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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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  • select distinct over specific columns

    - by Midhat
    A query in a system I maintain returns QID AID DATA 1 2 x 1 2 y 5 6 t As per a new requirement, I do not want the (QID, AID)=(1,2) pair to be repeated. We also dont care what value is selected from "data" column. either x or y will do. What I have done is to enclose the original query like this SELECT * FROM (<original query text>) Results group by QID,AID Is there a better way to go about this? The original query uses multiple joins and unions and what not, So I would prefer not to touch it unless its absolutely necesary

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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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  • 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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  • 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 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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  • 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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  • Should I change $_REQUEST to $_POST

    - by Scarface
    Hey guys quick question, I have a checkbox system where a list of items can be checked and deleted on the click of a button. I currently use request and it does the job but I was wondering if $_REQUEST was some sort of security risk or improper. If anyone has any advice I would appreciate it. Should I change to $_POST? If so, what is the best way to go about it? foreach ($_REQUEST as $key=>$value) { if (substr($key,0,3)==="img") { $id = substr($key,3); if(isset($_REQUEST['Delete'])) { $sql = 'SELECT file_name,username FROM images WHERE id=?'; $stmt = $conn->prepare($sql); $result=$stmt->execute(array($id)); while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { $image=$row['file_name']; $user=$row['username']; $myFile = "$user/images/$image"; unlink($myFile); } <input id=\"img".$id."\" name=\"img".$id."\" type=\"checkbox\">

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  • Can I do this in only one query ?

    - by Paté
    Merry christmas everyone, I Know my way around SQL but I'm having a hard time figuring this one out. First here are my tables (examples) User id name friend from //userid to //userid If user 1 is friend with user 10 then you a row with 1,10. User 1 cannot be friend with user 10 if user 10 is not friend with user 1 so you have 1,10 10,1 It may look weird but I need those two rows per relations. Now I'm trying to make a query to select the users that have the most mutual friend with a given user. For example User 1 is friend with user 10,9 and 7 and user 8 is friend with 10,9 and 7 too ,I want to suggest user 1 to invite him (like facebook). I want to get like the 10 first people with the most mutual friend. The output would be like User,NumOfMutualFriends I dont know if that can be done in a single query ? Thanks in advance for any help.

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  • What SQL query should I perform to get the result set expected?

    - by texai
    What SQL query should I perform to get the result set expected, giving the first element of the chain (2) as input data, or any of them ? table name: changes +----+---------------+---------------+ | id | new_record_id | old_record_id | +----+---------------+---------------+ | 1| 4| 2| | -- non relevant data -- | | 6| 7| 4| | -- non relevant data -- | | 11| 13| 7| | 12| 14| 13| | -- non relevant data -- | | 31| 20| 14| +----+---------------+---------------+ Result set expected: +--+ | 2| | 4| | 7| |13| |14| |20| +--+ I know I should consider change my data model, but: What if I couldn't? Thank you in advance!

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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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  • Is it expensive to hold on to PreparedStatements? (Java & JDBC)

    - by sbook
    I'm trying to figure out if it's efficient for me to cache all of my statements when I create my database connection or if I should only create those that are most used and create the others if/when they're needed.. It seems foolish to create all of the statements in all of the client threads. Any feedback would be greatly appreciated.

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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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  • 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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  • Tracking Votes and only allowing 1 vote per member

    - by MikeAdams
    What I'm trying to do is count the votes when someone votes on a "page". I think I lost myself trying to figure out how to track when a member votes or not. I can't seem to get the code to tell when a member has voted. //Generate code ID $useXID = intval($_GET['id']); $useXrank = $_GET['rank']; //if($useXrank!=null && $useXID!=null) { $rankcheck = mysql_query('SELECT member_id,code_id FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'" AND WHERE code_id="'.$useXID.'"'); if(!mysql_fetch_array($rankcheck) && $useXrank=="up"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit+="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } elseif(!mysql_fetch_array($rankcheck) && $useXrank=="down"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit-="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } // hide vote links since already voted elseif(mysql_fetch_array($rankcheck)){$voted="true";} //}

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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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  • Autopopulate from Select box from database

    - by Chris Spalton
    hope you can help, please forgive any poor coding or anytihng, I'm new to this and just hacking my way through to get things to work. That said, on one of my projects I have this code, which successfully populates the dropdown from a database when the page is loaded: <select name="Region" id="Region"> <option value="">-- Select Region --</option> <?php $region=$POST['Region']; if ($region); { $regionquery = "SELECT DISTINCT REGION FROM Sales_Execs "; $regionresult = mysql_query($regionquery); while($row = mysql_fetch_array($regionresult)) { echo "<option value=\"".$row['REGION']."\">".$row['REGION']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Region').value = <?php echo json_encode(trim($_POST['Region']));?>; </script> </select> On my next project that I'm working on now, I need to do the same thing, so I copied the above code amended, and placed in my new project: <select name="Sales_Exec" id="Sales_Exec"> <option value="">-- Select SE --</option> <?php $salesexec=$POST['Sales_Exec']; if ($salesexec); { $salesexecquery = "SELECT DISTINCT Assigned FROM Data "; $salesexecresult = mysql_query($salesexecquery); while($row = mysql_fetch_array($salesexecresult)) { echo "<option value=\"".$row['ASSIGNED']."\">".$row['ASSIGNED']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Sales_Exec').value = <?php echo json_encode(trim($_POST['Sales_Exec']));?>; </script> </select> This second chunk of code doesn't work... and I can't work out why as it seems I've copied it all and amended all the neccersary parts, can anyone spot what is wrong? Thankyou!

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  • PHP: How to get the days of the week?

    - by fwaokda
    I'm wanting to store items in my database with a DATE value for a certain day. I don't know how to get the current Monday, or Tuesday, etc. from the current week. Here is my current database setup. menuentry id int(10) PK menu_item_id int(10) FK day_of_week date message varchar(255) So I have a class setup that holds all the info then I was going to do something like this... foreach ( $menuEntryArray as $item ) { if ( $item->getDate() == [DONT KNOW WHAT TO PUT HERE] ) { // code to display menu_item information } } So I'm just unsure what to put in "[DONT KNOW WHAT TO PUT HERE]" to compare to see if the date is specified for this week's Monday, or Tuesday, etc. The foreach above runs for each day of the week - so it'll look like this... Monday Item 1 Item 2 Item 3 Tuesday Item 1 Wednesday Item 1 Item 2 ... Thanks!

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  • is Payment table needed when you have an invoice table like this?

    - by EBAGHAKI
    this is my invoice table: Invoice Table: invoice_id creation_date due_date payment_date status enum('not paid','paid','expired') user_id total_price I wonder if it's Useful to have a payment table in order to record user payments for invoices. payment table can be like this: payment_id payment_date invoice_id price_paid status enum('successful', 'not successful')

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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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  • What sort of schema can I use to accommodate manual date based data entries?

    - by meder
    I have an admin where users from multiple properties can enter in monthly statistics for twitter/facebook followers. We do not have access to the real data/db so this is why a manual entry. The form looks like this: Type ( radio, select **one** only ): - Twitter - Facebook Followers/Fans ( textfield ): Property (dropdown): Hotel A, Hotel B Date Start: mm/dd/yyyy (textfield) Date End: mm/dd/yyyy (textfield) Question 1.1: Since I am only keeping track of month per month, the date start/end fields which I have already created might be too specific. Would it be a better idea just to have a start month/year and and month/year if that's the only thing I care about? Question 1.2: What schema could I use for month to month statistics if I were to change the date start and end textfields to start month/year and end month/year dropdowns?

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