동일한 이전과 같은 단어로 다른 qstring을 읽으십시오.QString :: indexof
문제
i (기본 예제) 파일에서 두 개의 QStrings (Valesone, Valuetwo)를 읽습니다
int main(int argc, char *argv[]) {
QString x = ("yes "
"start ValueOne end"
"no"
"start ValueTwo end");
//try to read ValueOne
QString s = "start";
QString e = "end";
int start = x.indexOf(s, 0, Qt::CaseInsensitive);
int end = x.indexOf(e, Qt::CaseInsensitive);
if(start != -1){
QString y = x.mid(start + s.length(), (end - (start + s.length())));
qDebug() << y << (start + s.length()) << (end - (start + s.length()));
//try to read ValueTwo
QString s2 = "start";
QString e2 = "end";
int start2 = x.indexOf(s2, 0, Qt::CaseInsensitive);
int end2 = x.indexOf(e2, Qt::CaseInsensitive);
if(start2 != -1){
QString y2 = x.mid(start2 + s.length2(), (end2 - (start2 + s.length2())));
qDebug() << y2 << (start2 + s.length2()) << (end2 - (start2 + s.length2()));
}
}
.
소스 코드가 "시작"및 "끝"만으로 Valesone과 Valuetwo간에 차이가 없을 수 없으므로 QString :: mid (in mid) 메소드 (알고있는 한 줄로 줄을 진행됨)가 있으므로동일한 시작 위치 및 동일한 길이 ( http : // qt-project 참조).org / doc / qt-4.8 / qstring.html # 중순 ).그러므로 전체 문자열이 와 같은 한 줄이되었는지 생각했습니다.
QString x = "yes start ValueOne end no start ValueTwo end ";
.
I 값이 QSTRING S="YES START"와 QSTRING S2="시작되지 않음"과 차이가있을 수 있습니다.따라서 여러 줄 문자열을 한 줄 문자열로 변환하면 해결책이고 어떻게이 작업을 수행 할 수 있습니까?아니면 더 나은 해결책이 더 나은 것입니까? 인사말
해결책
As I already mention in your other question I would prefer QRegExp. It seems to be more readable.
If your first string is the 2n
value always and your second string is 2n+1
you could use the modulo operator:
#include <QDebug>
#include <QString>
#include <QStringList>
#include <QRegExp>
int main()
{
QString x = ("yes \nstart ValueOne end \nno \nstart ValueTwo end\n"
"yes \nstart ValueThree end \nno \nstart ValueFour end ");
QStringList y1;
QStringList y2;
// create regular expression
QRegExp rx("start\\s+(.+)\\s+end\\s+", Qt::CaseInsensitive);
// don't try to get the largest match (start ValueOne ... ValueFour end)
// minimal match should be (start ValueOne end)
rx.setMinimal(true);
int pos=0;
int i=0; // counter
// look for possible matches
QString match;
while ((pos=rx.indexIn(x, pos)) != -1) {
i+=1; // increase counter for every match
match=rx.cap(1); // get first match in (.+)
// use modulo to distinguish between y1/y2
if (i % 2) {
y1 << match;
} else {
y2 << match;
}
pos+=rx.matchedLength();
}
qDebug() << "y1:" << y1;
qDebug() << "y2:" << y2;
return 0;
}
다른 팁
With something similar to the code below you could find all the strings between "start" and "end". Put the search for "start" and "end" in a loop and use the offset parameter of indexOf to continue searching for new delimiters after the first one.
int main(int argc, char *argv[]) {
QString x = ("yes /nstart ValueOne end /nno /nstart ValueTwo end ");
QString s = "start";
QString e = "end";
// Look for all the strings between "start" and "end"
for(int offset(0); offset < x.length(); )
{
// Search for "start" starts from offset
int start = x.indexOf(s, offset, Qt::CaseInsensitive);
if(start < 0){
break;
}
// Search for "end" starts from the position of "start"
int end = x.indexOf(e, start, Qt::CaseInsensitive);
if(end < 0){
break;
}
// Next search for "start" will start from the current position of "end"
offset = end;
QString y = x.mid(start + s.length(), (end - (start + s.length())));
qDebug() << y << (start + s.length()) << (end - (start + s.length()));
}
}