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  • How to exclude result from 2 different table and show the result?

    - by Lucy Amalia Lusiana
    I have 2 tables: T1 (Total User) T2 (User_id_that_have_done_something_for_today) the mysql_query result from T1 are (Total User) ----- ID ----- | 1 | | 2 | | 3 | | 4 | | 5 | | 6 | | 7 | | 8 | the mysql_query from T2 are (logged) ----- user_id_log ----- | 1 | | 3 | | 4 | | 7 | How to compare T1 and T2 And only show the user that not approved? The end result that i wanted is (result) ----- ID ----- | 2 | | 5 | | 6 | | 8 | I spent 2 days to make this work but i cannot do it I have try LEFT JOIN and NOT IT but the result is not what i wanted Please help me, thank you

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  • Generating SQL for website

    - by Ethan
    I am working on a webapplication How can i create SQL for the following Database Information User information Username - String Password - String Admin or Client - boolean Last login – Date/Time LogItem typeLogItem – String (Page name?) hitCount – int View PageURL UserID Transaction User – String DateTimeStamp SKU – int Purchase-boolean TransactionID-int Inventory information Sku number - int Item description - String Price to customer - double Count - in

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  • % confuses python raw sql query

    - by Jonathan
    Following this SO question, I'm trying to "truncate" all tables related to a certain django application using the following raw sql commands in python: cursor.execute("set foreign_key_checks = 0") cursor.execute("select concat('truncate table ',table_schema,'.',table_name,';') as sql_stmt from information_schema.tables where table_schema = 'my_db' and table_type = 'base table' AND table_name LIKE 'some_prefix%'") for sql in [sql[0] for sql in cursor.fetchall()]: cursor.execute(sql) cursor.execute("set foreign_key_checks = 1") Alas I receive the following error: C:\dev\my_project>my_script.py Traceback (most recent call last): File "C:\dev\my_project\my_script.py", line 295, in <module> cursor.execute(r"select concat('truncate table ',table_schema,'.',table_name,';') as sql_stmt from information_schema.tables where table_schema = 'my_db' and table_type = 'base table' AND table_name LIKE 'some_prefix%'") File "C:\Python26\lib\site-packages\django\db\backends\util.py", line 18, in execute sql = self.db.ops.last_executed_query(self.cursor, sql, params) File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 216, in last_executed_query return smart_unicode(sql) % u_params TypeError: not enough arguments for format string Is the % in the LIKE making trouble? How can I workaround it?

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  • Is it possible for a Grails Domain to have no 'id'?

    - by firnnauriel
    Is it possible to create a table that has no 'id'? For example, this is my domain: class SnbrActVector { int nid String term double weight static mapping = { version false id generator: 'identity' } static constraints = { } } When I run this SQL statement, it fails: insert into snbr_act_vector values (5, 'term', 0.5) I checked the table and 'id' is already set to autoincrement. I'm thinking that another option is to remove the 'id' itself. Or is there another workaround for this? Please assume that it is not an option to change the givent SQL statement.

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  • How can I find all records for a model without doing a long list of "OR" conditions?

    - by gomezuk
    I'm having trouble composing a CakePHP find() which returns the records I'm looking for. My associations go like this: User -(has many)- Friends , User -(has many)- Posts I'm trying to display a list of all a user's friends recent posts, in other words, list every post that was created by a friend of the current user logged in. The only way I can think of doing this is by putting all the user's friends' user_ids in a big array, and then looping through each one, so that the find() call would look something like: $posts = $this->Post->find('all',array( 'conditions' => array( 'Post.user_id' => array( 'OR' => array( $user_id_array[0],$user_id_array[1],$user_id_array[2] # .. etc ) ) ) )); I get the impression this isn't the best way of doing things as if that user is popular that's a lot of OR conditions. Can anyone suggest a better alternative?

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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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  • check for several conditions when a user logs in

    - by paul
    I would like to accomplish the following: If a username or password field is null, notify the user. If user name already exists, do not insert into the database and notify user to create a different name. if the username is unique and password is not null, return the username to the user. As of now it always returns "Please enter a different user name." I believe the issue has to do with the database query but I am not sure. If anyone can have a look and see if I am making an error, I greatly appreciate it, thanks. if ($userName or $userPassword = null) { echo "Please enter a user name and password or return to the homepage."; } elseif (mysql_num_rows(mysql_query("SELECT count(userName) FROM logininfo WHERE userName = '$userName'")) ==1) { echo "Please enter a different user name."; } elseif ($userName and $userPassword != null) { echo "Your login name is: $userName"; }

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  • php session_start() error

    - by tooepic
    Hi, i've used a sample found on online and applied it to my code: <?php session_start(); if (isset($_REQUEST["email"])) { $_SESSION["name"] = true; $host = $_SERVER["HTTP_HOST"]; $path = dirname($_SERVER["PHP_SELF"]); $sid = session_name() . "=" . session_id(); header("Location: index.php?$sid"); exit; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> ... ... and rest of the html code When I open this page, I got an error: Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at /data/server/user/directory/sub-directory/login.php:1) in /data/server/user/directory/sub-directory/login.php on line 2 Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /data/server/user/directory/sub-directory/login.php:1) in /data/server/user/directory/sub-directory/login.php on line 2 I looked around to resolve this issue and saw few posts about this in this site also, but I just can't get a good grip on this...can't find the answer. Please help. Thanks.

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  • Why trigger query fails in the sqlite with Qt?

    - by dexterous_stranger
    I am a beginner in the SQL. I am using SQLite, Qt - on embedded systems. I want to put a trigger here. The trigger is that whenever the primary key Id is greater than 32145, then channelNum=101 should be set. I want to set the attrib name - text also, but I got the compilation issue. I believe that the setting of trigger is the part of DDL - Data definition language. Please let me know that if I am wrong here. Here is my create db code. I get the sql query error. Also please do suggest how to set the text attrib = "COmedy". /** associate db with query **/ QSqlQuery query ( m_demo_db ); /** Foreign keys are disabled by default in sqlite **/ /** Here is the pragma to turn them on first **/ query.exec("PRAGMA foreign_keys = ON;"); if ( false == query.exec()) { qDebug()<<"Pragma failed"; } /** Create Table for storing user preference LCN for DTT **/ qDebug()<<"Create Table postcode.db"; query.prepare(" CREATE TABLE dttServiceList (Id INTEGER PRIMARY KEY, attrib varchar(20), channelNum integer )" ); if ( false == query.exec()) { qDebug()<<"Create dttServiceList table failed"; } /** Try placing trigger here **/ triggerQuery = "CREATE TRIGGER upd_check BEFORE INSERT ON dttServiceList \ FOR EACH ROW \ BEGIN \ IF Id > 32145 THEN SET channelNum=101; \ END IF; \ END; "; query.prepare(triggerQuery); if ( false == query.exec()) { qDebug()<<"Trigger failed !!"; qDebug() << query.lastError(); } Also, how to set the text name in the trigger - I want to SET attrib = "Comedy".

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  • my sql insert query not working

    - by Piyush
    I am inserting userId.It is displaying correct but inserting 0 in spite of actual userId. mycode- If(! empty($userIDToCheck) || $userIDToCheck != '' ) { echo $userIDToCheck; $sql = "INSERT INTOpnpdb.ruser(userid) VALUES ('$userIDToCheck');"; mysql_query($sql)or die(mysql_error()); echo "Done"; } Output : pi203713 Done But is database it is inserting "0"???

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  • how do I copy value from one table and inserted to another in the same database??

    - by mathew
    I am having a tough time to do this... I have created two table say table-1 and table-2 in same database.what I want is I need to copy some values from table-1 and insert the same to table-2. I have tried many ways but it does not seems work. below is my code can any one tell me where I am missing?? $db = mysql_connect("localhost", "user", "pass") or die("Could not connect."); mysql_select_db("comdata",$db)or die(mysql_error()); $resultb = mysql_query("SELECT * FROM table-2")or die(mysql_error()); $row = mysql_fetch_array($resultb); $days = (strtotime(date("Y-m-d")) - strtotime($row['regtime'])) / (60 * 60 * 24); if($row > 0 && $days < 1){ $person = $row['person']; $catogr = $row['catog']; $position = $row['position']; $location = $row['location']; $rank = $row['rank']; mysql_close($db); }else{ $db = mysql_connect("localhost", "user", "pass") or die("Could not connect."); mysql_select_db("comdata",$db)or die(mysql_error()); $result = mysql_query("SELECT * FROM table-1 WHERE regtime = DATE(NOW()) ORDER BY rank ASC LIMIT 1;")or die(mysql_error()); $row = mysql_fetch_array($result); $person = $row['person']; $catogr = $row['catog']; $position = $row['position']; $location = $row['location']; $rank = $row['rank']; mysql_query("INSERT INTO table-2 (regtime,person,catog,position,location,rank) VALUES(NOW(),'$person','$catogr','$position','$location','$rank')"); mysql_close($db); }

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  • Select multiple records by one query

    - by kofto4ka
    Hello there. Please, give me advice, how to construct select query. I have table table with fields type and obj_id. I want to select all records in concordance with next array: $arr = array( 0 => array('type' => 1, 'obj_id' => 5), 1 => array('type' => 3, 'obj_id' => 15), 2 => array('type' => 4, 'obj_id' => 14), 3 => array('type' => 12, 'obj_id' => 17), ); I want to select needed rows by one query, is it real? Smth like select * from `table` where type in (1,3,4,12) and obj_id in (5,15,14,17) But this query returns also records with type = 3 and obj_id = 14, and for example type = 1 and obj_id = 17. p.s. moderators, please fix my title, I dont know how to describe my question.

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  • How to combine these three sql queries into one?

    - by lam3r4370
    How to combine these two sql queries into one? SELECT DISTINCT * FROM rss WHERE MATCH(content,title) AGAINST ('$filter') SELECT COUNT(content) FROM rss WHERE MATCH(content,title) AGAINST ('$filters') And if the result is 0 from the above query - SELECT DISTINCT * FROM rss WHERE content LIKE '%$filters%' OR title LIKE '%$filters%'; $filter .= $row['filter']; $filters = $row['filter']; $filters may be more than one keyword

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  • Worpress WorkFlow Modfications

    - by blgnklc
    Hi All WordPress Lovers, I would like to ask a help about Zensor which is a plugin that you publish a post then a moderator approves the post to be published on the wordpress blog site. When a post is awating for approval, each awaiting post is appearing "waiting moderation". But, I dont want any link appears before moderator approval. Actually I found the joing sentence below; 1- Must be added to the end of JOIN part of any query: LEFT JOIN wp_zensor ON ID = wp_zensor.post_id 2- Must be added to the end of WHERE condition : AND wp_zensor.moderation_status = 'approved' Could you please show me; where should I add these modification on the category link presentation below: <h2>Politics</h2> <?php $recent = new WP_Query("cat=31&showposts=1"); while($recent->have_posts()) : $recent->the_post();?> <b><a href="<?php the_permalink() ?>" rel="bookmark"><?php the_title(); ?></a></b> <?php the_content_limit(140, "devami &raquo;"); ?> <div class="hppostmeta"> <p><?php the_time('j F Y, H:i'); ?> | <?php the_author_posts_link(); ?></p> </div> <?php endwhile; ?> Or any general solutions will be welcomed. Thanks. BK

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  • PHP classes, parse syntax errors when using 'var' to declare variables

    - by jon
    I am a C# guy trying to translate some of my OOP understanding over to php. I'm trying to make my first class object, and are hitting a few hitches. Here is the beginning of the class: <?php require("Database/UserDB.php"); class User { private var $uid; private var $username; private var $password; private var $realname; private var $email; private var $address; private var $phone; private var $projectArray; public function _construct($username) { $userArray = UserDB::GetUserArray($username); $uid = $userArray['uid']; $username = $userArray['username']; $realname = $userArray['realname']; $email = $userArray['email']; $phone = $userArray['phone']; $i = 1; $projectArray = UserDB::GetUserProjects($this->GetID()); while($projectArray[$i] != null) { $projectArray[$i] = new Project($projectArray[$i]); } UserDB.php is where I have all my static functions interacting with the Database for this User Class. I am getting errors using when I use var, and I'm getting confused. I know I don't HAVE to use var, or declare the variables at all, but I feel it is a better practice to do so. the error is "unexpected T_VAR, expecting T_VARIABLE" When I simply remove var from the declarations it works. Why is this?

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  • Is it safe to modify CCK tables by hand?

    - by LanguaFlash
    I'm not intimately familiar with CCK but I have a one-time custom setup and know that I could get some performance gains if I created indexes and changed the field type and length of some of the fields in my CCK table. Is it save to modify this table at all or will I end up destroying something in the process? Thanks

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  • Binary string search on one field.

    - by CrazyJoe
    I have 300 boolean fields in one table, and im trying to do somithing like that: One string field: 10000010000100100100100100010001 Ha a simple way to do a simple search os this field like: select * from table where field xor "10000010000100100100000000010001" Im tring this but is to long: select * from teste where mid(info,2,1) and mid(info,3,1) :) Help!!

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  • IE8 and below <input type="image"> value work around

    - by kielie
    Hi guys, I have a slight problem, I am trying to capture the input of two buttons, one yes, one no, into a database but for some reason the database doesn't always show the value of the button clicked, it just shows up blank. <form action="refer.php" method="post" id="formID" > <div class="prompt_container" style="float: left;"> <span class="prompt_item"><input type="image" src="images/yes.jpg" alt="submit" value="yes" onclick="this.disabled=true,this.form.submit();" /></span> <input type="hidden" name="refer" value="yes"> </div> </form> <form action="thank_you.php" method="post" id="formID" > <div class="prompt_container" style="float: right;"> <span class="prompt_item"><input type="image" src="images/no.jpg" alt="submit" value="no" onclick="this.disabled=true,this.form.submit();" /></span> <input type="hidden" name="refer" value="no" > </div> </form> Apparently anything lower than IE8 will ignore the value attribute of all form inputs. How could I get this to work properly in all browsers? jQuery or Javascript maybe?

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  • retrieve value from db in php

    - by wangdoo
    I have a db in which 1 column value will be always null.I want it to pass to next page but it is passing all values of other column except this null value column. This is my code $row=mysql_query("SELECT * FROM `thirdpartycategorymaster` WHERE ISNULL(`delete`)"); while($row1=mysql_fetch_array($row)) { <td><b><a href="catdel.php?head1=<?php echo $row1['ThirdPartyCategoryName'] . '&&msg1=' . $row1['ThirdPartyCategoryID'] . '&&ctdel=' . $row1['delete']; ?> ">Delete</a></b></td> This my db table ThirdPartyCategoryID ThirdPartyCategoryName delete 29 ecommerce NULL

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