jitsi meet源码中,在登录界面提交登录信息的代码处理 5C

登录jitsimeet,在登录界面填写用户名和密码后,压下OK键,进入jitsimeet。
如下代码为压下OK按键的代码:

    _onLogin() {
        const { _conference: conference, dispatch } = this.props;
        const { password, username } = this.state;
        const jid = toJid(username, this.props._configHosts);
        let r;

        // If there's a conference it means that the connection has succeeded,
        // but authentication is required in order to join the room.
        if (conference) {
            r = dispatch(authenticateAndUpgradeRole(jid, password, conference));
        } else {
            r = dispatch(connect(jid, password));
        }

        return r;
    }
    
<pre>
    export function authenticateAndUpgradeRole(
    id: string,
    password: string,
    conference: Object) {
return (dispatch: Dispatch, getState: Function) => {
    const { password: roomPassword }
        = getState()['features/base/conference'];
    const process
        = conference.authenticateAndUpgradeRole({
            id,
            password,
            roomPassword,

            onLoginSuccessful() {
                // When the login succeeds, the process has completed half
                // of its job (i.e. 0.5).
                return dispatch(_upgradeRoleFinished(process, 0.5));
            }
        });

    dispatch(_upgradeRoleStarted(process));
    process.then(
        /* onFulfilled */ () => dispatch(_upgradeRoleFinished(process, 1)),
        /* onRejected */ error => {
            // The lack of an error signals a cancellation.
            if (error.authenticationError || error.connectionError) {
                logger.error('authenticateAndUpgradeRole failed', error);
            }

            dispatch(_upgradeRoleFinished(process, error));
        });

    return process;
};

}


在authenticateAndUpgradeRole函数中,如下代码段

        const process
            = conference.authenticateAndUpgradeRole({
                id,
                password,
                roomPassword,

                onLoginSuccessful() {
                    // When the login succeeds, the process has completed half
                    // of its job (i.e. 0.5).
                    return dispatch(_upgradeRoleFinished(process, 0.5));
                }
            });
    

其中的conference.authenticateAndUpgradeRole应该实现的是用户名和密码的验证功能,但是我在源码工程中没有找到该接口的定义,请大神指教上述代码段的功能,以及如何能够找到提交登录信息后,是如何验证用户名和密码的?

1个回答

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Problem Description Bob and Alice are quite closely friends. They are both intersted on Researching ciper. As everybody knows , they ,of course, are experienced computer users. So they always communicate their new reaserches on ciyper by Internet. But now, unfortunetly, their country is suffering from a cruel war caused by nucleus. However the enemy state is so powerful much more than their motherland. So after several months Bob and Alice's country is controled by enemy on many areas, including network. Any communications on the network are highly suspiciously. If anybody uncareful,he may lost his life! This situation is so dangerious to computer users. So Bob's mother and Alice's father make a decision together for their childern --- forbid them to use computer again. Bob and Alice can't disobey this dicesion ,but they also don't want to pause their new research. So they also make a decison together --- to meet at somewhere personally ,and , of coure,don't let their parents konw. Now , Bob and Alice fasten the last chance on network communication to confirm the place they will meet at in the next days. The city they lived , has many roads and hotels. Of course, they can't talk just on the roads, and they also can't enter any buildings except hotels. So they decide to meet in a certain hotel in their city. As they know,the city's building pattern is very regular like a rectangle with many griddings in it. And all hotels and houses are built in lines and columns, so are roads. So they can only walk straightly by lines or columns. While it is a mountainly city, every buliding may has a different height above sea level.If you walk form a hotel with a height x at an initial speed v0 to a hotel with a height y, then the speed on this road is surposed to be v0 × 2(x-y) . We already know that Bob lives in the city's left-up corner,and Alice lives in the right-down corner. The buliding in the grid among their houses are all hotels. and the length of the road between two neighbor( four directions connected) hotel is alway a same value L, and they set up at their own initial speed at the same time. Now , they want to pick a path which will take them a fastest time to meet each other. Pitfully for them ,can you help them to commpute the minmum time to meet ? You should always remember that ,they only meet at a certain hotel ,so maybe one of them will reach a hotel and don't walk again to wait the other one to arrive. Input There will be more than a test case. every case data goes like below: Line 1 : Four space-separated integers: N, Vb, Va, L. N is the size of the rectangle of Bob and Alice's city. Va is Alice's initial speed, while Vb is Bob's initial speed. L is the length between two neighbor hotels . Line 2 ~ Line N + 1 : every line has N space-separated integers,each one represents a hotel's height. N = 0 indicates the end of input and should not be processed. Output For each case , ouput an interger in a separated line which is the time they take to meet each other.The answer must round to three decimal places. Sample Input 3 1 1 1 1 2 3 3 2 1 1 2 1 0 0 0 0 Sample Output 4.000

When Can We Meet? 的思想思路

Description The ICPC committee would like to have its meeting as soon as possible to address every little issue of the next contest. However, members of the committee are so busy maniacally developing (possibly useless) programs that it is very difficult to arrange their schedules for the meeting. So, in order to settle the meeting date, the chairperson requested every member to send back a list of convenient dates by E-mail. Your mission is to help the chairperson, who is now dedicated to other issues of the contest, by writing a program that chooses the best date from the submitted lists. Your program should find the date convenient for the most members. If there is more than one such day, the earliest is the best. Input The input has multiple data sets, each starting with a line containing the number of committee members and the quorum of the meeting. N Q Here, N, meaning the size of the committee, and Q meaning the quorum, are positive integers. N is less than 50, and, of course, Q is less than or equal to N. N lines follow, each describing convenient dates for a committee member in the following format. M Date1 Date2 ... DateM Here, M means the number of convenient dates for the member, which is an integer greater than or equal to zero. The remaining items in the line are his/her dates of convenience, which are positive integers less than 100, that is, 1 means tomorrow, 2 means the day after tomorrow, and so on. They are in ascending order without any repetition and separated by a space character. Lines have neither leading nor trailing spaces. A line containing two zeros indicates the end of the input. Output For each data set, print a single line containing the date number convenient for the largest number of committee members. If there is more than one such date, print the earliest. However, if no dates are convenient for more than or equal to the quorum number of members, print 0 instead. Sample Input 3 2 2 1 4 0 3 3 4 8 3 2 4 1 5 8 9 3 2 5 9 5 2 4 5 7 9 3 3 2 1 4 3 2 5 9 2 2 4 3 3 2 1 2 3 1 2 9 2 2 4 0 0 Sample Output 4 5 0 2

如何在Android中捕获cocos的崩溃信息

需要手机Android项目中的崩溃信息到文件中,然后上传服务器。我现在知道怎么收集Android的崩溃信息到文件中,也知道如何上传服务器。但是我的项目中有cocos的代码,我怎么能收集到cocos的崩溃信息呢? 我看过http://www.cnblogs.com/lancidie/archive/2013/04/13/3019349.html这里面的内容,想试着做出来,但是在把jni部分的代码粘进去以后,出现了很多错误,代码如下: #include "cocos2d.h" #include <jni.h> #include "platform/android/jni/JniHelper.h" #include "BookManager.h" #include "jniIF.h" #include "LeFileManager.h" #include "StatisticalData.h" #include "ConfigFile.h" #include "StorePage.h" #include "signal.h" #include "unistd.h" #include "errno.h" using namespace cocos2d; extern "C" { //尝试捕获cocos的奔溃信息 //思路是,当发生崩溃的时候,在回调函数里面调用一个我们在Activity写好的静态函数。 //在这个函数里面通过执行命令获取logcat的输出信息(输出信息里面包含了jni的崩溃地址),然后上传这个崩溃信息。 //当我们获取到崩溃信息后,可以通过arm-linux-androideabi-addr2line(具体可能不是这个名字,在android ndk里面搜索*addr2line,找到实际的程序)解析崩溃信息。 //jni的崩溃回调函数如下 //#define sigaction(X, &handler, &old_sa[X]) void android_sigaction(int signal, siginfo_t *info, void *reserved) { JniMethodInfo t; if (!t.env) { return; } if (JniHelper::getStaticMethodInfo(t, CLASS_NAME, "onNativeCrashed", "()V")) { jmethodID methodID = t.env->GetStaticMethodID(t.classID, "onNativeCrashed", "()V"); } t.env->CallStaticVoidMethod(t.classID, t.methodID); old_sa[signal].sa_handler(signal); } //通过singal的设置,当崩溃发生的时候就会调用android_sigaction函数。 void InitCrashReport() { CCLOG("InitCrashReport"); // Try to catch crashes... struct sigaction handler; memset(&handler, 0, sizeof(struct sigaction)); handler.sa_sigaction = android_sigaction; handler.sa_flags = SA_RESETHAND; #define CATCHSIG(X) sigaction(X, &handler, &old_sa[X]) CATCHSIG(SIGILL); CATCHSIG(SIGABRT); CATCHSIG(SIGBUS); CATCHSIG(SIGFPE); CATCHSIG(SIGSEGV); CATCHSIG(SIGSTKFLT); CATCHSIG(SIGPIPE); } } 错误如下: ![图片说明](https://img-ask.csdn.net/upload/201605/24/1464074102_802173.png) ![图片说明](https://img-ask.csdn.net/upload/201605/24/1464074168_822967.png) 我是做Android的,对c++的代码基本不懂,希望各位能帮个忙,谢谢大家了

When Can We Meet?

Description The ICPC committee would like to have its meeting as soon as possible to address every little issue of the next contest. However, members of the committee are so busy maniacally developing (possibly useless) programs that it is very difficult to arrange their schedules for the meeting. So, in order to settle the meeting date, the chairperson requested every member to send back a list of convenient dates by E-mail. Your mission is to help the chairperson, who is now dedicated to other issues of the contest, by writing a program that chooses the best date from the submitted lists. Your program should find the date convenient for the most members. If there is more than one such day, the earliest is the best. Input The input has multiple data sets, each starting with a line containing the number of committee members and the quorum of the meeting. N Q Here, N, meaning the size of the committee, and Q meaning the quorum, are positive integers. N is less than 50, and, of course, Q is less than or equal to N. N lines follow, each describing convenient dates for a committee member in the following format. M Date1 Date2 ... DateM Here, M means the number of convenient dates for the member, which is an integer greater than or equal to zero. The remaining items in the line are his/her dates of convenience, which are positive integers less than 100, that is, 1 means tomorrow, 2 means the day after tomorrow, and so on. They are in ascending order without any repetition and separated by a space character. Lines have neither leading nor trailing spaces. A line containing two zeros indicates the end of the input. Output For each data set, print a single line containing the date number convenient for the largest number of committee members. If there is more than one such date, print the earliest. However, if no dates are convenient for more than or equal to the quorum number of members, print 0 instead. Sample Input 3 2 2 1 4 0 3 3 4 8 3 2 4 1 5 8 9 3 2 5 9 5 2 4 5 7 9 3 3 2 1 4 3 2 5 9 2 2 4 3 3 2 1 2 3 1 2 9 2 2 4 0 0 Sample Output 4 5 0 2

Bootstrap Modal中的JQuery验证表单在验证后不会提交

<div class="post-text" itemprop="text"> <p><strong>UPDATE:</strong> To put the question into perspective I created another <a href="https://jsfiddle.net/h6onufs5/12/" rel="nofollow noreferrer">Fiddle</a> that shows the same form outside of the modal. </p> <p>When it meet the right conditions i.e you type an email address and hit the Get Started button it submits properly display the PHP page, in this case it shows a 404 error, but it does what is supposed to do, <strong>SUBMIT!</strong> </p> <p><strong>ORIGINAL PROBLEM:</strong> Now back to the problem: I would like to submit my form inside a bootstrap modal, but when I open the modal and type an email, and press the get started button: <strong>NOTHING HAPPENS</strong> </p> <p><a href="https://i.stack.imgur.com/GxfF3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GxfF3.jpg" alt="php won't submit"></a></p> <p><strong>What am I doing wrong? Is there a JavaScript solution missing to submit the form correctly or are the validation errors interfering?</strong> </p> <p>I'm a novice in JavaScript so I can't seem to figure out the coding solution if it's in fact js based. </p> <p>Please help me figure this strange modal issue, thank you!</p> <p>Fiddle: <a href="https://jsfiddle.net/o5vpt6ck/3/" rel="nofollow noreferrer">https://jsfiddle.net/o5vpt6ck/3/</a></p> <p><strong>HTML:</strong></p> <pre><code>&lt;form id="signup-form" class="cd-signin-modal__form" action="confirm.php" method="post"&gt; &lt;h3 class="bigsentence black text-center font-weight-bold"&gt;Create&lt;/h3&gt; &lt;p class="cd-signin-modal__fieldset"&gt; &lt;label class="cd-signin-modal__label cd-signin-modal__label--email cd-signin-modal__label--image-replace" for="signup-email"&gt;Enter your email address&lt;/label&gt; &lt;input class="cd-signin-modal__input cd-signin-modal__input--full-width cd-signin-modal__input--has-padding cd-signin-modal__input--has-border signupfield" id="email" type="email" name="email" placeholder="Enter your email address"&gt; &lt;/p&gt; &lt;p class="cd-signin-modal__fieldset"&gt; &lt;input class="cd-signin-modal__input cd-signin-modal__input--full-width" name="submit" type="submit" value="GET STARTED"&gt; &lt;/p&gt; &lt;/form&gt; </code></pre> <p>JS:</p> <pre><code>$("#signup-form").validate ({ // validation rules for registration formd errorClass: "error-class", validClass: "valid-class", errorElement: 'div', errorPlacement: function(error, element) { if(element.parent('.input-group').length) { error.insertAfter(element.parent()); } else { error.insertAfter(element); } }, onError : function(){ $('.input-group.error-class').find('.help-block.form-error').each(function() { $(this).closest('.form-group').addClass('error-class').append($(this)); }); }, rules: { email: {email:true, required:true} }, messages: { email: { required: "Please enter your email address", }, highlight: function(element, errorClass) { $(element).removeClass(errorClass); } } }); </code></pre> </div>

鼠标放在新闻标题上显示完整的新闻信息

在主页上显示新闻标题,如果字符大于20,就只显示前20个字,后面用省略号,如果小于20,就显示完整标题,代码如下; [code="java"]<%if(notice_rs.getString("notice_title").length()>20){ %> <a href=/meet/it/noticeDetail.jsp?id=<%=notice_rs.getInt("notice_id")%> target="_blank"> <%=notice_rs.getString("notice_title").substring(0,20)+"..."%></a> <%}else{ %> <a href=/meet/it/noticeDetail.jsp?id=<%=notice_rs.getInt("notice_id")%> target="_blank"> <%=notice_rs.getString("notice_title")%></a><%} %>[/code] 我的问题是:我现在想实现鼠标放在省略了字符的标题上可以显示完整的标题信息,这个应该用什么实现,怎么实现呢?谢谢

freeswitch对接mrcp,想用lua脚本把结果取出来然后判断结果

也是从网上找的例子,但是我想解析这个xml,判断如果是yes,播放1.mp3,no播放2.MP3,该怎么做呢,有大神知道吗,我不会lua 代码如下: ``` tryagain = 1 while (tryagain == 1) do session:execute("play_and_detect_speech",welcome .. "detect:unimrcp {start-input-timers=false,no-input-timeout=" .. no_input_timeout .. ",recognition-timeout=" .. recognition_timeout .. "}" .. grammar) xl = session:getVariable('detect_speech_result') if (xl == nil) then freeswitch.consoleLog("CRIT","电话挂断\n") tryagain = 0 else freeswitch.consoleLog("CRIT","Result is '" .. xl .. "'\n") 这里取结果,判断 session:execute("playback", huifu2) tryagain = 0 end end ``` 得到的xml如下 ``` <result> <interpretation grammar="" confidence="1.0"> <instance> <callId>hostname-4418_35629.1554048033766857898-841227</callId> <logId>EDC18579-98CC-4649-9E60-6D5A1C71FEE0_-1</logId> <rolecategory>CLIENT</rolecategory> <extJson> <snStartTime>00:06.559</snStartTime> <snStopTime>00:07.519</snStopTime> <speed>1.04</speed></extJson> <categotyId>TXT</categotyId> </instance> <input mode="speech">yes,</input> </interpretation> </result> ```

在Golang中处理多个错误的最佳做法

<div class="post-text" itemprop="text"> <p>I am writing some code that parses a YAML file and validates the data inside. Each parser returns an error if the value found does not meet the requirements. Right now I am creating an array of errors and setting the return of each call to a different index in the array. My current implementation is working, but it seems wrong and I want to see if there is a better way to accomplish this.</p> <p>Here is an example</p> <pre><code>func createStruct(yamlMap map[interface{}]interface{}) (myStruct, error) { errs := make([]error, 6) s := myStruct{} s.Name, errs[0] = getString(yamlMap, "name", true) s.Tag, errs[1] = getIntValidRange(yamlMap, "tag", 1, 4094, true) s.TaggedPorts, errs[2] = getStringPortList(yamlMap, "tagged_ports", true) s.UntaggedPorts, errs[3] = getStringPortList(yamlMap, "untagged_ports", true) s.IP, errs[4] = getIPString(yamlMap, "ip", true) s.Netmask, errs[5] = getIPString(yamlMap, "netmask", true) return s, structCreateErrorChecker(errs) } </code></pre> <p>The reason I don't handle each error after each function, is because I want to try to parse everything first and then collect all errors and log them. That is what structCreateErrorChecker() does.</p> <p>Here is the YAML I am trying to parse (it's come up in the comments). When I say dynamic I mean that there can be any number of these controlling_bridge sections and each vlan section can have any number of vlans.</p> <pre><code>controlling_bridge_1: ip: "1.1.1.1" ports: ["1","2"] vlans: vlan01: name: "vlan1" tag: 1001 ports: ["1"] ip: "2.2.2.2" vlan02: name: "vlan02" tag: 1002 ports: ["3", "4"] ip: "3.3.3.1" controlling_bridge_2: ip: "1.1.1.1" ports: ["1","2"] vlans: vlan01: name: "vlan1" tag: 1001 ports: ["1"] ip: "2.2.2.2" vlan02: name: "vlan02" tag: 1002 ports: ["3", "4"] ip: "3.3.3.1" </code></pre> </div>

java超难编程题 挑战一下吧 大神帮吗发代码

描述 Alice writes an English composition with a length of N characters. However, her teacher requires that M illegal pairs of characters cannot be adjacent, and if 'ab' cannot be adjacent, 'ba' cannot be adjacent either. In order to meet the requirements, Alice needs to delete some characters. Please work out the minimum number of characters that need to be deleted. 输入 The first line contains the length of the composition N. The second line contains N characters, which make up the composition. Each character belongs to 'a'..'z'. The third line contains the number of illegal pairs M. Each of the next M lines contains two characters ch1 and ch2,which cannot be adjacent. For 20% of the data: 1 ≤ N ≤ 10 For 50% of the data: 1 ≤ N ≤ 1000 For 100% of the data: 1 ≤ N ≤ 100000, M ≤ 200. 输出 One line with an integer indicating the minimum number of characters that need to be deleted. ![图片说明](https://img-ask.csdn.net/upload/201610/10/1476100153_91801.png) 样例提示 Delete 'a' and 'd'.

Simple Scheduling Problem 代码编辑的思想

Problem Description Team YouAreSoStrong meet a very boring problem. They have to schedule n jobs on m machines. Each job needs exactly one unit time on one machine. At first, YouAreSoStrong thought ceil(n/m) units of time is enough. Soon, they find out there exist some ordering constraints between jobs. This is reasonable. For example, you have to complete "hand in hand" before you can "hug". And only after you've done "hug", can you "kiss". It's guaranteed that following 3 conditions are satisfied. 1. one job can only be relied by at most one job. 2. one job may rely more than one job. 3.There is no circular dependency. Such as a relies on b, and b relies on a. YouAreSoStrong thought for a long time, but do not have a clue. Can you help them? Input First line contains one integer T, indicates the number of test cases. For each case, the first line contains 3 integers : n, e, m, denotes number of jobs, number of ordering constraints, number of machines separately. Each line from 2 to (e+1)th contains one pair of integers (a,b), which stands for job a must be done before job b can start. T≤10. 0 < n, e , m≤100 000. Output Print one integer on one line for each case, which is the minimum units of time YouAreSoStrong can schedule all the jobs. Sample Input 2 4 2 1 1 3 2 3 4 2 2 1 3 2 3 Sample Output 4 2

上班一个月,后悔当初着急入职的选择了

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我说我不会算法,阿里把我挂了。

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抖音上很火的时钟效果

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记录下入职中软一个月(外包华为)

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手机经常收到"回复TD退订",回还是不回?今天总算是弄清楚了

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当HR压你价,说你只值7K,你该怎么回答?

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