我有解析 2 个命令行参数并打印参数的 C++ 代码。其中一个参数是谷歌搜索的 URL。我把代码贴在下面
int main(int argc, char* argv[])
{
std::cout << argv[1] << argv[2] << "\n";
}
当我编译后通过命令行传递 URL 时，如下所示，
./demo 1 https://www.google.co.in/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&client=ubuntu&q=size%20of%20unsigned%20char%20array%20c%2B%2B&oq=length%20of%20unsigned%20char*%20arra&aqs=chrome.4.69i57j0l5.13353j0j7
我得到的输出是，
[1] 8680
[2] 8681
[3] 8682
[4] 8683
[5] 8684
[6] 8685
[7] 8686
[2] Done ion=1
[3] Done espv=2
[4] Done ie=UTF-8
[6]- Done q=size%20of%20unsigned%20char%20array%20c%2B%2B
看起来字符串内部有一些 split 。有什么方法可以检索整个字符串吗？
提前谢谢你。
最佳答案
你必须引用它。否则 & 会被 shell 解释为“在后台调用 & 左边的内容”。
我有幸用 echo 替换了您的程序。
好:
$ echo "https://www.google.co.in/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&client=ubuntu&q=size%20of%20unsigned%20char%20array%20c%2B%2B&oq=length%20of%20unsigned%20char*%20arra&aqs=chrome.4.69i57j0l5.13353j0j7"
https://www.google.co.in/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&client=ubuntu&q=size%20of%20unsigned%20char%20array%20c%2B%2B&oq=length%20of%20unsigned%20char*%20arra&aqs=chrome.4.69i57j0l5.13353j0j7
差:
$ echo https://www.google.co.in/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&client=ubuntu&q=size%20of%20unsigned%20char%20array%20c%2B%2B&oq=length%20of%20unsigned%20char*%20arra&aqs=chrome.4.69i57j0l5.13353j0j7
[1] 21705
[2] 21706
https://www.google.co.in/search?sourceid=chrome-psyapi2
[3] 21707
[4] 21708
[5] 21709
[6] 21710
[7] 21711
[1] Done echo https://www.google.co.in/search?sourceid=chrome-psyapi2
[2] Done ion=1
[3] Done espv=2
[4] Done ie=UTF-8
[5] Done client=ubuntu
[6]- Done q=size%20of%20unsigned%20char%20array%20c%2B%2B
[7]+ Done oq=length%20of%20unsigned%20char*%20arra
关于c++ - 通过命令行传递 URL(C++)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36354743/