Java 8:获取选择性请求参数

Question

我试图找到所有MultiFileId_从a 开始的请求参数request

我正在迭代所有request键映射和checkig是否每个键都开始,MultiFileId_如果匹配则将其值添加到列表中

    List<String> uploadedFileIds = new ArrayList<String>();

    for (Object p : request.getParameterMap().keySet()) {
        if(p.toString().startsWith("MultiFileId_")) {
            String uploadedFileId = request.getParameter(p.toString());
            uploadedFileIds.add(uploadedFileId);
        }
    }

在Java 8中有一种简单的方法可以实现这一点吗？

Answer 1

如果您确定每个参数只有一个值:

import javax.servlet.http.HttpServletRequest;
import java.util.Collections;
import java.util.List;
import java.util.stream.Collectors;

public class Example1 {
    public void example(HttpServletRequest request) {

        List<String> uploadedFileIds = Collections
                .list(request.getParameterNames())
                .stream()
                .filter(parameterName -> parameterName.startsWith("MultiFileId_"))
                .map(request::getParameter)
                .collect(Collectors.toList());
    }
}

如果你不知道参数是否有多个值(即request.getParameterValues("MultiFileId_XXX")返回String[]与length > 1),你可以使用:

import javax.servlet.http.HttpServletRequest;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.stream.Collectors;

public class Example2 {
    public void example(HttpServletRequest request) {

        List<String> uploadedFileIds = Collections
                .list(request.getParameterNames())
                .stream()
                .filter(parameterName -> parameterName.startsWith("MultiFileId_"))
                .flatMap(parameterName -> Arrays.stream(
                        request.getParameterValues(parameterName)))
                .collect(Collectors.toList());
    }
}

Answer 2

这是另一种方式:

List<String> uploadedFileIds = request.getParameterMap().entrySet()
    .stream()
    .filter(e -> e.getKey().startsWith("MultiFileId_"))
    .flatMap(e -> Arrays.stream(e.getValue()))
    .collect(Collectors.toList());