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• TV Guide script - getting current date programmes to show

  - by whitstone86

  This is part of my TV guide script: //Connect to the database mysql_connect("localhost","root","PASSWORD"); //Select DB mysql_select_db("mytvguide"); //Select only results for today and future $result = mysql_query("SELECT programme, channel, episode, airdate, expiration, setreminder FROM mediumonair where airdate >= now()"); The episodes show up, so there are no issues there. However, it's getting the database to find data that's the issue. If I add a record for a programme that airs today this should show: Medium showing on TV4 8:30pm "Episode" Set Reminder Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder but this shows instead: Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder I almost have the SQL working; just not sure what the right code is here, to avoid the second mistake showing - as the record (which indicates a show currently airing) does not seem to work at present. Please can anyone help me with this? Thanks

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• How can I fake sql data while preserving statements without commenting my server-side code?

  - by Fedor

  I have to use hardcoded values for certain fields because at this moment we don't have access to the real data. When we do get access, I don't want to go through a lot of work uncommenting. Is it possible to keep this statement the way it is, except use '25' as the alias for ratecode? IF(special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode, I have about 8 or so IF statements similar to this and I'm just too lazy ( even with vim ) to re-append while commenting out each if statement line by line. I would have to do this: $sql = 'SELECT u.*,'; // IF ( special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode $sql.= '25 AS ratecode';

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• Why does this properly escaped SQL query fail?

  - by Jason Rhodes

  Here's the query: INSERT INTO jobemails (jobid, to, subject, message, headers, datesent) VALUES ('340', '[email protected]', 'We\'ve received your request for a photo shoot called \'another\'.', 'message', 'headers', '2010-04-22 15:55:06') The datatypes are all correct, it always fails at the subject, so it must be how I'm escaping the values, I assume. I'm sure one of you will see my idiot mistake right away. A little help?

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• Allow users to pull temporary data then delete table?

  - by JM4

  I don't know the best way to title this question but am trying to accomplish the following goal: When a client logs into their profile, they are presented with a link to download data from an existing database in CSV format. The process works, however, I would like for this data to be 'fresh' each time they click the link so my plan was - once a user has clicked the link and downloaded the CSV file, the database table would 'erase' all of its data and start fresh (be empty) until the next set of data populated it. My EXISTING CSV creation code: <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'database'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= '"'.$rowr[$j].'",'; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> any ideas?

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• Problem while redirecting user after registration

  - by Eternal Learner

  I am creating a simple website . My situation is like this. After registering an user, I want to redirect the user after say 3 seconds to a main page(if the registration succeeds) . The code I have now is as below $query = "INSERT INTO Privileges VALUES('$user','$password1','$role')"; $result = mysql_query($query, $dbcon) or die('Registration Failed: ' . mysql_error()); print 'Thanks for Registering , You will be redirected shortly'; ob_start(); echo "Test"; header("Location: http://www.php.net"); ob_flush() I get the error message Warning: Cannot modify header information - headers already sent by (output started at/home/srinivasa/public_html/ThanksForRegistering.php:27) in /home/srinivasa /public_html/ThanksForRegistering.php on line 35. What do I need to do now ?

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• problem during data fetch

  - by nectar

  here is my code $sql="SELECT * FROM $tbl_name WHERE ownerId='$UserId'"; $result=mysql_query($sql,$link)or die(mysql_error()); $row = mysql_fetch_array($result, MYSQL_ASSOC); <?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<tr>"; echo "<td>".$row['pinId']."</td>"; echo "<td>".$row['usedby']."</td>"; echo "<td>".$row['status']."</td>"; echo "</tr>"; } ?> it is ignoring the first record means if 4 rows are in $row its ignoring the 1st one rest three are coming on page. ownerId is not primary key.

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• How to structure data... Sequential or Hierarchical?

  - by Ryan

  I'm going through the exercise of building a CMS that will organize a lot of the common documents that my employer generates each time we get a new sales order. Each new sales order gets a 5 digit number (12222,12223,122224, etc...) but internally we have applied a hierarchy to these numbers: + 121XX |--01 |--02 + 122XX |--22 |--23 |--24 In my table for sales orders, is it better to use the 5 digital number as an ID and populate up or would it be better to use the hierarchical structure that we use when referring to jobs in regular conversation? The only benefit to not populating sequentially seems to be formatting the data later on in my view, but that doesn't sound like a good enough reason to go through the extra work. Thanks

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• Creating dynamic icons based on data entered into database from django forms

  - by John Hoke

  So I'm using Django to create a projects page with multiple forms for each project. Let's call them form 1, 2, 3, and 4. Once you create a project you can fill out any of these forms. I want to create "buttons" or links for each one of the forms that would show up on the main page. Now this is the part I need help with: Step 1. I want it so that if you click on a button for a form (say form 1) and none exists for that project yet a pop up would come up saying "This form does not exist yet, are you sure you want to create one?". And if you'd answer yes you would be directed to the form page. Step 2. But if that form does exist, I don't want any pop up to open and I want the link to take the user directly to that page. Step 3. My next problem is this. These forms are in order, so if you didn't create form 1 but created form 2, I don't want to give the user access to form 1. So in this scenario, if you click on form 1 I want a pop up to open and say "This form can no longer be created", and the link wouldn't function anymore. Basically the button will have 3 function. First it should look at the database and if data for that specific form exists it should do "Step 2", if data for that form and the proceeding forms don't exist it should do "Step 1", and if data for that form doesn't exist but data for proceeding form's does exist is should do "Step 3". Is this possible? Please help as I need to find a solution to this soon. Any help would be highly appreciated. Thank you

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• where id = multiple artists

  - by pixel

  Any time there is an update within my music community (song comment, artist update, new song added, yadda yadda yadda), a new row is inserted in my "updates" table. The row houses the artist id involved along with other information (what type of change, time and date, etc). My users have a "favorite artists" section where they can do just that -- mark artists as their favorites. As such, I'd like to create a new feature that shows the user the changes made to their various favorite artists. How should I be doing this efficiently? SELECT * FROM table_updates WHERE artist_id = 1 and artist_id = 500 and artist_id = 60032 Keep in mind, a user could have 43,000 of our artists marked as a favorite. Thoughts?

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• How to send the my sql result to controller in codeigniter as a variable

  - by udaya

  Hi In my mode I am selecting a field as $query1 = $this->db->query("SELECT dPassword FROM tbl_login WHERE dEmailID='[email protected]'"); How to return dpassword as a variable to my controller I tried this way return dpassword;

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• MULTIPLE CRITERIA TABLE JOIN

  - by user1447203

  I have a table listing clothing items (shirt, trousers, etc) named . Each item is identified with a unique CLOTHING.CLOTHING_ID. So a blue shirt is 01, a flowery shirt is 12 and jeans are 07 say. I have a second table identifying outfits with a column for shirts, for trousers, shoes etc. For example Outfit 1: shirt 01, trousers 07 (i.e. blue shirt with jeans) Outfit 2: shirt 12, trousers 07 (so flowery shirt with jeans). This table is named and each outfit is unique with OUTFIT_LIST.OUTFIT_ID. I want to produce a select statement that will list each outfit's contents, i.e. find the clothing specified in Outfit 1. Any help would be very much appreciated, and apologies in advance if I am missing a very simple solution. I have been playing with JOINS of all descriptions and CONCATS and so on with now luck - I am very new to this. Thanks.

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• PHP - How to display other values, when a query is limited by 3?

  - by Dodi300

  Hello. Can anyone tell me how to display the other values, when a query is limited my 3. In this question I asked how to order and limit values, but now I want to show the others in another query. How would I go about doing this? Here's the code I used before: $query = "SELECT gmd FROM account ORDER BY gmd DESC LIMIT 3"; $result = mysql_query($query); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { } Thanks!

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• How do you display a PHPmyadmin table onto a web browser/web page?

  - by user1390754

  Just a simple question, i've been searching around and for some reason i cannot find an answer to this. after creating a database/table in Phpmyadmin using xampp, what command do i need to put into my webpage (PHP) to show the table I made? I know the first step involves connecting to the database and i think i've done that properly. This is the code i found from somewhere about connecting (w3 tutorials I believe) $con = mysql_connect("localhost","xxxxxx,""); if (!$con) { die('Could not connect: ' . mysql_error()); }

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• How grouping and totaling are done into three tables using JOIN

  - by text

  Here are my tables respondents: field sample value respondentid : 1 age : 2 gender : male survey_questions: id : 1 question : Q1 answer : sample answer answers: respondentid : 1 question : Q1 answer : 1 --id of survey question I want to display all respondents who answered the certain survey, display all answers and total all the answer and group them according to the age bracket. I tried using this query: $sql = "SELECT res.Age, res.Gender, answer.id, answer.respondentid, SUM(CASE WHEN res.Gender='Male' THEN 1 else 0 END) AS males, SUM(CASE WHEN res.Gender='Female' THEN 1 else 0 END) AS females, CASE WHEN res.Age < 1 THEN 'age1' WHEN res.Age BETWEEN 1 AND 4 THEN 'age2' WHEN res.Age BETWEEN 4 AND 9 THEN 'age3' WHEN res.Age BETWEEN 10 AND 14 THEN 'age4' WHEN res.Age BETWEEN 15 AND 19 THEN 'age5' WHEN res.Age BETWEEN 20 AND 29 THEN 'age6' WHEN res.Age BETWEEN 30 AND 39 THEN 'age7' WHEN res.Age BETWEEN 40 AND 49 THEN 'age8' ELSE 'age9' END AS ageband FROM Respondents AS res INNER JOIN Answers as answer ON answer.respondentid=res.respondentid INNER JOIN Questions as question ON answer.Answer=question.id WHERE answer.Question='Q1' GROUP BY ageband ORDER BY res.Age ASC"; I was able to get the data but the listing of all answers are not present. What's wrong with my query. I want to produce something like this: ex: # of Respondents is 3 ages: 2,3 and 6 Question: what are your favorite subjects? Ages 1-4: subject 1: 1 subject 2: 2 subject 3: 2 total respondents for ages 1-4 : 2 Ages 5-10: subject 1: 1 subject 2: 1 subject 3: 0 total respondents for ages 5-10 : 1

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• How to properly design a simple favorites and blocked table?

  - by Nils Riedemann

  Hey, i am currently writing a webapp in rails where users can mark items as favorites and also block them. I came up two ways and wondered which one is more common/better way. 1. Separate join tables Would it be wise to have 2 tables for this? Like: users_favorites - user_id - item_id users_blocked - user_id - item_id 2. single table users_marks (or so) - users_id - item_id - type (["fav", "blk"]) Both ways seem to have advantages. Which one would you use and why?

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• How would I go about writing a conditional statement to check if visitor is coming from a particular

  - by Matthew

  Hello guys, What I have in mind is this... We are going to have people come from a particular site during a acquisition campaign and was wondering how I could conditionalize a certain section of my site to display a thank you message instead of the sign up form as they would have had the opportunity to fill this out before coming to my landing page. I have seen solutions like: $referal = mysql_real_escape_string($_SERVER['HTTP_REFERER']); I would like to know if this is the best way to get this to work??? - okay this is what i think might work. The third party website that is referring people to our landing page once the form on that site has been filled out can push into the record a hidden input value of "www.sample.com" or whatever... then I can have something check the for that particular value and fire off the conidtional. Does that even sound right?

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• select query related problem

  - by user222585

  i have interest rate and amount in a table where interest rate are ranged in different values like 4.5,4.6,5.2,5.6 etc. i want to get sum of amounts classified by interest rate where interest rate will be separated by .25. for example all amount having interest rate 1.25,1.3,1.4 will be in one group and 1.5,1.67,1.9 will be in another group how can i write the query?

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• PHP - displaying 1 random record for each week

  - by mike

  I want to display 1 random record from a database based on the week. I need to determine if it's a new, and if it is a new week, then select the record and display the new record. I'm thinking I can just use a single day of the week to generate the new record, either way will work. I'm really having a hard time conceptualizing how I'll store the record id and not select a new one when someone visits again the same day or refreshes the page. Any ideas? Let me know if I wasn't clear enough.

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• Can we use a sql data field as column name instead?

  - by Starx

  First a query SELECT * FROM mytable WHERE id='1' Gives me a certain number of rows For example id | context | cat | value 1 Context 1 1 value 1 1 Context 2 1 value 2 1 Context 1 2 value 3 1 Context 2 2 value 4 Now my problem instead of receiving the result in such way I want it is this way instead id | cat | Context 1 | Context 2 1 1 value 1 value 2 1 2 value 3 value 4

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• sql count function

  - by suryll

  Hi I have three tables and I want to know how much jobs with the wage of 1000 an employee has had The first SQL query gives me the names of all the employees that has recieved 1000 for a job SELECT distinct first_name FROM employee, job, link WHERE job.wage = 1000 AND job.job_id = link.job_id and employee.employee_id = link.employee_id; The second SQL query gives me the total number for all employees of how much jobs they have made for 1000 SELECT count(wage) FROM employee, job, link WHERE job.wage = 1000 AND job.job_id = link.job_id and employee.employee_id = link.employee_id; I was wondering if there was a way of joining both queries and also making the second for each specific employee???

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• Error during data INSERT in php

  - by nectar

  here my code- $sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES "; for($i=0; $i0) { $sql .= ", "; } $sql .= "('$pin[$i]', '$ownerid', 'Free', '1')"; } $sql .= ";"; echo $sql; mysql_query($sql); if(mysql_affected_rows() 0) { echo "done"; } else { echo "Fail"; } output: ** INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES ('13837927', 'admin', 'Free', '1'), ('59576082', 'admin', 'Free', '1'); Fail why it is not inserting values when $sql query is right?

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• How to get time from db depending upon conditions

  - by Somebody is in trouble

  I have a table in which the value are Table_hello date col2 2012-01-31 23:01:01 a 2012-06-2 12:01:01 b 2012-06-3 20:01:01 c Now i want to select date in days if it is 3 days before or less in hours if it is 24 hours before or less in minutes if it is 60 minutes before or less in seconds if it is 60 seconds before or less in simple format if it is before 3days or more OUTPUT for row1 2012-01-31 23:01:01 for row2 1 day ago for row3 1 hour ago UPDATE My sql query select case when TIMESTAMPDIFF(SECOND, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(SECOND, `date`,current_timestamp), ' seconds') when TIMESTAMPDIFF(DAY, `date`,current_timestamp) <= 3 then concat(TIMESTAMPDIFF(DAY, `date`,current_timestamp), ' days')end when TIMESTAMPDIFF(HOUR, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(HOUR, `date`,current_timestamp), ' hours') when TIMESTAMPDIFF(MINUTE, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(MINUTE, `date`,current_timestamp), ' minutes') from table_hello Only problem is i am unable to use break and default in sql like switch case in c++

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• Is this possible with sql?

  - by Andrew

  Is it possible to do something like this: INSERT INTO table(col1, col2) VALUES(something_from_another_table, value); With something_from_another_table being a SQL command? Like, is there something I can do that's equivelant to: INSERT INTO table(col1, col2) VALUES((SELECT value FROM table2 WHERE id = 3), value);

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• How to password protect my server for development?

  - by Camran

  I am about to upload my site to the Server, and I have ubuntu 9.1 as OS. Now, I have installed LAMP, and need to somehow password protect my website. How can I do this so only I have access to my website while developing it? Please be thorough... Thanks

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• php and SQL_CALC_FOUND_ROWS

  - by Lizard

  I am trying to add the SQL_CALC_FOUND_ROWS into a query (Please note this isn't for pagination) please note I am trying to add this to a cakePHP query the code I currently have is below: return $this->find('all', array( 'conditions' => $conditions, 'fields'=>array('SQL_CALC_FOUND_ROWS','Category.*','COUNT(`Entity`.`id`) as `entity_count`'), 'joins' => array('LEFT JOIN `entities` AS Entity ON `Entity`.`category_id` = `Category`.`id`'), 'group' => '`Category`.`id`', 'order' => $sort, 'limit'=>$params['limit'], 'offset'=>$params['start'], 'contain' => array('Domain' => array('fields' => array('title'))) )); Note the 'fields'=>array('SQL_CALC_FOUND_ROWS',' this obviously doesn't work as It tries to apply the SQL_CALC_FOUND_ROWS to the table e.g. SELECTCategory.SQL_CALC_FOUND_ROWS, Is there anyway of doing this? Any help would be greatly appreciated, thanks.

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