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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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  • 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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  • Insert into select and update in single query

    - by Ossi
    I have 4 tables: tempTBL, linksTBL and categoryTBL, extra on my tempTBL I have: ID, name, url, cat, isinserted columns on my linksTBL I have: ID, name, alias columns on my categoryTBL I have: cl_id, link_id,cat_id on my extraTBL I have: id, link_id, value How do I do a single query to select from tempTBL all items where isinsrted = 0 then insert them to linksTBL and for each record inserted, pickup ID (which is primary) and then insert that ID to categoryTBL with cat_id = 88. after that insert extraTBL ID for link_id and url for value. I know this is so confusing, put I'll post this anyhow... This is what I have so far: INSERT IGNORE INTO linksTBL (link_id,link_name,alias) VALUES(NULL,'tex2','hello'); # generate ID by inserting NULL INSERT INTO categoryTBL (link_id,cat_id) VALUES(LAST_INSERT_ID(),'88'); # use ID in second table I would like to add here somewhere that it only selects items where isinserted = 0 and iserts those records, and onse inserted, will change isinserted to 1, so when next time it runs, it will not add them again.

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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 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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  • Wordpress SQL_CALC fix causes PHP error

    - by ok1ha
    I'm looking for some followup on an older topic for Wordpress where SQL_CALC was found to slow things down when you have a large DB in Wordpress. I have been using the code, at the bottom of this post, to get around it but it does generate an error in my error log. How would I prevent this error? PHP Warning: Division by zero in /var/www/vhosts/domain.com/httpdocs/wp-content/themes/greatTheme/functions.php on line 19 The original thread: http://wordpress.org/support/topic/slow-queries-sql_calc_found_rows-bringing-down-site?replies=25 The code in my functions.php: add_filter('pre_get_posts', 'optimized_get_posts', 100); function optimized_get_posts() { global $wp_query, $wpdb; $wp_query->query_vars['no_found_rows'] = 1; $wp_query->found_posts = $wpdb->get_var( "SELECT COUNT(*) FROM wp_posts WHERE 1=1 AND wp_posts.post_type = 'post' AND (wp_posts.post_status = 'publish' OR wp_posts.post_status = 'private')" ); $wp_query->found_posts = apply_filters_ref_array( 'found_posts', array( $wp_query->found_posts, &$wp_query ) ); $wp_query->max_num_pages = ceil($wp_query->found_posts / $wp_query->query_vars['posts_per_page']); return $wp_query; }

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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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  • C# SQL SELECT Statement

    - by Feren6
    I have the following code: SqlCommand cmd2 = new SqlCommand("SELECT ClaimId FROM tblPayment WHERE PaymentId = " + PaymentID.ToString(), mvarDBConn); SqlDataReader reader = cmd2.ExecuteReader(); reader.Read(); Int32 ClaimId = reader.GetInt32(0); reader.Close(); If I run the SELECT statement in SQL it returns the number fine, but when I use ExecuteReader all it returns is 0. I've tried multiple methods including ExecuteScalar, ExecuteNonQuery, reader.GetString then casting that to an int, etc. What am I missing? Thanks.

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  • Looping Through Database Query

    - by DrakeNET
    I am creating a very simple script. The purpose of the script is to pull a question from the database and display all answers associated with that particular question. We are dealing with two tables here and there is a foreign key from the question database to the answer database so answers are associated with questions. Hope that is enough explanation. Here is my code. I was wondering if this is the most efficient way to complete this or is there an easier way? <html> <head> <title>Advise Me</title> <head> <body> <h1>Today's Question</h1> <?php //Establish connection to database require_once('config.php'); require_once('open_connection.php'); //Pull the "active" question from the database $todays_question = mysql_query("SELECT name, question FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Variable to hold $todays_question aQID $questionID = mysql_query("SELECT commentID FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Print today's question echo $todays_question; //Print comments associated with today's question $sql = "SELECT commentID FROM approvedQuestions WHERE status = active"; $result_set = mysql_query($sql); $result_num = mysql_numrows($result_set); for ($a = 0; $a < $result_num; $a++) { echo $sql; } ?> </body> </html>

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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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  • Running a sharded DB from a single machine

    - by ming yeow
    This sounds kinda dumb, but I have a sharded DB that I no longer think I need to run on 2 machines, and would like to run on one single machine instead. Any ideas on how that can potentially be done? There are lots of resources on how i can achieve the converse, but very little on how this can be done

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  • 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.

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  • how store date in myqsl database?

    - by Syom
    i have date in dd/mm/yyyy format. how can i store it in databse, if i fant to do some operations on them after? for example i must find out the rows, where date > something what type i must set to date field? thanks

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  • 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 :(

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  • 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"

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  • Django database - how to add this column in raw SQL.

    - by alex
    Suppose I have my models set up already. class books(models.Model): title = models.CharField... ISBN = models.Integer... What if I want to add this column to my table? user = models.ForeignKey(User, unique=True) How would I write the raw SQL in my database so that this column works?

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  • Adding to database. No repeat on refresh

    - by kevstarlive
    I have this code: Episode.php <?$feedback = new feedback; $articles = $feedback->fetch_all(); if (isset($_POST['name'], $_POST['post'])) { $cast = $_GET['id']; $name = $_POST['name']; $email = $_POST['email']; $post = nl2br ($_POST['post']); $ipaddress = $_SERVER['REMOTE_ADDR']; if (empty($name) or empty($post)) { $error = 'All Fields Are Required!'; }else{ $query = $pdo->prepare('INSERT INTO comments (cast, name, email, post, ipaddress) VALUES(?, ?, ?, ?, ?)'); $query->bindValue(1, $cast); $query->bindValue(2, $name); $query->bindValue(3, $email); $query->bindValue(4, $post); $query->bindValue(5, $ipaddress); $query->execute(); } }?> <div align="center"> <strong>Give us your feedback?</strong><br /><br /> <?php if (isset($error)) { ?> <small style="color:#aa0000;"><?php echo $error; ?></small><br /><br /> <?php } ?> <form action="episode.php?id=<?php echo $data['cast_id']; ?>" method="post" autocomplete="off" enctype="multipart/form-data"> <input type="text" name="name" placeholder="Name" /> / <input type="text" name="email" placeholder="Email" /><small style="color:#aa0000;">*</small><br /><br /> <textarea rows="10" cols="50" name="post" placeholder="Comment"></textarea><br /><br /> <input type="submit" onclick="myFunction()" value="Add Comment" /> <br /><br /> <small style="color:#aa0000;">* <b>Email will not be displayed publicly</b></small><br /> </form> </div> Include.php class feedback { public function fetch_all(){ global $pdo; $query = $pdo->prepare("SELECT * FROM comments"); $query->bindValue(1, $cast); $query->execute(); return $query->fetchAll(); } } This code updates to the database as it is suppose to. But after submission it reloads the current page as mentioned in the form action. But when I refresh the page to see the comment being added it asks to re submit. If I hit submit then the comment adds again. How can I stop this from happening? Maybe I could hide the comment box and display a thank you message but that would not stop a repeat entry. Please help. Thank you. Kev

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  • get value from database based on array in codeigniter

    - by Developer
    I have an array $user = array([0]=>1 [1]=>2 [2]=>3) which contains id's of certain users. I need to get the countries of these users from database. foreach($userid as $user){ $this->db->select('country'); $this->db->where('user_id',$user); $this->db->from('company'); $usercountry = $this->db->get(); $count = $usercountry->row(); $country = $count->country; } Suppose user1 has country ES, user2 has IN, user3 has US, user4 has UK. then if array contains 1,2,3. Then i need to get the countries ES,IN,US.

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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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  • Can't use method return value in write context; Not sure where to go from here

    - by Morgan Green
    This is my source for the variable. <?php if ($admin->get_permissions()=3) echo 'Welcome to the Admin Panel'; else echo 'Sorry, You do not have access to this page'; ?> And the code that I'm actually trying to call with the if statement is: public function get_permissions() { $username = $_SESSION['admin_login']; global $db; $info = $db->get_row("SELECT `permissions` FROM `user` WHERE `username` = '" . $db->escape($username) . "'"); if(is_object($info)) return $info->permissions; else return ''; } This should be a simple way to call my pages that the user is authorized for by using an else if statement. Or So I thought

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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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