Java:如何以优雅的方式将字符串数组从索引0压缩到-2?
我正在将一些订单页面文件(html)提取到JavaJava:如何以优雅的方式将字符串数组从索引0压缩到-2?,java,lambda,Java,Lambda,我正在将一些订单页面文件(html)提取到Javaorder类中,下面是这段代码: List<Order> orders = Files.walk(Paths.get(path, "orders", "html")) .map(Path::toFile) .map(this::readFileToString) .map(content -> { Order order = new Order();
order
类中,下面是这段代码:
List<Order> orders = Files.walk(Paths.get(path, "orders", "html"))
.map(Path::toFile)
.map(this::readFileToString)
.map(content -> {
Order order = new Order();
evaluateXPath("//*[@id='page']/div[2]/div[1]/div[3]/div[2]/span[2]/text()", content)
.ifPresent(x -> {
String[] results = x.split(" ");
if (results.length >= 3) {
StringBuilder stringBuilder = new StringBuilder();
for (int i = 0; i < results.length - 2; i++) {
stringBuilder.append(results[i]);
}
order.setConsignee(stringBuilder.toString());
order.setPhoneNumber(results[results.length - 2]);
order.setAddress(results[results.length - 1]);
}
});
return order;
}).collect(Collectors.toList());
//a exception free wrapper method for FileUtils.readFileToString
private String readFileToString(File file) {
try {
return FileUtils.readFileToString(file);
} catch (IOException e) {
LOGGER.error("read " + file + " failed.");
return "";
}
}
private Optional<String> evaluateXPath(String xpath, String content) {
//a mysterious implementation of evaluateXPath
}
List orders=Files.walk(path.get(path,“orders”,“html”))
.map(路径::toFile)
.map(此::readFileToString)
.map(内容->{
订单=新订单();
evaluateXPath(“//*[@id='page']/div[2]/div[1]/div[3]/div[2]/span[2]/text()”,内容)
.如果存在(x->{
字符串[]结果=x.split(“”);
如果(results.length>=3){
StringBuilder StringBuilder=新的StringBuilder();
对于(int i=0;i
我的问题是如何将这一部分(我认为它非常冗长)重写为一个更整洁的实现
StringBuilder stringBuilder = new StringBuilder();
for (int i = 0; i < results.length - 2; i++) {
stringBuilder.append(results[i]);
}
order.setConsignee(stringBuilder.toString());
StringBuilder StringBuilder=新建StringBuilder();
对于(int i=0;i
欢迎任何建议,即使是关于我代码的另一部分。您可以改变它:
Stream.of(results).limit(results.length-2).collect(Collectors.joining())
if (results.length >= 3) {
Deque<String> resultList = new ArrayDeque<>(Arrays.asList(results));
order.setAddress(resultList.removeLast());
order.setPhoneNumber(resultList.removeLast());
order.setConsignee(String.join("", resultList));
}
if(results.length>=3){
Deque resultList=newarraydeque(Arrays.asList(results));
setAddress(resultList.removeLast());
order.setPhoneNumber(resultList.removeLast());
set收货人(String.join(“,resultList));
}
您可以扭转局面:
if (results.length >= 3) {
Deque<String> resultList = new ArrayDeque<>(Arrays.asList(results));
order.setAddress(resultList.removeLast());
order.setPhoneNumber(resultList.removeLast());
order.setConsignee(String.join("", resultList));
}
if(results.length>=3){
Deque resultList=newarraydeque(Arrays.asList(results));
setAddress(resultList.removeLast());
order.setPhoneNumber(resultList.removeLast());
set收货人(String.join(“,resultList));
}
这真的很好。是IntStream.range(0,results.length-1).mapToObj(i->results[i]).collect(连接(“”)代码>“整洁”?是String.join(“,Arrays.asList(results).subList(0,results.length-1))
,“neater”?但无需在连接中指定”
。我也应该这样做。这真的很好。是IntStream.range(0,results.length-1).mapToObj(i->results[i]).collect(连接(“”)代码>“整洁”?是String.join(“,Arrays.asList(results).subList(0,results.length-1))
,“neater”?但无需在连接中指定”
。我们也应该这样做。