为什么不使用我的脚本 datatochart.push(data); 。请帮助我吗?
function function_name(){ $ .ajax({ 输入:“ POST”, 网址:“ / data / api / json.php?cmd = wipline”, contentType:“ application / json; charset = utf-8”, dataType:“ json”, 成功:功能(数据){ datatochart.push(data); }, 错误:function(errMsg){ 警报(errMsg); } }); } function_name();
int main()
{
int qid=msgget(IPC_PRIVATE, IPC_EXCL|IPC_CREAT|0600);
cout<<"\nThe message queue was created succesfully!\n";
cout<<"\nIts queue ID is "<<qid<<"\n";
pid_t cpid;
cpid=fork();
if(cpid==0)
{
cout<<execl("/receiver","receiver",(char *)0);
cout<<"The receiver child has process ID: "<<getpid()<<"\n";
exit(0);
}
cpid=fork();
if(cpid==0)
{
cout<<execl("/sender","sender",(char *)0);
cout<<"The sender child has process ID: "<<getpid()<<"\n";
exit(0);
}
while(wait(NULL) != -1);
}