急急急！将insert语句中的时间，格式化
如下：
insert into performdb.dbo.ta01_578(answ,day,hour_mi,nb2,nb3,nb4,ncct,ncit,noansw,ovfl,pegs,period,rerl,seiz,serl,succ,time,trk_name,year_to_day) values(0,31,"1000",0,0,0,31.0000,31,1,0,1,60,0.0000,1,0.0000,1,"2012-05-31 10:00:00","0","2012-05-31").
转换成
insert into performdb.dbo.ta01_578(answ,day,hour_mi,nb2,nb3,nb4,ncct,ncit,noansw,ovfl,pegs,period,rerl,seiz,serl,succ,time,trk_name,year_to_day) values(0,31,"1000",0,0,0,31.0000,31,1,0,1,60,0.0000,1,0.0000,1,to_date（2012-05-31 10:00:00，‘yyyy-mm-dd，hh24：mi：ss’）,"0",to_date（2012-05-31，‘yyyy-mm-dd’）).
[解决办法]
时间串自身不用单引号？
时间都是2012年的？
[解决办法]
仅供参考
#include <stdio.h>#include <string.h>char s[256];char *p;int r,n,i;int main() { while (1) { printf("请输入一行文字(空行结束),\"%%20\"将替换为\" \",\"你懂得\"将替换为\"XXXXXX\":\n"); fgets(s,256,stdin); if ('\n'==s[0]) break; p=s; while (1) { p=strstr(p,"%20"); if (p) { memmove(p+1,p+3,strlen(p)-3+1); p[0]=' '; } else break; } p=s; while (1) { p=strstr(p,"你懂得"); if (p) { memmove(p+6,p+6,strlen(p)-6+1); for (i=0;i<6;i++) p[i]='X'; } else break; } printf("%s",s); } return 0;}//请输入一行文字(空行结束),"%20"将替换为" ","你懂得"将替换为"XXXXXX"://abcdefg%20helloworld%20something.pdf//abcdefg helloworld something.pdf//请输入一行文字(空行结束),"%20"将替换为" ","你懂得"将替换为"XXXXXX"://这是测试文字你懂得，在这个你懂的地方，就得做你懂得的事//这是测试文字XXXXX，在这个你懂的地方，就得做XXXXX的事//请输入一行文字(空行结束),"%20"将替换为" ","你懂得"将替换为"XXXXXX"://