C++ 如何查找和替换字符串？

How to find and replace string?

提问by neuromancer

If sis a std::string, then is there a function like the following?

如果s是 a std::string，那么是否有如下函数？

s.replace("text to replace", "new text");

回答by Mateen Ulhaq

Try a combination of std::string::findand std::string::replace.

尝试组合std::string::find和std::string::replace。

This gets the position:

这得到了位置：

std::string s;
std::string toReplace("text to replace");
size_t pos = s.find(toReplace);

And this replaces the firstoccurrence:

这取代了第一次出现：

s.replace(pos, toReplace.length(), "new text");

---

---

Now you can simply create a function for your convenience:

现在您可以简单地创建一个函数来方便您：

std::string replaceFirstOccurrence(
    std::string& s,
    const std::string& toReplace,
    const std::string& replaceWith)
{
    std::size_t pos = s.find(toReplace);
    if (pos == std::string::npos) return s;
    return s.replace(pos, toReplace.length(), replaceWith);
}

回答by Czarek Tomczak

Do we really need a Boost library for seemingly such a simple task?

我们真的需要一个 Boost 库来完成看似如此简单的任务吗？

To replace all occurences of a substring use this function:

要替换所有出现的子字符串，请使用此函数：

std::string ReplaceString(std::string subject, const std::string& search,
                          const std::string& replace) {
    size_t pos = 0;
    while ((pos = subject.find(search, pos)) != std::string::npos) {
         subject.replace(pos, search.length(), replace);
         pos += replace.length();
    }
    return subject;
}

If you need performance, here is an optimized function that modifies the input string, it does not create a copy of the string:

如果您需要性能，这里是一个修改输入字符串的优化函数，它不会创建字符串的副本：

void ReplaceStringInPlace(std::string& subject, const std::string& search,
                          const std::string& replace) {
    size_t pos = 0;
    while ((pos = subject.find(search, pos)) != std::string::npos) {
         subject.replace(pos, search.length(), replace);
         pos += replace.length();
    }
}

Tests:

测试：

std::string input = "abc abc def";
std::cout << "Input string: " << input << std::endl;

std::cout << "ReplaceString() return value: " 
          << ReplaceString(input, "bc", "!!") << std::endl;
std::cout << "ReplaceString() input string not modified: " 
          << input << std::endl;

ReplaceStringInPlace(input, "bc", "??");
std::cout << "ReplaceStringInPlace() input string modified: " 
          << input << std::endl;

Output:

输出：

Input string: abc abc def
ReplaceString() return value: a!! a!! def
ReplaceString() input string not modified: abc abc def
ReplaceStringInPlace() input string modified: a?? a?? def

回答by phooji

Yes: replace_allis one of the boost string algorithms:

是：replace_all是增强字符串算法之一：

Although it's not a standard library, it has a few things on the standard library:

虽然它不是标准库，但它在标准库上有一些东西：

1. More natural notationbased on ranges rather than iterator pairs. This is nice because you can nest string manipulations (e.g., replace_allnested inside a trim). That's a bit more involved for the standard library functions.
2. Completeness.This isn't hard to be 'better' at; the standard library is fairly spartan. For example, the boost string algorithms give you explicit control over how string manipulations are performed (i.e., in place or through a copy).

1. 基于范围而不是迭代器对的更自然的表示法。这很好，因为您可以嵌套字符串操作（例如，replace_all嵌套在 a 中trim）。这对于标准库函数来说涉及更多。
2. 完整性。这并不难“更好”；标准库相当简陋。例如，boost 字符串算法使您可以明确控制如何执行字符串操作（即，就地或通过副本）。

回答by Mark Tolonen

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

int main ()
{
    string str("one three two four");
    string str2("three");
    str.replace(str.find(str2),str2.length(),"five");
    cout << str << endl;
    return 0;
}

Output

输出

one five two four

回答by Sylvain Rochette

like some say boost::replace_all

就像有人说 boost::replace_all

here a dummy example:

这是一个虚拟示例：

    #include <boost/algorithm/string/replace.hpp>

    std::string path("file.gz");
    boost::replace_all(path, ".gz", ".zip");

回答by Nawaz

Not exactly that, but std::stringhas many replaceoverloaded functions.

不完全是这样，但std::string有许多replace重载的功能。

Go through this linkto see explanation of each, with examples as to how they're used.

通过此链接查看每个链接的解释，以及有关如何使用它们的示例。

Also, there are several versions of string::findfunctions (listed below) which you can use in conjunction with string::replace.

此外，还有多个版本的string::find函数（如下所列）可以与string::replace.

• find
• rfind
• find_first_of
• find_last_of
• find_first_not_of
• find_last_not_of

• 找
• 找到
• find_first_of
• find_last_of
• find_first_not_of
• find_last_not_of

---

---

Also, note that there are several versions of replacefunctions available from <algorithm>which you can also use (instead of string::replace):

另外，请注意，您还可以使用多个版本的replace函数<algorithm>（而不是string::replace）：

• replace
• replace_if
• replace_copy
• replace_copy_if

• 代替
• 替换_如果
• 替换_复制
• replace_copy_if

回答by Bayram AKGüL

// replaced text will be in buffer.
void Replace(char* buffer, const char* source, const char* oldStr,  const char* newStr)
{
    if(buffer==NULL || source == NULL || oldStr == NULL || newStr == NULL) return; 

    int slen = strlen(source);
    int olen = strlen(oldStr);
    int nlen = strlen(newStr);

    if(olen>slen) return;
    int ix=0;

    for(int i=0;i<slen;i++)
    {
        if(oldStr[0] == source[i])
        {
            bool found = true;
            for(int j=1;j<olen;j++)
            {
                if(source[i+j]!=oldStr[j])
                {
                    found = false;
                    break;
                }
            }

            if(found)
            {
                for(int j=0;j<nlen;j++)
                    buffer[ix++] = newStr[j];

                i+=(olen-1);
            }
            else
            {
                buffer[ix++] = source[i];
            }
        }
        else
        {
            buffer[ix++] = source[i];
        }
    }
}

回答by anisoptera

Here's the version I ended up writing that replaces all instances of the target string in a given string. Works on any string type.

这是我最终编写的版本，它替换了给定字符串中目标字符串的所有实例。适用于任何字符串类型。

template <typename T, typename U>
T &replace (
          T &str, 
    const U &from, 
    const U &to)
{
    size_t pos;
    size_t offset = 0;
    const size_t increment = to.size();

    while ((pos = str.find(from, offset)) != T::npos)
    {
        str.replace(pos, from.size(), to);
        offset = pos + increment;
    }

    return str;
}

Example:

例子：

auto foo = "this is a test"s;
replace(foo, "is"s, "wis"s);
cout << foo;

Output:

输出：

thwis wis a test

Note that even if the search string appears in the replacement string, this works correctly.

请注意，即使搜索字符串出现在替换字符串中，这也能正常工作。

回答by snb

void replace(char *str, char *strFnd, char *strRep)
{
    for (int i = 0; i < strlen(str); i++)
    {
        int npos = -1, j, k;
        if (str[i] == strFnd[0])
        {
            for (j = 1, k = i+1; j < strlen(strFnd); j++)
                if (str[k++] != strFnd[j])
                    break;
            npos = i;
        }
        if (npos != -1)
            for (j = 0, k = npos; j < strlen(strRep); j++)
                str[k++] = strRep[j];
    }

}

int main()
{
    char pst1[] = "There is a wrong message";
    char pfnd[] = "wrong";
    char prep[] = "right";

    cout << "\nintial:" << pst1;

    replace(pst1, pfnd, prep);

    cout << "\nfinal : " << pst1;
    return 0;
}

回答by Shubham Agrawal

void replaceAll(std::string & data, const std::string &toSearch, const std::string &replaceStr)
{
    // Get the first occurrence
    size_t pos = data.find(toSearch);
    // Repeat till end is reached
    while( pos != std::string::npos)
    {
        // Replace this occurrence of Sub String
        data.replace(pos, toSearch.size(), replaceStr);
        // Get the next occurrence from the current position
        pos =data.find(toSearch, pos + replaceStr.size());
    }
}

More CPP utilities: https://github.com/Heyshubham/CPP-Utitlities/blob/master/src/MEString.cpp#L60

更多 CPP 实用程序：https: //github.com/Heyshubham/CPP-Utilities/blob/master/src/MEString.cpp#L60