我正在研究一些 Java Servlets,基本上我正在将sql查询的结果输出到表中.我在html代码中有一些表格的基本格式,但我也想链接一个css文件.
每当我链接样式表(即使是空白样式表,或者与表中的html标记具有相同属性的样式表)时,它只会破坏表中的任何格式,并将结果作为一个连续列表输出.
任何建议都将是一个很大的帮助.
这是我的servlet代码:
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String category = request.getParameter("categoryname");
AlbumDAO albumData = new AlbumDAO();
ArrayList<AlbumBean> albums = albumData.findFromCategory(category);
PrintWriter out = response.getWriter();
response.setContentType("text/html");
String title = category + " albums";
String stylesheet = "<link rel=\"stylesheet\" type=\"text/css\" href=\"/CSS/stylesheet.css>";
out.println("<!DOCTYPE html><html>");
out.println("<head>");
out.println("<title>" + title + "</title>");
out.println(stylesheet);
out.println("</head>");
out.println("<body>");
out.println("<Center><H1>" + category + " albums</Center>");
out.println("<table border=\"1\" cellspacing=\"5\" cellpadding=\"5\">"
+ "<tr><th>ID</th><th>Artist</th><th>Title</th><th>Image Name</th><th>Tracks</th><th>Price</th><th>In Stock</th></tr>");
for (AlbumBean a : albums){
out.println("<tr><td> "+ a.getRecording_id() + "</td>");
out.println("<td>" + a.getArtist_name() + "</td>");
out.println("<td> " + a.getTitle() + "</td>");
out.println("<td> " + a.getCategory() + "</td>");
out.println("<td> " + a.getImage_name() + "</td>");
out.println("<td> " + a.getPrice() + "</td>");
out.println("<td> " + a.getStock_count() + "</td>");
out.println("</tr>");
}
out.println("</table>");
out.println("</body>");
out.println("<footer><a href = \"index.html\"> let's go home</a></footer>");
out.println("</html>");
}
最佳答案 看起来您没有关闭链接语句中的引号
String stylesheet = "<link rel=\"stylesheet\" type=\"text/css\" href=\"/CSS/stylesheet.css\">";