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  • In SQL, a Join is actually an Intersection? And it is also a linkage or a "Sideway Union"?

    - by Jian Lin
    I always thought of a Join in SQL as some kind of linkage between two tables. For example, select e.name, d.name from employees e, departments d where employees.deptID = departments.deptID In this case, it is linking two tables, to show each employee with a department name instead of a department ID. And kind of like a "linkage" or "Union" sideway". But, after learning about inner join vs outer join, it shows that a Join (Inner join) is actually an intersection. For example, when one table has the ID 1, 2, 7, 8, while another table has the ID 7 and 8 only, the way we get the intersection is: select * from t1, t2 where t1.ID = t2.ID to get the two records of "7 and 8". So it is actually an intersection. So we have the "Intersection" of 2 tables. Compare this with the "Union" operation on 2 tables. Can a Join be thought of as an "Intersection"? But what about the "linking" or "sideway union" aspect of it?

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  • Zip multiple database PDF blob files

    - by Michael
    I have a database table that contains numerous PDF blob files. I am attempting to combine all of the files into a single ZIP file that I can download and then print. Please help! <?php include '../config.php'; include '../connect.php'; $session = $_GET['session']; $query = "SELECT $tbl_uploads.username, $tbl_uploads.description, $tbl_uploads.type, $tbl_uploads.size, $tbl_uploads.content, $tbl_members.session FROM $tbl_uploads LEFT JOIN $tbl_members ON $tbl_uploads.username = $tbl_members.username WHERE $tbl_members.session = '$session'"; $result = mysql_query($query) or die('Error, query failed'); $files = array(); while(list($username, $description, $type, $size, $content) = mysql_fetch_array($result)) { $files[] = "$username-$description.pdf"; } $zip = new ZipArchive; $zip->open('file.zip', ZipArchive::CREATE); foreach ($files as $file) { $zip->addFile($file); } $zip->close(); header('Content-Type: application/zip'); header('Content-disposition: attachment; filename=filename.zip'); header('Content-Length: ' . filesize($zipfilename)); readfile($zipname); exit(); ?>

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  • How can I get all children from a parent row in the same table?

    - by Johnny Freeman
    Let's say I have a table called my_table that looks like this: id | name | parent_id 1 | Row 1 | NULL 2 | Row 2 | NULL 3 | Row 3 | 1 4 | Row 4 | 1 5 | Row 5 | NULL 6 | Row 6 | NULL 7 | Row 7 | 8 8 | Row 8 | NULL 9 | Row 9 | 4 10 | Row 10 | 4 Basically I want my final array in PHP to look like this: Array ( [0] => Array ( [name] => Row 1 [children] => Array ( [0] => Array ( [name] => Row 3 [children] => ) [1] => Array ( [name] => Row 4 [children] => Array ( [0] => Array ( [name] => Row 9 [children] => ) [1] => Array ( [name] => Row 10 [children] => ) ) ) ) ) [1] => Array ( [name] => Row 2 [children] => ) [2] => Array ( [name] => Row 5 [children] => ) [3] => Array ( [name] => Row 6 [children] => ) [4] => Array ( [name] => Row 8 [children] => Array ( [0] => Array ( [name] => Row 7 [children] => ) ) ) ) So, I want it to get all of the rows where parent_id is null, then find all nested children recursively. Now here's the part that I'm having trouble with: How can this be done with 1 call to the database? I'm sure I could do it with a simple select statement and then have PHP make the array look like this but I'm hoping this can be done with some kind of fancy db joining or something like that. Any takers?

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  • Select where a value present

    - by Roy
    First a database example: id, product_id, cat, name, value -------------------------------- 1,1,Algemeen,Processor,2 Ghz 2,1,Algemeen,Geheugen,4 GB 3,2,Algemeen,Processor,3 Ghz 4,2,Algemeen,Geheugen,4 GB 5,3,Beeldscherm,Inch,22" 6,3,Beeldscherm,Kleur,Zwart 7,3,Algemeen,Geheugen,3 GB 8,3,Algemeen,Processor,3 Ghz I want with one query to select the follow id's: 1,2,3,4,7,8 Because the cat = algemeen and the name = processor by these products. ID 5,6 are only present by product 3. So, the entry's (cat and name) which are present by all products (product_id) have to be selected. The database contains 80.000 entry's with a lot of diffrent cat's, name's and value's. Is this possible with one query or is some php necessary? How do I do this? My apologies for the bad English.

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  • Broken count(*) after adding LEFT JOIN

    - by Iain Urquhart
    Since adding the LEFT JOIN to the query below, the count(*) has been returning some strange values, it seems to have added the total rows returned in the query to the 'level': SELECT `n`.*, exp_channel_titles.*, round((`n`.`rgt` - `n`.`lft` - 1) / 2, 0) AS childs, count(*) - 1 + (`n`.`lft` > 1) + 1 AS level, ((min(`p`.`rgt`) - `n`.`rgt` - (`n`.`lft` > 1)) / 2) > 0 AS lower, (((`n`.`lft` - max(`p`.`lft`) > 1))) AS upper FROM `exp_node_tree_6` `n` LEFT JOIN `exp_channel_titles` ON (`n`.`entry_id`=`exp_channel_titles`.`entry_id`), `exp_node_tree_6` `p`, `exp_node_tree_6` WHERE `n`.`lft` BETWEEN `p`.`lft` AND `p`.`rgt` AND ( `p`.`node_id` != `n`.`node_id` OR `n`.`lft` = 1 ) GROUP BY `n`.`node_id` ORDER BY `n`.`lft` I'm totally stumped... Thank you!

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  • Open Source PHP search engine

    - by Ravi Gupta
    I am looking for an open source search engine plugin written in php for my website(eCommerce). Before anybody answer that I have a doubt regarding the search engine. Usually search engine crawl web pages, create indexes and then use them while looking for contents. But will the same model work for eCommerce websites? Yeah, it can crawl products pages, index them but don't you think it would be better if it crawls the database directly and index the products stored in the database? And when a user search for any product, it will simply give us the rows of the table which matches the user query? May be what I am asking is a stupid question but I am new to web development, so kindly help me to understand the concept. I have looked at a search engine called Sphider but didn't get what all I have to do to make it work with an eCommerce website.

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  • SQL query for fetching friend list

    - by Bhavyanshu
    I need help with SQL query. I have two tables. One is users and other one is userfriends users table: aid email firstname 1 [email protected] example 2 [email protected] example2 3 [email protected] example3 4 [email protected] example4 userfriends tables: reqid email friendemail status 1 [email protected] [email protected] 1 (example1 is frnds with example2) 2 [email protected] [email protected] 2 (example2 request pending) 3 [email protected] [email protected] 1 (example1 is frnds with example3) 4 [email protected] [email protected] 1 (example1 is frnds with example4) So when status is 2 the add request is pending and at status 1 they are friends. What i want is that i want to retrieve the complete friendlist for user example1. I want to pull out names from users table for corresponding output from previous query to display as friendlist.

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  • Moving from dedicated to shared cpanel - any scripts to do all / some of the install tasks ?

    - by mbbcat
    Hi, I have a few hundred phpld sites to move - each has its own cpanel, ( & the target may have shared cpanel) & I can do a full cpanel backup on the original server, but I don't have whm on the current host - the backups are fairly easy to organize but the installs so far means picking through files & setting up db's & mail etc by hand - I am thinking there ought to be an easier ie scripted way to do the installs or at least some parts - can anyone please suggest something ? I would like to migrate the stats at the same time Thanks M

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  • show count of rows between 2 dates

    - by hello
    I am trying to show the number of rows that have a created_at date between 2 dates. Here is my code: $result=mysql_query("select * from payments where created_at between '2013/10/01 00:00:00' and '2013/10/30 00:00:00'") or die('You need to add an administrator ' ); $counter = mysql_query("select * from payments where created_at between '2013/10/01 00:00:00' and '2013/10/30 00:00:00'"); $row = mysql_fetch_array($result); $id = $row['id']; $num = mysql_fetch_array($counter); $countjan = $num["id"]; However when i echo (<?php echo"$jan";?>)this shows as 0 any idea how i can get this to work P.s there is 1 row within this date range

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  • PhpMyAdmin; Should I disable root login?

    - by Camran
    I have this setup in Phpmyadmin: USER HOST PASSW PRIVILEGES GRANT debian-sys-maint localhost Yes ALL PRIVILEGES YES phpmyadmin localhost Yes USAGE NO root 127.0.0.1 Yes ALL PRIVILEGES YES root localhost Yes ALL PRIVILEGES YES root my_hostname Yes ALL PRIVILEGES YES username localhost Yes ALL PRIVILEGES YES Where "username" is my username and "my_hostname" is my hostname. I am currently only logging in as the last one (username, localhost). Also, I have php which also uses the last ones login details. Should I disable the other ones? And, what other security measures should I take? BTW: My server is Linux and I have root access. Thanks

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  • The best way to structure this database?

    - by James P
    At the moment I'm doing this: gems(id, name, colour, level, effects, source) id is the primary key and is not auto-increment. A typical row of data would look like this: id => 40153 name => Veiled Ametrine colour => Orange level => 80 effects => +12 sp, +10 hit source => Ametrine (Some of you gamers might see what I'm doing here :) ) But I realise this could be sorted a lot better. I have studied database relationships and secondary keys in my A-Level computing class but never got as far as to set one up properly. I just need help with how this database should be organised, like what tables should have what data with what secondary and foreign keys? I was thinking maybe 3 tables: gem, effects, source. Which then have relationships to each other? Can anyone shed some light on this? Is a complex way like I'm proposing really the way to go or should I just carry on with what I'm doing? Cheers.

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  • Where to place web server root?

    - by nefo_x
    Hello everybody, I've just made an upgrade and now partly thinking on web-server directory structure for local workstation for web-development on linux platform. Running multiple hosts and different projects required. Where is it better to put all the server's docroots? /var/www? /srv? /www? I plan to make it as separate partition - could it be good for backups? :) I'm looking forward to your thoughts on this.

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  • Always get the correct datetime?

    - by Jesper Mansa
    I was wandering if there is a way/site/link whith the correct time? Not the servers datetime or the clients datetime. I'm using datetime to count down on my gaming site for when it is the users time to make a play. Users come from all ower the world so using the users client time would not match if its from the US to Europe. Then normally I would use the servers time, but somehow it skips 1.2 hours sometimes? I would like to make sure that everbody makes a timestamp from the same source and that source always is correct! Hoping for help and thanks in advance.

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  • How to check if a checkbox/ radio button is checked in php

    - by user225269
    I have this html code: <tr> <td><label><input type="text" name="id" class="DEPENDS ON info BEING student" id="example">ID</label></td> </tr> <tr> <td> <label> <input type="checkbox" name="yr" class="DEPENDS ON info BEING student"> Year</label> </td> </tr> But I don't have any idea on how do I check this checkboxes if they are checked using php, and then output the corresponding data based on the values that are checked. Please help, I'm thinking of something like this. But of course it won't work, because I don't know how to equate checkboxes in php if they are checked: <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $id = mysql_real_escape_string($_POST['idnum']); if($_POST['id'] == checked & $_POST['yr'] ==checked ){ $result2 = mysql_query("SELECT * FROM student WHERE IDNO='$id'"); echo "<table border='1'> <tr> <th>IDNO</th> <th>YEAR</th> </tr>"; while($row = mysql_fetch_array($result2)) { echo "<tr>"; echo "<td>" . $row['IDNO'] . "</td>"; echo "<td>" . $row['YEAR'] . "</td>"; echo "</tr>"; } echo "</table>"; } mysql_close($con); ?>

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  • How do I return the numeric value from a database query in PHP?

    - by Luke
    Hello, I am looking to retreive a numerical value from the database function adminLevel() { $q = "SELECT userlevel FROM ".TBL_USERS." WHERE id = '$_SESSION[id]'"; return mysql_query($q, $this->connection); } This is the SQL. I then wrote the following php/html: <?php $q = $database->adminLevel(); if ($q > 7) { ?> <a href="newleague.php">Create a new league</a> <? } ?> The problem I have is that the userlevel returned isn't affecting the if statement. It is always displayed. How do i get it to test the value of userlevel is greater than 7? Thanks

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  • GROUP BY on multiple columns

    - by Tams
    I have a table that looks like the following - Id Reference DateAttribute1 DateAttribute2 1 MMM005 2011-09-11 2012-09-10 2 MMM005 2012-06-13 2012-09-10 3 MMM006 2012-08-22 2012-09-10 4 MMM006 2012-08-22 2012-09-11 I have handle to the id values. I would like to query such that I get the following result Id Reference DateAttribute1 DateAttribute2 2 MMM005 2012-06-13 2012-09-10 4 MMM006 2012-08-22 2012-09-11 I would like my result to be grouped by reference and then 'DateAttribute1' and then 'DateAttribute2' as such - DateAttribute1 has a priority over DateAttribute2 as you can see above in the result. How should I write my query to fetch the results in the above manner? Any solution?

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  • inserting facebook app users details to database

    - by fusion
    i'm trying to insert user details, who authorize the app, into the database, but nothing seems to be happening. the data is null and no record is being inserted. is there something wrong with the code? function insertUser($user_id,$sk,$conn) { //$info = $facebook->api_client->users_getInfo($user_id, 'first_name, last_name', 'name', 'sex'); $info = $facebook->api_client->fql_query("SELECT uid, first_name, last_name, name, sex FROM user WHERE uid = $user_id"); for ($i=0; $i < count($info); $i++) { $record = $info[$i]; $first_name=$record['first_name']; $last_name=$record['last_name']; $full_name=$record['name']; $gender=$record['sex']; } $data= mysql_query("select uid from users where uid='{$user_id}'",$conn); if(mysql_num_rows($data)==0) { $sql = "INSERT INTO users (uid,sessionkey, active, fname, lname, full_name, gender) VALUES('{$user_id}','{$sk}','1', '{$first_name}', '{$last_name}', '{$full_name}', '{$gender}')"; mysql_query($sql,$conn); return true; } return false; }

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  • Problem with joining to an empty table

    - by Imran Omar Bukhsh
    I use the following query: select * from A LEFT JOIN B on ( A.t_id != B.t_id) to get all the records in A that are not in B. The results are fine except when table B is completely empty, but then I do not get any records, even from table A. Later It wont work yet! CREATE TABLE IF NOT EXISTS T1 ( id int(11) unsigned NOT NULL AUTO_INCREMENT, title varchar(50) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL, t_id int(11) NOT NULL, PRIMARY KEY (id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; -- -- Dumping data for table T1 INSERT INTO T1 (id, title, t_id) VALUES (1, 'apple', 1), (2, 'orange', 2); -- -- Table structure for table T2 CREATE TABLE IF NOT EXISTS T2 ( id int(11) NOT NULL AUTO_INCREMENT, title varchar(50) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL, t_id int(11) NOT NULL, PRIMARY KEY (id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; -- -- Dumping data for table T2 INSERT INTO T2 (id, title, t_id) VALUES (1, 'dad', 2); Now I want to get all records in T1 that do not have a corresponding records in T2 I try SELECT * FROM T1 LEFT OUTER JOIN T2 ON T1.t_id != T2.t_id and it won't work

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  • Beginner having difficulty with SQL query

    - by Vulcanizer
    Hi, I've been studying SQL for 2 weeks now and I'm preparing for an SQL test. Anyway I'm trying to do this question: For the table: 1 create table data { 2 id int, 3 n1 int not null, 4 n2 int not null, 5 n3 int not null, 6 n4 int not null, 7 primary key (id) 8 } I need to return the relation with tuples (n1, n2, n3) where all the corresponding values for n4 are 0. The problem asks me to solve it WITHOUT using subqueries(nested selects/views) It also gives me an example table and the expected output from my query: 01 insert into data (id, n1, n2, n3, n4) 02 values (1, 2,4,7,0), 03 (2, 2,4,7,0), 04 (3, 3,6,9,8), 05 (4, 1,1,2,1), 06 (5, 1,1,2,0), 07 (6, 1,1,2,0), 08 (7, 5,3,8,0), 09 (8, 5,3,8,0), 10 (9, 5,3,8,0); expects (2,4,7) (5,3,8) and not (1,1,2) since that has a 1 in n4 in one of the cases. The best I could come up with was: 1 SELECT DISTINCT n1, n2, n3 2 FROM data a, data b 3 WHERE a.ID <> b.ID 4 AND a.n1 = b.n1 5 AND a.n2 = b.n2 6 AND a.n3 = b.n3 7 AND a.n4 = b.n4 8 AND a.n4 = 0 but I found out that also prints (1,1,2) since in the example (1,1,2,0) happens twice from IDs 5 and 6. Any suggestions would be really appreciated.

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  • insert into table where if not in list

    - by jim smith
    Can anybody help me with the syntax? insert into history (company,partnumber,price) values ('blah','IFS0090','0.00') if company NOT IN ('blah','blah2','blah3','blah4','blah4') and partnumber='IFS0090'; Background: I have a history table which stores daily company, products and prices. But sometimes a company will remove itself for a few days. Complicating the issue is because I'm only saving daily CHANGES to prices only and not snapshotting the entire days list (the data would be huge) when I display the data the company will still come up for the previous days price. So I need to do something like this, where a 0.00 price means they're no longer there.

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  • duplicate record

    - by user349953
    Insert into Attendancemst ( emp_code , name, date , timetable , on_duty,out_duty,clockin , clockout, late, early, mis_in , mis_out , absent , halfday, total_time ) values (pemp_code,pname,pdate,ptimetable,pon_duty,pout_duty ,pclockin,pclockout,plate, pearly, pmis_in,pmis_out,pabsent,phalfday,ptotal_time )ON DUPLICATE KEY UPDATE emp_code=pemp_code and date = pdate;

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