将.txt中的数据输入结构数组 我的C++作业，我有一个.txt文档，里面包含了类似于：的建筑信息。

建筑物名称 建成年份 横向坐标 lon坐标

前

我必须将其读入我创建的结构的数组中：

struct list 
{
    char name[50];
    int yearBuilt;
    double latCoord;
    double lonCoord;
} building;

现在，我已经创建了一个for循环，以便将数据读入我创建的类型列表数组：

list buildingnumber[SIZE]; //array for buildings

但是，当我试图打印出“k”最早建造的建筑物时，数组中没有显示任何数据

以下是我当前的代码：

#include <iostream>
#include <fstream>
#include <istream>
#include <cstdlib>
#include <string>

using namespace std;
const int SIZE = 5000;

//struct for building type
struct list 
{
    char name[50];
    int yearBuilt;
    double latCoord;
    double lonCoord;
}building;

list buildingnumber[SIZE]; //array for buildings

void InsertionSort(list buildingnumber[], int buildingsloaded)
{
    int key = 0, i = 0;
    for(int j = 1; j < buildingsloaded; j++)
    {
        key=buildingnumber[j].yearBuilt;
        i=j-1;
        while(buildingnumber[i].yearBuilt > key && i >= 0)
        {
            buildingnumber[i+1] = buildingnumber[i];
            i--;
        }
        buildingnumber[i+1].yearBuilt = key;
    }
}

int main()
{
    char filePath[50];
    ifstream openFile;
    cout << "Enter the path of the building file: ";
    cin.getline(filePath, 50);
    openFile.open(filePath);

    //verify if file is opened + report buildings loaded
    int buildingsloaded = 0;
    if(!openFile.fail())
    {
        while(openFile >> building.name >> building.yearBuilt >> building.latCoord >> building.lonCoord)
        {
            buildingsloaded++;
        }
        cout << buildingsloaded << " buildings have been loaded." << endl;
    }

    // get how many buildings user wants
    int k = 0;
    cout << "How many buildings are you interested in?: ";
    cin >> k;

    //create array
    // loadKbuildings(building, buildingsloaded);
    for(int i = 0; i < k; i++)
    {
        openFile >> buildingnumber[i].name >> buildingnumber[i].yearBuilt >> buildingnumber[i].latCoord >> buildingnumber[i].lonCoord;  
    }

    // insertion sort
    InsertionSort(buildingnumber, buildingsloaded);

    // display earliest k buildings
    cout << "The " << k << " oldest buildings are: " << endl;
    int i = 0;
    while ( i < k )
    {
        cout << buildingnumber[i].name << endl;
        cout << "Year Built: " << buildingnumber[i].yearBuilt << endl;
        cout << "Coordinates: (" << buildingnumber[i].latCoord << "," << buildingnumber[i].lonCoord << ")" << endl;
        cout << endl;
        i++;
    }
}

#包括
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#包括
使用名称空间std；
常数int SIZE=5000；
//建筑类型结构
结构列表
{
字符名[50]；
国际年建成；
双拉特库德；
双隆库德；
}建筑；
列表buildingnumber[大小]//建筑物阵列
void InsertionSort（列表buildingnumber[]，int BuildingLoaded）
{
int键=0，i=0；
对于（int j=1；jkey&&i>=0）
{
建筑编号[i+1]=建筑编号[i]；
我--；
}
buildingnumber[i+1]。YearBuild=key；
}
}
int main（）
{
char文件路径[50]；
ifstreamopenfile；
cout>building.name>>building.yearbuilded>>building.latCoord>>building.lonCoord）
{
buildingsload++；
}
cout buildingnumber[i]。名称>>buildingnumber[i]。年建>>buildingnumber[i]。latCoord>>buildingnumber[i]。lonCoord；
}
//插入排序
插入排序（buildingnumber，BuildingLoaded）；
//显示最早的k建筑物
cout问题在于：

if(!openFile.fail())
{
    while(openFile >> building.name >> building.yearBuilt >> building.latCoord >> building.lonCoord)
    {
        buildingsloaded++;
    }
    cout << buildingsloaded << " buildings have been loaded." << endl;
}

您应该使用
std:：string
和
std:：vector
而不是C样式的数组。由于字符串是如何从输入文本流中读取的，您的名称读取将只提取多单词名称的第一部分。一旦发生这种情况，流的其余部分将被回避，以便将字段与流位置进行匹配。建议你设计一个
getline（）
计划。它似乎工作得更好，因为我已经获得了建筑名称、lat和long坐标。但循环似乎并没有增加。如果我对5栋建筑感兴趣，它会显示同一栋建筑的5栋。很抱歉，在键入上一栋之前，我没有听到你的cin.getline（）评论，我会尝试一下，然后返回you@kxf951即使没有更改getline，它应该仍然适用于以下测试用例：Parking/1993/34.2252/37.5563/Admin/1985/34.2356/37.5734/并键入2。或者可能是你的插入运动。试着把它拿出来？我想我的插入排序可能有问题。正如你可能猜到的，我试图报告“k”最早建造的建筑。你是对的，是我的插入排序导致相同的建筑重复出现。我试着看看我做错了什么
if(!openFile.fail())
{
    while(openFile >> building.name >> building.yearBuilt >> building.latCoord >> building.lonCoord)
    {
        buildingsloaded++;
    }
    cout << buildingsloaded << " buildings have been loaded." << endl;
}

int c = 0;
if(!openFile.fail())
{
    // *
    while(openFile >> buildingnumber[c].name >> buildingnumber[c].yearBuilt >> buildingnumber[c].latCoord >> buildingnumber[c].lonCoord)
    {
        buildingsloaded++;
        c++;
    }
    cout << buildingsloaded << " buildings have been loaded." << endl;
}

//create array
// loadKbuildings(building, buildingsloaded);
for(int i = 0; i < k; i++)
{
    openFile >> buildingnumber[i].name >> buildingnumber[i].yearBuilt >> buildingnumber[i].latCoord >> buildingnumber[i].lonCoord;  
}