我刚在家用电脑上安装了PHP和Apache.当我尝试调用函数时,mysql_connect我得到:
fatal error: call to undefined function mysql_connect.
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我已经加载了php.ini,其中我有未注释的行extension=php_mysql.dll,
extension=php_mysqli.dll并将扩展目录更改为extension_dir = "C:\php\ext"- 这是文件php_mysql.dll和php_mysqli.dll所在的目录.我该如何解决这个问题?
phpinfo()的 输出:http ://jsfiddle.net/MMTwA/
我想要打印代码:
B B
A B
B A
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但它打印
Item Item
Item Item
Item Item
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代码:
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
struct Item {
Item(){}
virtual void method1 (Item x, Item y) {cout << "Item Item\n";}
};
struct A : public Item {
A(){}
};
struct B : public Item {
B(){}
virtual void method1 (B x, B y) {cout << "B B\n";}
virtual void method1 (A x, B y) {cout << "A B\n";}
virtual void method1 …Run Code Online (Sandbox Code Playgroud) 我一直在想是否有可能说出State课程实施IEnumerable<Person>,IEnumerable<City>所以我可以通过foreach以及所有城市让所有居住在该州的人们.它甚至不会编译说:( Error 1 'ConsoleApplication1.City' does not implement interface member 'System.Collections.IEnumerable.GetEnumerator()'怪异)......这是代码:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Collections;
namespace ConsoleApplication1
{
class Person
{
}
class City : IEnumerable<Person>
{
// City has citizens:
Person[] citizens;
IEnumerator<Person> IEnumerable<Person>.GetEnumerator()
{
foreach (Person p in citizens)
yield return p;
}
}
class State : IEnumerable<Person>, IEnumerable<City>
{
// State has cities:
City[] cities;
IEnumerator<Person> IEnumerable<Person>.GetEnumerator()
{
foreach (City c in cities)
foreach (Person …Run Code Online (Sandbox Code Playgroud) 可能重复:
嵌套类型问题
假设我有这段代码:
public class Tree
{
private readonly int nodeCapacity;
public int NodeCapacity { get { return nodeCapacity; } }
public Tree(int nodeCapacity)
{
this.nodeCapacity = nodeCapacity;
}
private class Node
{
object[] objects;
Node()
{
objects = new object[nodeCapacity];
}
}
}
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这不编译并给出此错误:
无法访问外部类型的非静态成员...
有没有办法访问外部类型的非静态成员(来自嵌套类)而不通过构造函数参数传递变量,也不使用"static"关键字?
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