Jump to content

Search the Community

Showing results for tags 'join'.

  • Search By Tags

    Type tags separated by commas.
  • Search By Author

Content Type


Forums

  • Welcome to PHP Freaks
    • Announcements
    • Introductions
  • PHP Coding
    • PHP Coding Help
    • Regex Help
    • Third Party Scripts
    • FAQ/Code Snippet Repository
  • SQL / Database
    • MySQL Help
    • PostgreSQL
    • Microsoft SQL - MSSQL
    • Other RDBMS and SQL dialects
  • Client Side
    • HTML Help
    • CSS Help
    • Javascript Help
    • Other
  • Applications and Frameworks
    • Applications
    • Frameworks
    • Other Libraries
  • Web Server Administration
    • PHP Installation and Configuration
    • Linux
    • Apache HTTP Server
    • Microsoft IIS
    • Other Web Server Software
  • Other
    • Application Design
    • Other Programming Languages
    • Editor Help (PhpStorm, VS Code, etc)
    • Website Critique
    • Beta Test Your Stuff!
  • Freelance, Contracts, Employment, etc.
    • Services Offered
    • Job Offerings
  • General Discussion
    • PHPFreaks.com Website Feedback
    • Miscellaneous

Find results in...

Find results that contain...


Date Created

  • Start

    End


Last Updated

  • Start

    End


Filter by number of...

Joined

  • Start

    End


Group


AIM


MSN


Website URL


ICQ


Yahoo


Jabber


Skype


Location


Interests


Age


Donation Link

Found 4 results

  1. I've got two tables user table - id - username - fname - lname - group_id event Table - id - user_id - event_time - event_date - event ('Arrival','Departure','Break',etc..) My problem is getting all user by group and joining other table even date, but it should be base on 'event_date' I use `LEFT JOIN` and `LEFT OUTER JOIN` but because I have to specify the `event_date` not all user can be shown because some are absent. I made this SELECt * FROM user as u LEFT OUTER JOIN event as e ON u.id = e.user_id WHERE u.group_id = 6 AND ( e.event_date = '2016-07-05' AND e.event = 'Arrival' ) GROUP BY u.id but it only show the user who has an event on the date specified. I tried to make an VIEW table but it still has the condition base on event_date. Although I made this on PHP by condition but it takes time to load as I select event on every user. is there a way to get it using SQL? it should be like this:
  2. I want to fetch the titles from the Courses table, but I only want the Course Title to return once when I join Events. Courses CID | CourseTitle 1 | Course A 2 | Course B 3 | Course C Events EID | CID | EventDate 1 | 1 | 2016-02-22 2 | 1 | 2016-02-23 3 | 2 | 2016-02-24 4 | 3 | 2016-02-25 5 | 3 | 2016-02-26 If I use a JOIN, SELECT Courses.CourseTitle FROM Courses LEFT JOIN Events on (Events.CID = Courses.CID) then I get Course A Course A Course B Course C Course C But what I want is Course A Course B Course C Because ultimately, I'm going to select an Event date range, and I want to see just the courses with the event date range. Thanks!
  3. Hello all. I dont know how to go about this. I have a table (Transactions) that contains transactions of users. Another table (Confirmed) contains details of every confirmed user. I want to do a select statement that will display all the confirmed user with only the last of their transaction. But so far all it does is replicate the user and their date of transaction and that is not what i want. My intention is to get something like: Firstname Surname Date Registered Last Transaction andrews john 12-12-2014 10-10-2015 doe andy 12-12-2010 12-12-2014 But i'm getting something like: Firstname Surname Date Registered Last Transaction andrews john 12-12-2014 10-10-2015 andrews john 12-12-2014 10-11-2015 doe andy 12-12-2010 12-12-2014 doe andy 12-12-2010 01-12-2014 doe andy 12-12-2010 12-12-2013 Thanks $stm=$pdo->query("select * from confirmed left join transaction on confirmed.user_id = transaction.user_id where confirmed.status='confirmed' order by date"); while($row = $stmt->fetch(PDO::FETCH_ASSOC)) { echo $row['firstname']; echo $row['surname']; echo $row['regDate']; echo $row['lastTrans']; }
  4. Hello. I'm trying to get this piece of code finished but it's not going my way, if anyone could help me out that would be great. Here whats I'm trying to do: Table1 Table2 ---------------- --------------------------------- ID | | PlayerID | PlayerName | ---------------- -------------------------------- 1393 | | 1393 | Player1 | 2097 | | 2097 | Player2 | 3888 | | 3888 | Player3 | 3888 | | 4017 | Player4 | 3888 | --------------------------------- 4017 | 4017 | 4017 | ---------------- I Want to Count the entries in Table1 (3888 has 3 entries so it will display like "3888: 3") Then I want to join Table1 and Table2 using the ID so I can get the players name (3888=Player3 so it would display like Player3 : 3) Here's the code I'm using: <?php //connect to db $query = "SELECT * FROM Table1 INNER JOIN Table2 WHERE Table1.PlayerID = Table2.ID"; $query2 = "SELECT ID, SUM(ID=0) AS n0, SUM(ID=1) AS n1, COUNT(*) AS total FROM Table1 GROUP BY ID"; $result = mysql_query($query) or die($query."<br/><br/>".mysql_error()); $result2 = mysql_query($query2) or die($query2."<br/><br/>".mysql_error()); echo "<table border='1'> <tr> <th>PlayerName</th> <th>Entries</th> </tr>"; while($row2 = mysql_fetch_array($result2)) while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ID'] . "</td>"; echo "<td>" . $row['PlayerName'] . "</td>"; echo "<td>" . $row2['Total'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($link); ?> If I use $query on its own it joins perfectly, If I use $query2 on its own, it displays the results exactly how I want them listed but with the ID instead of the PlayerName, I tried putting them both together as shown above but I can't get them to work together. What I want my end Result to be Player1 1 Player2 1 Player3 3 Player4 3 How it keeps coming out; 1393 1 2097 1 3888 3 4017 3 Can anyone see where I'm going wrong? Thank you!!
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.