Group all from and to messages per user

Group all from and to messages per user

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Group all from and to messages per user
Tag : mysql , By : Jesse
Date : November 24 2020, 01:01 AM

help you fix your problem I was wondering if someone could help me out. I am working on a Private Message center which will make users able to send PM's to each other through front end. The inboxes are working perfectly fine, however I would like to add a page where you are able to click on a users name and then you can see all messages send to and from with this user and the current logged in user. , Try this:
SELECT t1.id, t1.user_nicename, t2.id, t2.from_user, t2.message_title 
FROM table1 t1
INNER JOIN table2 t2 ON t1.id IN (t2.from_user, t2.to_user)
ORDER BY t1.id;
SELECT t1.id, t1.user_nicename, t2.id, t2.from_user, t2.message_title 
FROM table1 t1
INNER JOIN table2 t2 ON t1.id IN (t2.from_user, t2.to_user)
WHERE t1.id = 1;

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Display latest messages from messages table, group by user

Tag : ruby-on-rails , By : Anna
Date : March 29 2020, 07:55 AM
may help you . I'm trying to create an inbox for messaging between users. , This should be rather efficient:
SELECT u.name, sub.*
          m.message_from AS user_id
        , m.message AS last_message
   FROM   users    u
   JOIN   messages m ON m.message_to = u.id
   WHERE  u.name = 'Paul'   -- must be unique
   ORDER  BY 1, m.id DESC
   ) sub
JOIN  users u ON sub.user_id = u.id;

Group messages if the last was from the same user

Tag : php , By : Naveen
Date : March 29 2020, 07:55 AM
like below fixes the issue I have a chat and the messages are printed more or less like this: , A little example to get you started:

$messages = array(
        "From" => "A",
        "To" => "B",
        "Msg" => "A first message"
        "From" => "A",
        "To" => "B",
        "Msg" => "A second message"
        "From" => "B",
        "To" => "A",
        "Msg" => "B first message"

$previousSender = null;
foreach($messages as $message) {
    if($previousSender == null || $previousSender != $message["From"])
        printf("%s: %s", $message["From"], $message["Msg"]);
        printf("%s", $message["Msg"]);  

    $previousSender = $message["From"];


Group messages from json file by user id

Tag : java , By : tjh0001
Date : March 29 2020, 07:55 AM
I wish this help you First off, the JSON posted is invalid. You need to add enclosing braces and remove the last comma in the user section. So change
   "created_at":"Thu Apr 30 10:47:49 +0000 2015",
      "user_date":"Thu May 17 10:47:49 +0000 2010",
    "created_at": "Thu Apr 30 10:47:49 +0000 2015",
    "id": 593728455901990900,
    "user": {
        "id": 12,
        "name": "GiGi",
        "user_date": "Thu May 17 10:47:49 +0000 2010"
        Map<Long, List<Message>> userGroup = new HashMap<>();
        for (Message message : messages) {
            List<Message> userMessages = userGroup.get(message.author.getId());
            if (userMessages == null) {
                userMessages = new ArrayList<>();
                userGroup.put(message.author.getId(), userMessages);
        for (Entry<Long, List<Message>> entry : userGroup.entrySet()) {
            for (Message message : entry.getValue()) {
                System.out.println(entry.getKey() + " : " + message.date);
public int hashCode() {
    int hash = 7;
    hash = 23 * hash + (int) (this.id ^ (this.id >>> 32));
    return hash;

public boolean equals(Object obj) {
    if (obj == null) {
        return false;
    if (getClass() != obj.getClass()) {
        return false;
    final Author other = (Author) obj;
    if (this.id != other.id) {
        return false;
    return true;
Map<Author, List<Message>> userGroup = new HashMap<>();
for (Message message : messages) {
    List<Message> userMessages = userGroup.get(message.author);
    if (userMessages == null) {
        userMessages = new ArrayList<>();
        userGroup.put(message.author, userMessages);
for (Entry<Author, List<Message>> entry : userGroup.entrySet()) {
    for (Message message : entry.getValue()) {
        System.out.println("\t" + message.date);

How to group chat messages per user?

Tag : javascript , By : ap.
Date : March 29 2020, 07:55 AM
should help you out I solved this problem for myself pretty recently. Here's a full example.
The core business logic for grouping messages, in the above example, can be found under src/store.js in the addMessage function.
addMessage({ state }, { msg, selfHash }) {
  let addAsNewMsg = true;
  const lastMessage = state.messages.slice(-1)[0];
  if (lastMessage && lastMessage.userHash === msg.userHash) {
    // The last message was also sent by the same user
    // Check if it arrived within the grouping threshold duration (60s)
    const lastMessageTime = moment(lastMessage.time);
    const msgTime = moment(msg.time);
    const diffSeconds = msgTime.diff(lastMessageTime, "seconds");
    if (diffSeconds <= 60) {
      addAsNewMsg = false; // We're going to be appending.. so
      // We're going to now update the timestamp and (any) other fields.
      delete msg.message; // Since, we've already added this above
      Object.assign(lastMessage, msg); // Update with any remaining properties
  if (addAsNewMsg) {

Send messages to telegram group without user input

Tag : python-3.x , By : jamerson
Date : March 29 2020, 07:55 AM
seems to work fine I'm trying to build a bot which automatically sends a message whenever there is an update in the latest news using python. Following is what I did. , Using the python-telegram-bot api, you can send a message like this
bot.send_message(id, text='Message')
mybots = {}

def start(bot, update):
    """Send a message when the command /start is issued."""
      mybots[update.message.chat_id] = bot

def send_later():
    for id, bot in mybots.items():
        bot.send_message(id, text='Beep!')
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