2015-12-16 96 views
-1
#include <QGuiApplication> 
#include <QQmlApplicationEngine> 
#include <QProcess> 
#include <QFile> 
#include <QDebug> 
#include <stdio.h> 

int main(int argc, char *argv[]) 
{ 
    QGuiApplication app(argc, argv); 

    FILE* file1 = popen ("make", "r"); 

    char buff[5122]; 

    while(fgets(buff, sizeof(buff), file1)!=NULL) 
    { 
     qDebug() << "from here: " << buff; 
    } 


    QQmlApplicationEngine engine; 
    engine.load(QUrl(QStringLiteral ("qrc:/main.qml"))); 
    return app.exec(); 
} 

的輸出隨着make命令的輸出是:POPEN不捕獲所有的命令

QML debugging is enabled. Only use this in a safe environment.
from here: make: Nothing to be done for first'.`

隨着ping命令的輸出是:

QML debugging is enabled. Only use this in a safe environment. 
Usage: ping [-aAbBdDfhLnOqrRUvV] [-c count] [-i interval] [-I interface] 
      [-m mark] [-M pmtudisc_option] [-l preload] [-p pattern] [-Q tos] 
      [-s packetsize] [-S sndbuf] [-t ttl] [-T timestamp_option] 
      [-w deadline] [-W timeout] [hop1 ...] destination 

正如您所看到的,通過make命令,輸出被捕獲並顯示qDebug但是,ping並非如此。

無論它是一個錯誤還是什麼,我希望每個輸出都可以通過我的程序通過qDebug被捕獲和顯示。

我現在應該做什麼?

回答

5

在您的代碼:

FILE* file1 = popen ("make", "r"); 

,你可以使用任何外殼命令是在機器上有用。

FILE* file1 = popen ("make 2>&1", "r"); 

延伸閱讀:

雖然在技術上是可以打開多個管道的子進程,它比單行調用popen要複雜得多: