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  • checkbox checked with php form post?

    - by Patrick
    how do I check a checkbox? I've tried 1, On, Yes that doesn't work. putting the worked "checked" alone works but then how do I check with php after form post of the checkbox is checked? <input type="checkbox" class="inputcheckbox" id="newmsg" name=chk[newmsg2] value="1" />

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  • Get the sum by comparing between two tables

    - by Ismail Gunes
    I have to tables ProdBiscuit As tb and StockData As sd , I have to get the sum of the quantity in StockData (quantite) with the condition of if (sd.status0 AND sd.prodid = tb.id AND sd.matcuisine = 3) Here is my sql query SELECT tb.id, tb.nom, tb.proddate, tb.qty, tb.stockrecno FROM ProdBiscuit AS tb JOIN (SELECT id, prodid, matcuisine, status, SUM(quantite) AS rq FROM StockData) AS sd ON (tb.id = sd.prodid AND sd.status > 0 AND sd.matcuisine = 3) LIMIT 25 OFFSET @Myid This gives me no rows at all ? There is only 3 rows in ProdBiscuit and 11 rows in Stockdata and there is only 2 rows in StockData good with the condition. And as shown in the picture there is only two rows which give the condition. What is wrong in my query ? PS: The green lines on the image shows the condition in my query.

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  • I can't delete record in Codeigniter

    - by jomblo
    I'm learning CRUD in codeigniter. I have table name "posting" and the coloumns are like this (id, title, post). I successed to create a new post (both insert into database and display in the view). But I have problem when I delete my post in the front-end. Here is my code: Model Class Post_Model extends CI_Model{ function index(){ //Here is my homepage code } function delete_post($id) { $this->db->where('id', $id); $this->db->delete('posting'); } } Controller Class Post extends CI_Controller{ function delete() { $this->load->model('Post_Model'); $this->Post_Model->delete_post("id"); redirect('Post/index/', 'refresh'); } } After click "delete" in the homepage, there was nothing happens. While I'm looking into my database, my records still available. Note: (1) to delete record, I'm following the codeigniter manual / user guide, (2) I found a message error (Undefined variable: id) after hiting the "delete" button in the front-end Any help or suggestion, please

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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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  • Easy way to compute how close an auto_increment is to its maximum value?

    - by David M
    So yesterday we had a table that has an auto_increment PK for a smallint that reached its maximum. We had to alter the table on an emergency basis, which is definitely not how we like to roll. Is there an easy way to report on how close each auto_increment field that we use is to its maximum? The best way I can think of is to do a SHOW CREATE TABLE statement, parse out the size of the auto-incremented column, then compare that to the AUTO_INCREMENT value for the table. On the other hand, given that the schema doesn't change very often, should I store information about the columns' maximum values and get the current AUTO_INCREMENT with SHOW TABLE STATUS?

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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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  • 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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  • login/logout problem in PHP

    - by user356170
    i have a question. I have a website in which i am giving security like login id and password( as usual). No what i want is that, 1) I don't want to allow a single user to login in different machine at the same time. 2) For this i am using a column in database which is keeping the current status of user(i.e. loging/logout). I am allowing user to login only when has session has not closed and status is login. 3) So my problem is that when i am logging out manually. it is closing the session as well as updating the database with status "logout". 4) but when i am closing the window from Cross buttonat top right corner. it is closing the ssion but table data is still "login". so later on i can't be able to login into the same user. 5) So how could i solve this problem. Please help me!

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  • SELECT GROUP BY latest entry in INBOX msg sent by user

    - by Mohatir S
    i am building a simple PM for my personal website where a user can join and send message to the registered users on my website, i am stuck in grouping and selecting latest user (by latest msg time) in INBOX page my table : id msg_from msg_to msg date in_del out_del i want to show latest user's id each in a group here is my SQL query : SELECT ttalk.id, ttalk.msg_from, users.first_name, users.last_name FROM ttalk INNER JOIN users ON ttalk.msg_from = users.id WHERE ttalk.msg_to = '$_SESSION[user_id]' AND ttalk.in_del='0' GROUP BY ttalk.msg_from DESC LIMIT 500 Thanks in advance :-)

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  • MOD_REWRITE HELP!

    - by shahinkian
    I want to use mode rewrite to display the following: mydomain.com/Florida/Tampa/ instead of mydomain.com/place.php?state=Florida&city=Tampa I've akready done this: (since I think it might make a difference!) mydomain.com/[name].html instead of mydomain.com/profile?user=[name] Here is the code! Options +FollowSymLinks Options +Indexes RewriteEngine On RewriteBase / RewriteCond %{SCRIPT_FILENAME}! !-f RewriteCond %{SCRIPT_FILENAME}! !-d RewriteRule (.*).html profile.php?user=$1 [QSA.L]

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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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  • Does UNIQ constraint mean also an index on that field(s)?

    - by Gremo
    As title, should i defined a separate index on email column (for searching purposes) or the index is "automatically" added along with UNIQ_EMAIL_USER constraint? CREATE TABLE IF NOT EXISTS `customer` ( `id` int(11) NOT NULL AUTO_INCREMENT, `user_id` int(11) NOT NULL, `first` varchar(255) NOT NULL, `last` varchar(255) NOT NULL, `slug` varchar(255) NOT NULL, `email` varchar(255) NOT NULL, `created_at` datetime NOT NULL, `updated_at` datetime NOT NULL, PRIMARY KEY (`id`), UNIQUE KEY `UNIQ_SLUG` (`slug`), UNIQUE KEY `UNIQ_EMAIL_USER` (`email`,`user_id`), KEY `IDX_USER` (`user_id`) ) ENGINE=InnoDB;

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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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  • 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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  • 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!

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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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  • Problem with sending "SetCookie" first in php code

    - by Camran
    According to this manual: http://us2.php.net/setcookie I have to set the cookie before anything else. Here is my cookie code: if (isset($_COOKIE['watched_ads'])){ $expir = time()+1728000; //20 days $ad_arr = unserialize($_COOKIE['watched_ads']); $arr_elem = count($ad_arr); if (in_array($ad_id, $ad_arr) == FALSE){ if ($arr_elem>10){ array_shift($ad_arr); } $ad_arr[]=$ad_id; setcookie('watched_ads', serialize($ad_arr), $expir, '/'); } } else { $expir = time()+1728000; //20 days $ad_arr[] = $ad_id; setcookie('watched_ads', serialize($ad_arr), $expir, '/'); } As you can see I am using variables in setting the cookie. The variables comes from a mysql_query and I have to do the query first. But then, if I do, I will get an error message: Cannot modify header information - headers already sent by ... The error points to the line where I set the cookie above. What should I do?a

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  • Is this a secure way to structure a mysql_query in PHP

    - by Supernovah
    I have tried and tried to achieve an SQL injection by making custom queries to the server outside of firefox. Inside the php, all variables are passed into the query in a string like this. Note, by this stage, $_POST has not been touched. mysql_query('INSERT INTO users (password, username) VALUES(' . sha1($_POST['password']) . ',' . $_POST['username'] . ')); Is that a secure way to make a change?

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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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  • SQL with codition on calculated value

    - by user619893
    I have a table with products, their amount and their price. I need to select all entries where the average price per article is between a range. My query so far: SELECT productid,AVG(SUM(price)/SUM(amount)) AS avg FROM stock WHERE avg=$from AND avg<=$to GROUP BY productid If do this, it tells me avg doesnt exist. Also i obviously need to group by because the sum and average need to be per wine

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