Java crawler4j的实现
我正在尝试运行crawler4j的基本形式,如图所示。我通过定义rootFolder和numberOfCrawlers修改了前几行,如下所示:Java crawler4j的实现,java,string,web-crawler,Java,String,Web Crawler,我正在尝试运行crawler4j的基本形式,如图所示。我通过定义rootFolder和numberOfCrawlers修改了前几行,如下所示: public class BasicCrawlController { public static void main(String[] args) throws Exception { if (args.length != 2) { System.out.println("Nee
public class BasicCrawlController {
public static void main(String[] args) throws Exception {
if (args.length != 2) {
System.out.println("Needed parameters: ");
System.out.println("\t rootFolder (it will contain intermediate crawl data)");
System.out.println("\t numberOfCralwers (number of concurrent threads)");
return;
}
/*
* crawlStorageFolder is a folder where intermediate crawl data is
* stored.
*/
String crawlStorageFolder = args[0];
args[0] = "/data/crawl/root";
/*
* numberOfCrawlers shows the number of concurrent threads that should
* be initiated for crawling.
*/
int numberOfCrawlers = Integer.parseInt(args[1]);
args[1] = "7";
CrawlConfig config = new CrawlConfig();
config.setCrawlStorageFolder(crawlStorageFolder);
无论我如何定义它,我仍然收到错误
Needed parameters:
rootFolder (it will contain intermediate crawl data)
numberOfCralwers (number of concurrent threads)
我认为我需要“在运行配置中设置参数”窗口,但我不知道这意味着什么。如何正确配置此基本爬虫程序以使其启动并运行?使用javac关键字编译程序后,需要通过键入以下内容来运行它: java BasicCrawler控制器“arg1”“arg2” 错误告诉您在运行程序时没有指定arg[0]或arg[1]。另外,在您已经收到爬虫数量参数之后,这个“args[1]=“7”;”是什么意思
对于您试图执行的操作,请删除前5行,因为您仍在尝试使用硬编码值。然后将crawlForStorage字符串设置为目录路径,将numberOfCrawlers设置为7。这样就不必指定命令行参数。如果要使用命令行参数,请去掉上面的硬编码值,并在CL中指定它们,我必须对目录进行硬编码,然后一起去掉异常捕获。谢谢