Jump to content

Search the Community

Showing results for tags 'sum'.

  • 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. Hi people, I'm having a problem with my SQL query which is not doing what I would like. I have this query: SELECT a. id, a.kataloski_broj, SUM(a.quantity) AS zbroj, entry_type_desc, posting_date, item_category, product_group, b.nabavna_cijena as neto_nabavna, b.neto_VPC, b.preporucena_VPC, c.nabavna_cijena as rabat_nabavna, c.VPC, d.VPC as NAV_VPC, d.PVPC as NAV_PVPC FROM prodaja_zaliha a LEFT JOIN kalkulacija_stavke b ON a.kataloski_broj = b.kataloski_broj LEFT JOIN kalkulacija_stavke_rabat c ON a.kataloski_broj = c.kataloski_broj LEFT JOIN katalog_pribora_item d ON a.kataloski_broj = d.kataloski_broj_NAV WHERE a.entry_type_desc = 'Prodaja' AND YEAR(a.posting_date) = '2015' AND a.kataloski_broj = 'M1200' GROUP BY a.id ASC The problem is that it returns SUM(a.quantity) AS zbroj in 1432 quantity, while there is actually 182 quantity in DB. What am I doing wrong?
  2. Hi, I am trying to add values together in my while loop, so far I have: $q->query($DB, $SQL); while($avrating=$q->getrow()): $averageratings = $avrating['sum(rating)'] / $noofreviews; $overallaverage = round($averageratings, 2) + round($averageratings, 2); echo '<p>'.$avrating['rating_name'].': '.round($averageratings, 2).'</p>'; endwhile; echo 'x'.$overallaverage; $overallaverage += $overallaverage / 10; echo 'The Average: '.$overallaverage; from this, I get: x7.34The Average: 8.074 the total shold be: 42.34 then dived by 10 should be 4.23
  3. 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!!
  4. I'm making statistic for 5 tables. I have made the example with one client data. loan: payment_schedule: payment_schedule_row payment_schedule_cover: payment_schedlue_delay: And the query is: SELECT period, loan_sum, covers, delay FROM (SELECT MAX(EXTRACT(YEAR_MONTH FROM psc.date)) AS period, (SELECT SUM(psr2.payment) FROM payment_schedule_row AS psr2 WHERE psr.payment_schedule_id = psr2.payment_schedule_id) AS loan_sum, (SELECT SUM(psc2.sum) FROM payment_schedule_cover AS psc2 WHERE psc.payment_schedule_id = psc2.payment_schedule_id) AS covers, (SELECT SUM(psd2.delay) FROM payment_schedule_delay AS psd2 WHERE psr.id = psd2.payment_schedule_row_id) AS delay FROM loan INNER JOIN payment_schedule AS ps ON ps.loan_id = loan.id INNER JOIN payment_schedule_row AS psr ON psr.payment_schedule_id = ps.id INNER JOIN payment_schedule_cover AS psc ON psc.payment_schedule_id = ps.id WHERE loan.status = 'payed' GROUP BY ps.id) AS sum_by_id GROUP BY period Sqlfiddle link: http://sqlfiddle.com/#!2/21585/2/0 Result for the query: period | loan_sum | covers | delay --------------------------------------- 201407 | 384 | 422 | 0.07 Everything is right except the delay. It should be 0.11 (0.07 + 0.03 + 0.01) So I have been trying to find the error from the query for days now. Maybe someone can tell me what I'm doing wrong.
×
×
  • 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.