run Code Online (Sandbox Code Playgroud) 我有一个JavaScript字符串,例如:
var test1 = "aaaaaa@bbbbb.temp" ; // Function must return true in this case
var test2 = "ccccc@ddddd.fr " ; // Function must return false in this case
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我想构建一个函数,如果String包含@字符,则返回true,后跟任何字符"bbbbbb"后跟".temp"扩展名.
我怎样才能做到这一点 ?
我正在尝试为java中的一个类做一些功课我正在尝试做的部分是在homework.class中创建一个新的人,它扩展了.name类
名称类
public class Homework extends Object implements Comparable<Homework> {
private int id;
private Name name;
private int section;
private Files files;
private int dateSubmitted;
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......部分鹰派
public Homework(String first, String last, int section, int dateSubmitted){ //fix me
this.id = nextAvailableUid();
this.section = section;
this.dateSubmitted = dateSubmitted;
this.name.Name(first,last); //error is here in the Name call telling me Name is not a method of Name
}
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我得到的错误是在Name中,它告诉我在Name类中创建一个方法Name,我知道它有...从name.class看到的exert
public class Name implements Comparable<Name>{
private String Fname;
private String Lname;
public Name(String first, String …
Run Code Online (Sandbox Code Playgroud) 嗨,我正在使用php模块添加新产品.一切都很好但这个代码由于某种原因没有设定产品的数量.
$id_product = (int)Db::getInstance()->getValue("SELECT id_product FROM `$product` WHERE reference = '$product_xml[id]'");
$p = new Product();
$p->reference = $product_xml['id'];
$p->price = (float)$price;
$p->active = 1;
$p->quantity = (int)$product_xml['count'];
$p->minimal_quantity = 1;
$p->available_for_order=1;
$p->id_category = array(26);
$p->id_category_default = 26;
$p->name[1] = $product_xml['name'];
$p->description[1] = utf8_encode($product_xml->Description);
$p->description_short[1] = utf8_encode($product_xml->Short_Description);
$p->link_rewrite[1] = Tools::link_rewrite($product_xml['name']);
if (!isset($p->date_add) || empty($p->date_add))
$p->date_add = date('Y-m-d H:i:s');
$p->date_upd = date('Y-m-d H:i:s');
$p->save();
$id_product ? $p->updateCategories(array(26)) : $p->addToCategories(array(26));
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