q=%26%2340%E7%BE%8E%26%2341%E6%89%98%E9%A9%AC%E6%96%AF%C2%B7%E6%96%AF%E5%9D%8E%E4%BC%A6%26%2340Thomas%2BM.%2BScanlon%26%2341&searchType=standard&isFacet=true&view=standard&searchWay=author&rows=10&sortWay=score&sortOrder=desc&searchWay0=marc&logical0=AND
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(美)托马斯·斯坎伦(Thomas+M.+Scanlon)