2013-03-12 138 views
1

我在windows中使用wget,它对基本的http请求工作得很好,但是当我运行下面的命令时,基本上这是snom手机上的一个动作URL。它不会运行它并在链接下给我提示信息。有谁知道如何解决这个问题?在windows的批处理文件中运行wget命令行

动作网址:

http://Phone-IP-Address/dummy.htm?settings=save&setting_server=http://WEB-Server-IP/settings.xml&store_settings&=save 

结果

syswgetrc = C:\Program Files (x86)\GnuWin32/etc/wgetrc 
--2013-03-12 10:10:53-- 
Connecting to `192.168.1.100:80`... connected. 
HTTP request sent, awaiting response... 200 Ok 
Length: 0 [text/html] 
Saving to: `[email protected]=save' 

[ <=>         ] 0   --.-K/s in 0s 

2013-03-12 10:10:53 (0.00 B/s) - `[email protected]=save' saved [0/0] 

'setting_server' is not recognized as an internal or external command, operable program or batch file. 
'store_settings' is not recognized as an internal or external command, operable program or batch file. 

回答

0

&字符in Windows是一个特殊字符告诉壳在一个行到执行多条命令。所以,你必须附上网址在双引号:

"http://Phone-IP-Address/dummy.htm?settings=save&setting_server=http://WEB-Server-IP/settings.xml&store_settings&=save" 

这种方式,外壳不会尝试分析它作为命令的一部分,但将把它作为一个参数,字符串的单个斑点,因此&字符不被视为命令分隔符太...

它在Linux too特殊的意义:

  • 一个人会异步运行的命令,在后台
  • &&意思是逻辑AND
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