在C++中使用正则解析字符串的例子

2011-02-27 wcdj

 

问题来源：
http://topic.csdn.net/u/20110227/13/ee7a3998-5170-4c6e-b5e2-d20f5462968e.html?seed=788428640&r=71838505#r_71838505
问题描述：
实现一个电话簿的功能（电话簿的信息用结构体实现）：
通过名字（区分大小写）找到相应的人的所有信息。（例如输入Li，就可以得到含有Li这个词的所有名字：LiYi，LiEr，ZhangLiang，LinKen, LingEr,， 还要找到LongKui，因为LongKui中含有L和i）

注意：由于使用了TR1的扩展库，下面的代码要使用VS2010或sp1的VS2008才能编译。主要使用正则库中的regex_search函数来实现这个例子。

 

#include<iostream>#include <string>#include <regex>#include <iomanip>using namespace std;struct PhoneNameT {char name[22];int phonenum; int postcode;char birthday[11];};struct LinkRec {PhoneNameT p;LinkRec *next;};int main(){ const int LEN =15;PhoneNameT n[LEN] = { {"LiYi",12387400,200240,"1990-1-1"}, {"LiEr",54741414,200240,"1989-12-9"},{"ZhangLiang",46328821,200240,"1990-7-7"},{"ZhangJiao",23001031,100000,"1989-11-11"},{"ZhangZhiDong",12345643,200000,"1989-4-1"},{"LinKen",12812818,300000,"1989-4-5"},{"ZuoYou",52013140,100000,"1989-10-1"},{"LongKui",10254102,200240,"1990-7-29"},{"JiangMing",44052422,200000,"1989-4-4"},{"JingTian",54515171,200000,"1990-4-1"},{"GongAn",77714756,100000,"1990-5-9"},{"GangYi",54745815,300000,"1989-8-8"},{"HongKui",44471741,200240,"1990-7-29"},{"GuangGang",55478155,400000,"1989-4-1"},{"LingEr",25412623,200240,"1989-9-8"}}; cout << "通讯录内容:" << endl;cout << setiosflags(ios::left);PhoneNameT *pp = n;LinkRec *head, *rear,*a;for (int i=0; i<sizeof(n)/sizeof(PhoneNameT); ++i)cout << setw(20) << (pp+i)->name << (pp+i)->phonenum << '/t' << setw(10) << (pp+i)->postcode << '/t' << (pp+i)->birthday << endl;cout << "还需要添加几个人?" << endl;int m=0;cin >> m;PhoneNameT *NewPerson = NULL;if (m>0){NewPerson = new PhoneNameT[m];// new 动态创建数组 head = rear=new LinkRec;// new 创建头节点for (int i=0; i<m; ++i) { cout << "请输入第" << i+1 << "个人的信息:" << endl;cout << "姓名 电话号码 邮编 生日"<< endl;cin >> NewPerson[i].name >> NewPerson[i].phonenum >> NewPerson[i].postcode >> NewPerson[i].birthday;a = new LinkRec;// new 创建节点a->p = NewPerson[i];rear->next = a;rear = a;}rear->next = NULL;// 链表尾节点}LinkRec *head1, *rear1,*a1;head1= rear1 = new LinkRec;// new 创建头节点for (int i=0; i<LEN;++i){ a1 = new LinkRec;a1->p = n[i];rear1->next = a1;rear1 = a1;}rear1->next=head->next;// 跳过新插入联系人链表的头结点cout << "1.获得人物信息" << endl << "2.通过电话反查人物"<< endl << "3.寻找同城人" << endl << "4.最年轻的人" << endl;cout << "请选择方式:" << endl;int choose;cin >> choose;if (choose == 1){ char key [22];LinkRec *h; cout << "输入关键词:" << endl;cin >> key;bool bFind = false;string str,sKey;vector<LinkRec*> lvec;tr1::regex rx;char *p = key;sKey = ".*";while (*p){sKey.append(1,*p);// 使用append连接单个字符到字符串sKey += ".*";++p;}rx = sKey;// 保存模式匹配串if(strcmp(key,""))// 输入关键字非空{ for (h = head1->next; h; h=h->next)// 初始化h = head1->next跳过头结点，因为头结点未初始化没有保存有效信息{str = h->p.name;if (regex_search(str.begin(), str.end(), rx))// 使用正则regex_search函数{bFind = true;lvec.push_back(h);// 保存匹配到的节点指针}}}if (bFind)// 输出匹配的信息 { //cout << setiosflags(ios::left);for (vector<LinkRec*>::const_iterator iter=lvec.begin(); iter!=lvec.end();++iter){cout << setw(20) << (*iter)->p.name << (*iter) ->p.phonenum << '/t' << setw(10) << (*iter)->p.postcode << '/t' << (*iter)->p.birthday << endl;}}elsecout <<"查无此人"<<endl;}// 释放动态分配的资源delete []NewPerson;LinkRec *p1 = head1, *p2;// 第2个链表while (p1){p2 = p1->next;delete p1;p1 = p2;}delete head;// 第1个链表，新插入联系人链表，删除头结点即可return 0;}