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6 answers

Windows 2000 does not have the discovery feature like XP. You need to use the software that came with your wireless card. If the card is built-in (like on a laptop), then it's probably pre-installed. If the software is missing, you need to download the software for the card. to see if it's installed, goto your start>programs> and browse for any software that maybe for a wireless card , such as Microsoft Wireless, Belkin Wireless, Linksys Wireless, D-Link Wireless, etc...and click the main application icon. It will start a program or put an icon by the clock, bottom right of your screen (icons look different for each model). And you can double click that icon, then you can finally select a "view available networks" type of option. If you don't know where to ge the software, goto the manufacturers site and search for your model number.

2007-03-02 05:00:54 · answer #1 · answered by Oguz1 3 · 0 0

The ZCFG service is not available in Windows 2000 as it is in XP, you would need to use the wireless card utility that is associated with the wireless card

2007-03-02 05:52:25 · answer #2 · answered by Anonymous · 0 0

Start > Control Panel > Newtwork Connections > Right-click in the Wireless Network adapter and choose 'View Available..."

Make sure the device is not disabled.... if it is , enable it.

2007-03-02 04:57:32 · answer #3 · answered by blndchik 5 · 0 0

There might want to were a setup disk that got here with the cardboard. set up it there should be an app fairly for the prompt card, which will allow you to see attainable networks. solid success

2016-11-27 00:18:45 · answer #4 · answered by ? 4 · 0 0

In all versions of windows since 98, it's the icon in the taskbar, if not, you need to install drivers.

2007-03-02 04:54:12 · answer #5 · answered by Adrian 2 · 0 0

http://www.microsoft.com/technet/community/columns/cableguy/cg0703.mspx
http://www.utexas.edu/its/wireless/install/config_winall.html

Try that if not let me know ill send you the microsoft tech net solution

2007-03-02 04:56:50 · answer #6 · answered by Anonymous · 0 0

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