问题
I need to search a string and edit the formatting of it.
So far I can replace the first occurrence of the string, but I am unable to do so with the next occurrences of this string.
This is what I have working, sort of:
if(chartDataString.find("*A") == string::npos){ return;}
else{chartDataString.replace(chartDataString.find("*A"), 3,"[A]\n");}
If it doesn't find the string, nothing prints at all, so that's not good.
I know I need to loop through the entire string chartDataString and replace all occurrences. I know there are a lot of similar posts to this but I don't understand (like this Replace substring with another substring C++)
I've also tried to do something like this to loop over the string:
string toSearch = chartDataString;
string toFind = "*A:";
for (int i = 0; i<toSearch.length() - toFind.length(); i++){
if(toSearch.substr(i, toFind.length()) == toFind){
chartDataString.replace(chartDataString.find(toFind), 3, "[A]\n");
}
}
EDIT taking into consideration suggestions, this in theory should work, but I don't know why it doesn't
size_t startPos=0;
string myString = "*A";
while(string::npos != (startPos = chartDataString.find(myString, startPos))){
chartDataString.replace(chartDataString.find(myString, startPos), 3, "*A\n");
startPos = startPos + myString.length();
}
回答1:
try the following
const std::string s = "*A";
const std::string t = "*A\n";
std::string::size_type n = 0;
while ( ( n = chartDataString.find( s, n ) ) != std::string::npos )
{
chartDataString.replace( n, s.size(), t );
n += t.size();
}
回答2:
In case boost is available, you can use the following:
std::string origStr = "this string has *A and then another *A";
std::string subStringToRemove = "*A";
std::string subStringToReplace = "[A]";
boost::replace_all(origStr , subStringToRemove , subStringToReplace);
To perform the modification on the original string, OR
std::string result = boost::replace_all_copy(origStr , subStringToRemove , subStringToReplace);
To perform the modifications without modifying the original string.
回答3:
The find function takes an optional second argument: the position from which to begin searching. By default this is zero.
A good position to begin searching for the next match is the position where the previous replacement was inserted, plus that replacement's length. For instance if we insert a string of length 3 at position 7, then the next find should begin at position 10.
If the search string happens to be a substring of the replacement, this approach will avoid an infinite loop. Imagine if you try to replace all occurrences of log with analog, but don't skip over the replacement.
回答4:
It's fairly awkward (and probably not too efficient) to do it in place. I usually use a function along the lines of:
std::string
replaceAll( std::string const& original, std::string const& from, std::string const& to )
{
std::string results;
std::string::const_iterator end = original.end();
std::string::const_iterator current = original.begin();
std::string::const_iterator next = std::search( current, end, from.begin(), from.end() );
while ( next != end ) {
results.append( current, next );
results.append( to );
current = next + from.size();
next = std::search( current, end, from.begin(), from.end() );
}
results.append( current, next );
return results;
}
Basically, you loop as long as you can find an instance of
from, appending the intermediate text and to, and advancing
to the next instance of from. At the end, you append any text
after the last instance of from.
(If you're going to do much programming in C++, it's probably
a good idea to get used to using iterators, like the above,
rather than the special member functions of std::string.
Things like the above can be made to work with any of the C++
container types, and for this reason, are more idiomatic.)
回答5:
/// Returns a version of 'str' where every occurrence of
/// 'find' is substituted by 'replace'.
/// - Inspired by James Kanze.
/// - http://stackoverflow.com/questions/20406744/
std::string replace_all(
const std::string & str , // where to work
const std::string & find , // substitute 'find'
const std::string & replace // by 'replace'
) {
using namespace std;
string result;
size_t find_len = find.size();
size_t pos,from=0;
while ( string::npos != ( pos=str.find(find,from) ) ) {
result.append( str, from, pos-from );
result.append( replace );
from = pos + find_len;
}
result.append( str, from , string::npos );
return result;
/*
This code might be an improvement to James Kanze's
because it uses std::string methods instead of
general algorithms [as 'std::search()'].
*/
}
int main() {
{
std::string test = "*A ... *A ... *A ...";
std::string changed = "*A\n ... *A\n ... *A\n ...";
assert( changed == replace_all( test, "*A", "*A\n" ) );
}
{
std::string GB = "My gorila ate the banana";
std::string gg = replace_all( GB, "gorila", "banana" );
assert( gg == "My banana ate the banana" );
gg = replace_all( gg, "banana", "gorila" );
assert( gg == "My gorila ate the gorila" );
std::string bb = replace_all( GB, "banana", "gorila" );
assert( gg == "My gorila ate the gorila" );
bb = replace_all( bb, "gorila" , "banana" );
assert( bb == "My banana ate the banana" );
}
{
std::string str, res;
str.assign( "ababaabcd" );
res = replace_all( str, "ab", "fg");
assert( res == "fgfgafgcd" );
str="aaaaaaaa"; assert( 8==str.size() );
res = replace_all( str, "aa", "a" );
assert( res == "aaaa" );
assert( "" == replace_all( str, "aa", "" ) );
str = "aaaaaaa"; assert( 7==str.size() );
res = replace_all( str, "aa", "a" );
assert( res == "aaaa" );
str = "..aaaaaa.."; assert( 10==str.size() );
res = replace_all( str, "aa", "a" );
assert( res == "..aaa.." );
str = "baaaac"; assert( 6==str.size() );
res = replace_all( str, "aa", "" );
assert( res == "bc" );
}
}
回答6:
If ever the strings you need to invert are not of the same size:
void Replace::replace(std::string & str, std::string const & s1, std::string const & s2)
{
size_t pos;
pos = 0;
while ((pos = str.find(s1, pos)) != std::string::npos)
{
str.erase(pos, s1.length());
str.insert(pos, s2);
pos += s2.length();
}
return ;
}
来源：https://stackoverflow.com/questions/20406744/how-to-find-and-replace-all-occurrences-of-a-substring-in-a-string