java.lang.IllegalArgumentException: Illegal character in query at index ...解决办法
2013-04-16 19:32
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url转换问题
String url = baseUrl + "?" + "name=" + name + "&age=" + age;
url = url.replaceAll("&", "%26");
url = url.replaceAll(" ", "%20");
另外：
？ %3F
& %26
| %124
= %3D
# %23
/ %2F
+ %2B
% %25
空格 %20
String url = baseUrl + "?" + "name=" + name + "&age=" + age;
url = url.replaceAll("&", "%26");
url = url.replaceAll(" ", "%20");
另外：
？ %3F
& %26
| %124
= %3D
# %23
/ %2F
+ %2B
% %25
空格 %20
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