Question

我想首先说我还在学习，有些人可能认为我的代码看起来很糟糕，但现在就这样了。

所以我有这个文本文件，我们可以调用example.txt。

example.txt中的一行如下所示：

randomstuffhereitem=1234randomstuffhere

我希望我的程序接收item =旁边的数字=我已经使用以下代码开始了一些。

#include <iostream>
#include <fstream>
#include <string>
using namespace std;

    string word;

int main()
{
    ifstream readFile("example.txt", ios::app);
    ofstream outfile("Found_Words.txt", ios::app);
    bool found = false; 

    long int price;
    cout << "Insert a number" << endl;
    cout << "number:";
    cin >> number;
    system("cls");
    outfile << "Here I start:";
    while( readFile >> word )
    {
        if(word == "item=")

这是问题所在;首先它只搜索“item =”但是为了找到它，它不能包含在其他字母中。它必须是一个独立的词。

它不会找到：

helloitem=hello

它会找到：

hello item= hello

必须用空格分隔，这也是一个问题。

其次我想找到item =旁边的数字。就像我希望它能够找到item = 1234并且请注意1234可以是任何数字，如6723。

而且我不希望它找到数字之后的内容，所以当数字停止时，它将不再接收数据。喜欢item = 1234hello必须是item = 1234

            {
            cout <<"The word has been found." << endl;
            outfile << word << "/" << number;
            //outfile.close();
                if(word == "item=")
                {
        outfile << ",";
                }

        found = true;
            }
    }
    outfile << "finishes here" ;
    outfile.close();
    if( found = false){
    cout <<"Not found" << endl;
    }
    system ("pause");
}

Answer 1

您可以使用以下代码：

bool get_price(std::string s, std::string & rest, int & value)
{
    int pos = 0; //To track a position inside a string
    do //loop through "item" entries in the string
    {
        pos = s.find("item", pos); //Get an index in the string where "item" is found
        if (pos == s.npos) //No "item" in string
            break;
        pos += 4; //"item" length
        while (pos < s.length() && s[pos] == ' ') ++pos; //Skip spaces between "item" and "="
        if (pos < s.length() && s[pos] == '=') //Next char is "="
        {
            ++pos; //Move forward one char, the "="
            while (pos < s.length() && s[pos] == ' ') ++pos; //Skip spaces between "=" and digits
            const char * value_place = s.c_str() + pos; //The number
            if (*value_place < '0' || *value_place > '9') continue; //we have no number after =
            value = atoi(value_place); //Convert as much digits to a number as possible
            while (pos < s.length() && s[pos] >= '0' && s[pos] <= '9') ++pos; //skip number
            rest = s.substr(pos); //Return the remainder of the string
            return true; //The string matches 
        }
    } while (1);
    return false; //We did not find a match
}

请注意，您还应该更改从文件中读取字符串的方式。你可以读取换行符（std :: getline）或流的末尾，如下所述：stackoverflow question

在它之后找到Word + X字母

1 个答案: