我需要搜索一个字符串并编辑它的格式。
到目前为止,我可以替换第一次出现的字符串,但是我无法在下一次出现此字符串时这样做。
这就是我的工作,有点像:
if(chartDataString.find("*A") == string::npos){ return;}
else{chartDataString.replace(chartDataString.find("*A"), 3,"[A]\n");}
如果找不到字符串,则根本不打印任何内容,因此不好。
我知道我需要遍历整个字符串chartDataString并替换所有出现的事件。我知道有很多类似的帖子,但我不明白(像这个Replace substring with another substring C++)
我也尝试过这样的事情来循环遍历字符串:
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");
}
}
编辑考虑到建议,这在理论上应该有效,但我不知道为什么不这样做
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();
}
尝试以下方法
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();
}
如果boost
可用,您可以使用以下内容:
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);
要对原始字符串执行修改,请执行
std::string result = boost::replace_all_copy(origStr , subStringToRemove , subStringToReplace);
在不修改原始字符串的情况下执行修改。
find
函数采用可选的第二个参数:开始搜索的位置。默认情况下,这是零。
开始搜索下一场比赛的好位置是插入先前替换的位置,以及替换的长度。例如,如果我们在位置7处插入长度为3的字符串,那么下一个find
应该从位置10开始。
如果搜索字符串碰巧是替换的子字符串,则此方法将避免无限循环。想象一下,如果你试图用log
替换所有出现的analog
,但不要跳过替换。
这样做是相当尴尬(也可能不是太有效)。我通常使用一个函数:
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;
}
基本上,只要你能找到from
的实例,附加中间文本和to
,然后前进到from
的下一个实例,你就会循环。最后,在from
的最后一个实例之后附加任何文本。
(如果您要在C ++中进行大量编程,那么习惯使用迭代器(如上所述)而不是std::string
的特殊成员函数可能是一个好主意。上面的内容可以用于任何C ++容器类型,因此更加惯用。)
/// 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" );
}
}
如果您需要反转的字符串大小不同:
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 ;
}