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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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  • 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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  • 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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  • 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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  • 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.

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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 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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  • 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 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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  • 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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  • 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 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 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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  • 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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  • 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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  • Is this SQL is valid?

    - by Beck
    UPDATE polls_options SET `votes`=`votes`+1, `percent`=ROUND((`votes`+1) / (SELECT voters FROM polls WHERE poll_id=? LIMIT 1) * 100,1) WHERE option_id=? AND poll_id=? Don't have table data yet, to test it properly. :) And by the way, in what type % integers should be stored in database? Thanks for the help!

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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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  • Error during data UPDATE in php

    - by Piyush
    $sql = "UPDATE tblprofile SET name = '$membername' , f_h_name = '$fathername', maritalS = '$mstatus' , dob = '$dob' , occupation = '$occupation' , nominee = '$nominee' , address1 = '$address1' , address2 = '$address2', city = '$city', district = '$district', state = '$state', pin = '$areapin', mobile = '$mobileno', email = '$email', PANno = '$panno', bankname = '$bankname', branch = '$branch', accountno = '$accountno' WHERE userId = '$_SESSION['UserId']' "; //line 212 if(mysql_query($sql)) { echo "Updation Done."; } Error comes in browser : Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\xampp\htdocs\303\saveEditProfile.php on line 212

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  • How can I remove this query from within a loop?

    - by Chris
    I am currently designing a forum as a personal project. One of the recurring issues I've come across is database queries in loops. I've managed to avoid doing that so far by using table joins or caching of data in arrays for later use. Right now though I've come across a situation where I'm not sure how I can write the code in such a way that I can use either of those methods easily. However I'd still prefer to do at most 2 queries for this operation rather than 1 + 1 per group of forums, which so far has resulted in 5 per page. So while 5 isn't a huge number (though it will increase for each forum group I add) it's the principle that's important to me here, I do NOT want to write queries in loops What I'm doing is displaying forum index groupings (eg admin forums, user forums etc) and then each forum within that group on a single page index, it's the combination of both in one page that's causing me issue. If it had just been a single group per page, I'd use a table join and problem solved. But if I use a table join here, although I can potentially get all the data I need it'll be in one mass of results and it needs displaying properly. Here's the code (I've removed some of the html for clarity) <?php $sql= "select * from forum_groups"; //query 1 $result1 = $database->query($sql); while($group = mysql_fetch_assoc($result1)) //first loop {?> <table class="threads"> <tr> <td class="forumgroupheader"> <?php echo $group['group_name']; ?> </td> </tr> <tr> <td class="forumgroupheader2"> <?php echo $group['group_desc']; ?> </td> </tr> </table> <table> <tr> <th class="thforum"> Forum Name</th> <th class="thforum"> Forum Decsription</th> <th class="thforum"> Last Post </th> <tr> <?php $group_id = $group['id']; $sql = "SELECT forums.id, forums.forum_group_id, forums.forum_name, forums.forum_desc, forums.visible_rank, forums.locked, forums.lock_rank, forums.topics, forums.posts, forums.last_post, forums.last_post_id, users.username FROM forums LEFT JOIN users on forums.last_post_id=users.id WHERE forum_group_id='{$group_id}'"; //query 2 $result2 = $database->query($sql); while($forum = mysql_fetch_assoc($result2)) //second loop {?> So how can I either a) write the SQL in such a way as to remove the second query from inside the loop or b) combine the results in an array either way I need to be able to access the data as an when so I can format it properly for the page output, ie within the loops still.

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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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  • 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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