相关疑难解决方法(0)

"PKIX路径构建失败"和"无法找到请求目标的有效证书路径"

我正在尝试使用twitter4j库为我的java项目发送推文.在我第一次运行时,我收到了关于证书sun.security.validator.ValidatorException和错误的错误sun.security.provider.certpath.SunCertPathBuilderException.然后我添加了twitter证书:

C:\Program Files\Java\jdk1.7.0_45\jre\lib\security>keytool -importcert -trustcacerts -file PathToCert -alias ca_alias -keystore "C:\Program Files\Java\jdk1.7.0_45\jre\lib\security\cacerts"
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但没有成功.这是获得twitters的过程:

public static void main(String[] args) throws TwitterException {
    ConfigurationBuilder cb = new ConfigurationBuilder();
    cb.setDebugEnabled(true)
        .setOAuthConsumerKey("myConsumerKey")
        .setOAuthConsumerSecret("myConsumerSecret")
        .setOAuthAccessToken("myAccessToken")
        .setOAuthAccessTokenSecret("myAccessTokenSecret");

    TwitterFactory tf = new TwitterFactory(cb.build());
    Twitter twitter = tf.getInstance();

    try {
        Query query = new Query("iphone");
        QueryResult result;
        result = twitter.search(query);
        System.out.println("Total amount of tweets: " + result.getTweets().size());
        List<Status> tweets = result.getTweets();

        for (Status tweet : tweets) {
            System.out.println("@" + tweet.getUser().getScreenName() + " …
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java jsp servlets twitter4j

350
推荐指数
18
解决办法
99万
查看次数

javax.naming.CommunicationException:简单绑定失败

当尝试使用简单的LDAP应用程序连接到LDAP服务器时,我收到一条错误,其中显示"简单绑定失败".我假设这与某种BIND有关.我在其中一个属性文件中有一个绑定属性用于不同的应用程序,但我不知道如何将该属性传递给该程序.

我是否需要添加更多详细信息?

import javax.naming.directory.*;   
import javax.naming.*;   
import java.util.Vector;   
import java.util.Enumeration;   
import java.util.Properties;   
public class SearchLDAP {   
    public static void main(String[] args) {   
        String base = "";   

        String filter = "(objectclass=*)";   

        Properties env = new Properties();   

        env.put(DirContext.INITIAL_CONTEXT_FACTORY,"com.sun.jndi.ldap.LdapCtxFactory");   
        env.put(DirContext.PROVIDER_URL,"ldaps://misguided.com.au:343"); 

        try {   

            System.out.println("11");
            DirContext dc = new InitialDirContext(env);
            System.out.println("22");

            SearchControls sc = new SearchControls();   
            sc.setSearchScope(SearchControls.OBJECT_SCOPE);   
            NamingEnumeration ne = null;   

            ne = dc.search(base, filter, sc);   

            while (ne.hasMore()) {   
                SearchResult sr = (SearchResult) ne.next();   
                System.out.println(sr.toString()+"\n");   
            }   
            dc.close();   
        } catch (NamingException nex) {   
            System.err.println("Error: " …
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java jndi ldap

16
推荐指数
1
解决办法
7万
查看次数

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java ×2

jndi ×1

jsp ×1

ldap ×1

servlets ×1

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