使用 URI 构建器构建 URL

Question

编码URL-

https://spreadsheets.google.com/tq?tqx=out:JSON&tq=select%20*%20where%20((A%20%3D%20201))&key=1424bS7kU8nJbHdu4QdoAFdIdWDSnmEnj2NqfMb6rPTU

解码URL-

https://spreadsheets.google.com/tq?tqx=out:JSON&tq=select * where ((A= 201))&key=1424bS7kU8nJbHdu4QdoAFdIdWDSnmEnj2NqfMb6rPTU

在 URI 构建器中 URL 我们应该使用编码还是解码？

此外，给定的代码无效 :-

String xyz="1424bS7kU8nJbHdu4QdoAFdIdWDSnmEnj2NqfMb6rPTU";
    String X="201";


    final String STUDENT_RECORD_URL="https://spreadsheets.google.com/tq?tqx=out:JSON&tq=select*where";
    final String Frag1 = "A";
    final String Frag2 =  "key";
    Builder builder = Uri.parse(STUDENT_RECORD_URL).buildUpon();
    builder.appendQueryParameter(Frag1,X);
    builder.appendQueryParameter(Frag2,xyz);
    String finalUri=builder.build().toString();

Answer 1

我认为您误解了查询参数。它们都应该是您 url 中的键值对。对于您的 url https://spreadsheets.google.com/tq?tqx=out:JSON&tq=select * where ((A= 201))&key=1424bS7kU8nJbHdu4QdoAFdIdWDSnmEnj2NqfMb6rPTU，构建它看起来像：

final String STUDENT_RECORD_URL="https://spreadsheets.google.com/tq";

Builder builder = Uri.parse(STUDENT_RECORD_URL).buildUpon();
builder.appendQueryParameter("txq", "out:JSON");
builder.appendQueryParameter("tq", "select * where ((A= 201))");
builder.appendQueryParameter("key", "1424bS7kU8nJbHdu4QdoAFdIdWDSnmEnj2NqfMb6rPTU");

String finalUri = builder.build().toString();

appendQueryParameter 方法将自动 url 编码您的参数。

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