Search Results

Search found 14142 results on 566 pages for 'mysql workbench'.

Page 384/566 | < Previous Page | 380 381 382 383 384 385 386 387 388 389 390 391  | Next Page >

  • how to increase or decrease the quantity in a field in my sql when we enter the new quantity

    - by madhu
    I'm doing a project I'm encountering a problem in my project......... the problem is i would like to increase or decrease the quantity in my sql quantity field when i pass the issued or delivered quantity i.e.... initially the quantity in the quantity filed is 1 when i intake the product of a quantity 10 then automatically it should update the quantity as 10+1=11 so it must update as 11, if i remove a 1 quantity it should update as 0........ how to write a code in jsp.......... pls do help

    Read the article

  • 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

    Read the article

  • Are Triggers Based On Queries Atomic?

    - by David
    I have a table that has a Sequence number. This sequence number will change and referencing the auto number will not work. I fear that the values of the trigger will collide. If two transactions read at the same time. I have ran simulated tests on 3 connections @ ~1 million records each and no collisions. CREATE TABLE `aut` ( `au_id` int(10) NOT NULL AUTO_INCREMENT, `au_control` int(10) DEFAULT NULL, `au_name` varchar(50) DEFAULT NULL, `did` int(10) DEFAULT NULL, PRIMARY KEY (`au_id`), KEY `Did` (`did`) ) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=latin1 TRIGGER `binc_control` BEFORE INSERT ON `aut` FOR EACH ROW BEGIN SET NEW.AU_CONTROL = (SELECT COUNT(*)+1 FROM aut WHERE did = NEW.did); END;

    Read the article

  • num_rows is 0 when it should be >0 for php mysqli code

    - by jpporterVA
    My num_rows is coming back as 0, and I've tried calling it several ways, but I'm stuck. Here is my code: $conn = new mysqli($dbserver, "dbuser", "dbpass", $dbname); // get the data $sql = 'SELECT AT.activityName, AT.createdOn FROM userActivity UA, users U, activityType AT WHERE U.userId = UA.userId and AT.activityType = UA.activityType and U.username = ? order by AT.createdOn'; $stmt = $conn->stmt_init(); $stmt->prepare($sql); $stmt->bind_param('s', $requestedUsername); $stmt->bind_result($activityName, $createdOn); $stmt->execute(); // display the data $numrows = $stmt->num_rows; $result=array("user activity report for: " . $requestedUsername . " with " . $numrows . " rows:"); $result[]="Created On --- Activity Name"; while ($stmt->fetch()) { $msg = " " . $createdOn . " --- " . $activityName . " "; $result[] = $msg; } $stmt->close(); There are multiple rows found, and the fetch loop process them just fine. Any suggestions on what will enable me to get the number of rows returned in the query? Suggestions are much appreciated. Thanks in advance.

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

  • 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!

    Read the article

  • 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?

    Read the article

  • Php random row help...

    - by Skillman
    I've created some code that will return a random row, (well, all the rows in a random order) But i'm assuming its VERY uneffiecent and is gonna be a problem in a big database... Anyone know of a better way? Here is my current code: $count3 = 1; $count4 = 1; //Civilian stuff... $query = ("SELECT * FROM `*Table Name*` ORDER BY `Id` ASC"); $result = mysql_query($query); while($row = mysql_fetch_array($result)) { $count = $count + 1; $civilianid = $row['Id']; $arrayofids[$count] = $civilianid; //echo $arrayofids[$count]; } while($alldone != true) { $randomnum = (rand()%$count) + 1; //echo $randomnum . "<br>"; //echo $arrayofids[$randomnum] . "<br>"; $currentuserid = $arrayofids[$randomnum]; $count3 += 1; while($count4 < $count3) { $count4 += 1; $currentarrayid = $listdone[$count4]; //echo "<b>" . $currentarrayid . ":" . $currentuserid . "</b> "; if ($currentarrayid == $currentuserid){ $found = true; //echo " '" .$found. "' "; } } if ($found == true) { //Reset array/variables... $count4 = 1; $found = false; } else { $listdone[$count3] = $currentuserid; //echo "<u>" . $count3 .";". $listdone[$count3] . "</u> "; $query = ("SELECT * FROM `*Tablesname*` WHERE Id = '$currentuserid'"); $result = mysql_query($query); $row = mysql_fetch_array($result); $username = $row['Username']; echo $username . "<br>"; $count4 = 1; $amountdone += 1; if ($amountdone == $count) { //$count $alldone = true; } } } Basically it will loop until its gets an id (randomly) that hasnt been chosen yet. -So the last username could take hours :P Is this 'bad' code? :P :(

    Read the article

  • 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.

    Read the article

  • indexing question

    - by user522962
    I have a table w/ 3 columns: name, phone, date. I have 3 indexes: 1 on phone, 1 on date and 1 on phone and date. I have the following statement: SELECT * FROM ( SELECT * FROM people WHERE phone IS NOT NULL ORDER BY date DESC) as t GROUP BY phone Basically, I want to get all unique phone numbers ordered by date. This table has about 2.5 million rows but takes forever to execute....are my indexes right? UPDATE: My EXPLAIN statement comes back with 2 rows: 1 for primary table and 1 for derived table. It says I am using temporary and using filesort for my primary table. For my derived table, it says my possible keys are (phone), and (phone, date) but it is using filesort.

    Read the article

  • Inserting Record into multiple tables No Common ID

    - by the_
    OK so I have two tables, MEDIA and BUSINESS. I want it set up so the forms to input into them are on the same page. MEDIA has a row that is biz_id that is the id of BUSINESS. So MEDIA is really a part of BUSINESS. HOW do I insert/add these into their tables without a common ID because I haven't yet made the record for business? I'm sorry I didn't really word this very much... You might need more clarification to answer properly and I'll be glad to give any more info. Any help would be greatly appreciated, thanks!

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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()); }

    Read the article

  • PHP & HTML Purifier Error: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given

    - by TaG
    I'm trying to Integrate HTML Purifier http://htmlpurifier.org/ to filter my user submitted data but I get the following error below. And I was wondering how can I fix this problem? I get the following error. on line 22: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given line 22 is. if (mysqli_num_rows($dbc) == 0) { Here is the php code. if (isset($_POST['submitted'])) { // Handle the form. require_once '../../htmlpurifier/library/HTMLPurifier.auto.php'; $config = HTMLPurifier_Config::createDefault(); $config->set('Core.Encoding', 'UTF-8'); // replace with your encoding $config->set('HTML.Doctype', 'XHTML 1.0 Strict'); // replace with your doctype $purifier = new HTMLPurifier($config); $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.*, profile.* FROM users INNER JOIN contact_info ON contact_info.user_id = users.user_id WHERE users.user_id=3"); $about_me = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['about_me'])); $interests = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['interests'])); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO profile (user_id, about_me, interests) VALUES ('$user_id', '$about_me', '$interests')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE profile SET about_me = '$about_me', interests = '$interests' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { // There was an error...do something about it here... print mysqli_error($mysqli); return; } }

    Read the article

  • 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?

    Read the article

  • 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++

    Read the article

  • Testing a SQL Query for True or False

    - by KickingLettuce
    $sql = "SELECT # FROM users WHERE onduty = 1 AND loc_id = '{$site}';"; $result = mysql_query($sql); I simply want to test if this is true or false. If it returns 0 rows, I want next line to be something like: if (!$result) { //do this; } However, in my test, I am getting false when I know it should be true. Is this sound logic here? (note, yes I know I should be using mysqli_query, that is not what I am asking here)

    Read the article

  • 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

    Read the article

  • 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

    Read the article

  • Want to avoid the particular rows from select join query... See description

    - by OM The Eternity
    I have a Select Left Join Query whis displays me the rows for the latest changedone(its a time) column name ("field" should not be equal) column name ("trackid" should not be equal), and column name "Operation should be "UPDATE" ", below is the query I am talking about... SELECT j1. * FROM jos_audittrail j1 LEFT OUTER JOIN jos_audittrail j2 ON ( j1.trackid != j2.trackid AND j1.field != j2.field AND j1.changedone < j2.changedone ) WHERE j1.operation = 'UPDATE' AND j2.id IS NULL Now here I don't want a row to be displayed with a two particular column's value i.e. "field's value" the value is "LastvisitDate" and "hits" Now if if append the condition in the above query that " AND j1.field != 'lastvistDate' AND j1.field != 'hits' " theni do not get any result... The table structure is jos_audittrail: id trackid operation oldvalue newvalue table_name live changedone(its a time) I hope i have given the details properly If u still find something missing I will try to provide it more better way... Pls help me to avoid those two rows with those to mentioned value of "field"

    Read the article

  • 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

    Read the article

< Previous Page | 380 381 382 383 384 385 386 387 388 389 390 391  | Next Page >