Question

我想编写一个程序来查找用户输入的单词，我认为我的解决方案是正确的，但是当我运行它时，该程序在控制台中什么也没显示有人可以解决吗？

$distance = "st_distance_sphere(Point({$lat}, {$lon}), Point({$lat}, {$lon})) * 0.000621371192";

$query = "SELECT table.*, {$distance} as distance
WHERE {$distance} < 10 
   AND postcode IN ( 'XX12 3XX', 'XX12 3XX', 'XX12 3XX' )
ORDER BY distance asc
LIMIT 100";

Answer 1

没有输出的主要原因是

if (maybedel == del)  // <<< this will *never* be true
  cout << maybedel;   // will never run

由于比较数组中的“字符串”需要std::strcmp(maybedel,del) == 0的帮助会更好。

更新：

另一种攻击方法是避免原始循环，并利用STL对您有利。这是一个更强大的解决方案：

#include <algorithm>
#include <iostream>
#include <iterator>
#include <sstream>
#include <vector>
using namespace std;
int main() {
    cout << "enter sentence :\n";
    string sen;
    if (!getline(cin, sen)) throw std::runtime_error("Unable to read sentence");
    cout << "which word do you want to delete ? ";
    string del;
    if (!(cin >> del)) throw std::runtime_error("Unable to read delete word");
    istringstream stream_sen(sen);
    vector<string> arrayofkeptwords;
    remove_copy_if(istream_iterator<string>(stream_sen), istream_iterator<string>(),
                   back_inserter(arrayofkeptwords),
                   [&del](auto const &maybedel) { return maybedel == del; });
    copy(begin(arrayofkeptwords), end(arrayofkeptwords),
         ostream_iterator<string>(cout, " "));
    cout << '\n';
}

Answer 2

除非您的代码以（''）开头，否则if（sen [i] ==''）行，即代码的第12行，阻止代码进入代码块！我稍微修改了代码，现在可以正常工作了。

char sen[200], del[200], maybedel[200];
cout << "enter sentence :" << endl;
cin.getline(sen, 200);
cout << "which word do you want to delete ?" << endl;
cin.getline(del, 200);

int len = strlen(sen);
int t = 0;
for(int i = 0; i <= len; i++) {

    if(sen[i] == ' ' || sen[i] == '\0') {
        maybedel[t] = '\0';
        t = 0;

        if(strcmp(del,maybedel)==0) {
            cout << maybedel << endl;
        }
    }
    else
    {
        maybedel[t] = sen[i];
        t++;
    }
}

在C ++中找到句子中的某个单词

2 个答案: