| 发表于:2007-08-28 02:34:48 楼主 |
192.168.58.60:135 192.168.58.50:135 192.168.58.40:135 192.168.58.30:135 192.168.58.20:135 192.168.58.10:135 如何将字符串中的:135去掉?也就是说如何将:包括:后面的东西去掉呢? 我是这样实现的, textzl = " " dim q as integer, w as string, e() as string e() = split(ip_jieguo, vbcrlf) for q = 0 to ubound(e) w = e(q) if w = " " then exit for else r = left(w, instrrev(w, ": ")) + vbcrlf r = replace(r, ": ", " ") textzl = textzl + r ip_jieguo = textzl end if next q 但结果是点一次,确实是实现了去掉:135了,但在点一次就都变回车了... textzl = " " 我觉得不是这个原因,因为我是用两个text来实现的,我知道我的代码很垃圾,如果有高手会的请告诉下小弟了,先谢谢了.. 还有个问题,就是处理数据的, 有一个文本文件里有着这样的内容.. ------------------------------------------------------------------------------- performing time: 8/26/2007 4:5:23 --> normal scan: about to scan 65536 ip using 512 threads lastip scanned: 218.25.20.23:135 scan 5142 ips complete in 0 hours 0 minutes 33 seconds. found 0 hosts ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- performing time: 8/27/2007 5:22:34 --> normal scan: about to scan 65535 ports using 512 thread 127.0.0.1 80 open 127.0.0.1 110 open 127.0.0.1 135 open 127.0.0.1 443 open lastip scanned: 127.0.0.1:518 scan 127.0.0.1 complete in 0 hours 0 minutes 3 seconds. found 4 open ports ------------------------------------------------------------------------------- 我只想要其中的 127.0.0.1 80 open 127.0.0.1 110 open 127.0.0.1 135 open 127.0.0.1 443 open 这个部分,然后处理成 127.0.0.1:80 127.0.0.1:110 127.0.0.1:135 127.0.0.1:443 如上这个样子,我刚刚接触vb很多东西都得学习,请大家帮帮忙。 |
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