Search Results

Search found 34274 results on 1371 pages for 'mysql table'.

Page 206/1371 | < Previous Page | 202 203 204 205 206 207 208 209 210 211 212 213  | Next Page >

  • Query to find all bars that sell three different beers at the same price

    - by Eternal Learner
    Query to find "All bars that sell three different beers at the same price?" My Tables are Sells(bar,beer,price) - bar - foreign Key.. Bars(name,addr) - name primary key. I thought of something like this but that dosent seem to work ... Select A.bar As bar , B.bar as bar From Sells AS A, Sells AS B Where A.bar = B.bar and A.beer <> B.beer Group By(A.beer) Having Count(Distinct A.beer) >= 2 Is this the correct SQL query ?

    Read the article

  • Evaluate column value into rows

    - by Hugo Palma
    I have a column whose value is a json array. For example: [{"att1": "1", "att2": "2"}, {"att1": "3", "att2": "4"}, {"att1": "5", "att2": "6"}] What i would like is to provide a view where each element of the json array is transformed into a row and the attributes of each json object into columns. Keep in mind that the json array doesn't have a fixed size. Any ideas on how i can achieve this ?

    Read the article

  • How to insert form data into MySQL database table

    - by Richard
    So I have this registration script: The HTML: <form action="register.php" method="POST"> <label>Username:</label> <input type="text" name="username" /><br /> <label>Password:</label> <input type="text" name="password" /><br /> <label>Gender:</label> <select name="gender"> <optgroup label="genderset"> <option value="Male">Male</option> <option value="Female">Female</option> <option value="Hermaphrodite">Hermaphrodite</option> <option value="Not Sure!!!">Not Sure!!!</option> </optgroup> </select><br /> <input type="submit" value="Register" /> </form> The PHP/SQL: <?php $username = $_POST['username']; $password = $_POST['password']; $gender = $_POST['gender']; mysql_query("INSERT INTO registration_info (username, password, gender) VALUES ('$username', '$password', '$gender') ") ?> The problem is, the username and password gets inserted into the "registration_info" table just fine. But the Gender input from the select drop down menu doesn't. Can some one tell me how to fix this, thanks.

    Read the article

  • How can this SQL be wrong? What am I not seeing?

    - by ropstah
    Can anybody please spot my error, this should be a legal query in SQL shouldn't it?? Unknown column u.usr_auto_key in the ON clause This is the database schema: User: (usr_auto_key, name, etc...) Setting: (set_auto_key, name etc..) User_Setting: (usr_auto_key, set_auto_key, value) And this is the query... SELECT `u`.`usr_auto_key` AS `u__usr_auto_key`, `s`.`set_auto_key` AS `s__set_auto_key`, `u2`.`usr_auto_key` AS `u2__usr_auto_key`, `u2`.`set_auto_key` AS `u2__set_auto_key`, `u2`.`value` AS `u2__value` FROM `User` `u`, `Setting` `s` LEFT JOIN `User_Setting` `u2` ON `u`.`usr_auto_key` = `u2`.`usr_auto_key` WHERE (`s`.`sct_auto_key` = 1 AND `u`.`usr_auto_key` = 1 AND admin_property is null)

    Read the article

  • one query instead of four - is it possible?

    - by Syom
    i must get data from four tables. i wrote the script with four queries, but i use it in ajax, and i wan't to do it by one query. here is queries... $query1 = "SELECT `id`,`name_ar` FROM `tour_type` ORDER BY `order`"; $query2 = "SELECT `id`,`name_ar` FROM `hotel_type` ORDER BY `order`"; $query3 = "SELECT `id`,`name_ar` FROM `food_type` ORDER BY `order`"; $query4 = "SELECT `id`,`name_ar` FROM `cities` WHERE `id_parrent` = '$id_parrent' ORDER BY `name_ar`"; is it possible to write in one query? thanks

    Read the article

  • Example of user-defined integrity rule in database systems?

    - by Pavel
    Hey everyone. I'm currently preparing for my exams and would like to know some examples of user-defined integrity rule in database systems. As far as I understand, it means that I can set up certain conditions for the columns and when data is inserted it needs to fulfill these conditions. For example: if I set up a rule that an ID needs to consist of 5 integers ONLY then when I insert a row with ID which is made up of integers and some chars then it won't accept it and return an error. Could someone confirm and give me some opinion on that? Thank you very much in advance!

    Read the article

  • how do I copy value from one table and inserted to another in the same database??

    - by mathew
    I am having a tough time to do this... I have created two table say table-1 and table-2 in same database.what I want is I need to copy some values from table-1 and insert the same to table-2. I have tried many ways but it does not seems work. below is my code can any one tell me where I am missing?? $db = mysql_connect("localhost", "user", "pass") or die("Could not connect."); mysql_select_db("comdata",$db)or die(mysql_error()); $resultb = mysql_query("SELECT * FROM table-2")or die(mysql_error()); $row = mysql_fetch_array($resultb); $days = (strtotime(date("Y-m-d")) - strtotime($row['regtime'])) / (60 * 60 * 24); if($row > 0 && $days < 1){ $person = $row['person']; $catogr = $row['catog']; $position = $row['position']; $location = $row['location']; $rank = $row['rank']; mysql_close($db); }else{ $db = mysql_connect("localhost", "user", "pass") or die("Could not connect."); mysql_select_db("comdata",$db)or die(mysql_error()); $result = mysql_query("SELECT * FROM table-1 WHERE regtime = DATE(NOW()) ORDER BY rank ASC LIMIT 1;")or die(mysql_error()); $row = mysql_fetch_array($result); $person = $row['person']; $catogr = $row['catog']; $position = $row['position']; $location = $row['location']; $rank = $row['rank']; mysql_query("INSERT INTO table-2 (regtime,person,catog,position,location,rank) VALUES(NOW(),'$person','$catogr','$position','$location','$rank')"); mysql_close($db); }

    Read the article

  • NHibernate and MySql is inserting and Selecting, not updating

    - by Chris Brandsma
    Something strange is going on with NHibernate for me. I can select, and I can insert. But I can't do and update against MySql. Here is my domain class public class UserAccount { public virtual int Id { get; set; } public virtual string UserName { get; set; } public virtual string Password { get; set; } public virtual bool Enabled { get; set; } public virtual string FirstName { get; set; } public virtual string LastName { get; set; } public virtual string Phone { get; set; } public virtual DateTime? DeletedDate { get; set; } public virtual UserAccount DeletedBy { get; set; } } Fluent Mapping public class UserAccountMap : ClassMap<UserAccount> { public UserAccountMap() { Table("UserAccount"); Id(x => x.Id); Map(x => x.UserName); Map(x => x.Password); Map(x => x.FirstName); Map(x => x.LastName); Map(x => x.Phone); Map(x => x.DeletedDate); Map(x => x.Enabled); } } Here is how I'm creating my Session Factory var dbconfig = MySQLConfiguration .Standard .ShowSql() .ConnectionString(a => a.FromAppSetting("MySqlConnStr")); FluentConfiguration config = Fluently.Configure() .Database(dbconfig) .Mappings(m => { var mapping = m.FluentMappings.AddFromAssemblyOf<TransactionDetail>(); mapping.ExportTo(mappingdir); }); and this is my NHibernate code: using (var trans = Session.BeginTransaction()) { var user = GetById(userId); user.Enabled = false; user.DeletedDate = DateTime.Now; user.UserName = "deleted_" + user.UserName; user.Password = "--removed--"; Session.Update(user); trans.Commit(); } No exceptions are being thrown. No queries are being logged. Nothing.

    Read the article

  • Drupal - db_fetch_array returns NULL for every row

    - by darudude
    I'm using Drupal's db_fetch_array to fetch rows from my db_query. However, every row returned is equal to NULL. Typing the query into PHP myadmin works, so I have no idea whats going on. db_num_rows returns the number of rows as well. Here is the code: if(count($rebuild_ids)) { $ids=implode(",",$rebuild_ids); $type_stmt = "SELECT * from {" . ItemType::$type_table_name . "} where id IN ($ids)"; $new_items=db_query($type_stmt); if(!$new_items || db_num_rows($new_items) == 0) { return; } while($row = db_fetch_array($new_items)); { if ($row!=NULL) { echo "I work!" $game_items[] = $row['id']; ItemInstance::$nid_to_item_type_code[$row['nid']] = $row['id']; } } } However, it never gets into the third if statement (i.e. never echos "I work!") Any ideas?

    Read the article

  • restrict duplicate rows in specific columns in mysql

    - by JPro
    I have a query like this : select testset, count(distinct results.TestCase) as runs, Sum(case when Verdict = "PASS" then 1 else 0 end) as pass, Sum(case when Verdict <> "PASS" then 1 else 0 end) as fails, Sum(case when latest_issue <> "NULL" then 1 else 0 end) as issues, Sum(case when latest_issue <> "NULL" and issue_type = "TC" then 1 else 0 end) as TC_issues from results join testcases on results.TestCase = testcases.TestCase where platform = "T1_PLATFORM" AND testcases.CaseType = "M2" and testcases.dummy <> "flag_1" group by testset order by results.TestCase The result set I get is : testset runs pass fails issues TC_issues T1 66 125 73 38 33 T2 18 19 16 16 15 T3 57 58 55 55 29 T4 52 43 12 0 0 T5 193 223 265 130 22 T6 23 12 11 0 0 My problem is, this is a result table which has testcases running multiple times. So, I am able to restrict the runs using the distinct TestCases but when I want the pass and fails, since I am using case I am unable to eliminate the duplicates. Is there any way to achieve what I want? any help please? thanks.

    Read the article

  • Similar SQL queries returning different results...

    - by Pablo
    Here are the SQL Queries: $sql1 = "SELECT count(thread) AS total FROM comments WHERE thread=1 AND parent_id=0 "; $sql2 = "SELECT count(thread) AS total FROM comments, users WHERE thread=1 AND parent_id=0 AND users.user_id=comments.user_id "; $sql3 = "SELECT comments.*, users.username AS username FROM comments, users WHERE thread=1 AND parent_id=0 AND users.user_id=comments.user_id ORDER BY date LIMIT 10, 5 "; My question is why would $sql1 and $sql2 would return two different results? $sql1 returns 61 rows $sql2 returns 56 rows The 5th line in $sql2 is just for testing, is not required, is just a variation of $sql1 which gets the total rows for a pagination.

    Read the article

  • How to return null value if the query has no corresponding value?

    - by Holicreature
    Hi i've a query select c.name as companyname, u.name,u.email,u.role,a.date from useraccount u, company c, audittrial a where u.status='active' and u.companyid=c.id and (u.companyid=a.companyID and a.activity like 'User activated%' and a.email=u.email) order by u.companyid desc limit 10 So if the following part doesnt't satisfy, (u.companyid=a.companyID and a.activity like 'User activated%' and a.email=u.email) no rows will be returned.. but i want to return the result of the following query select c.name as companyname, u.name,u.email,u.role,a.date from useraccount u, company c, audittrial a where u.status='active' and u.companyid=c.id order by u.companyid desc limit 10 but to add that, i should return the date if available and return null value if date is not available.. how can i do this?

    Read the article

  • jQuery Reference First Column in HTML Table

    - by Vic
    I have a table where all of the cells are INPUT tags. I have a function which looks for the first input cell and replaces it with it's value. So this: <tr id="row_0" class="datarow"> <td><input class="tabcell" value="Injuries"></td> <td><input class="tabcell" value="01"></td> becomes this: <tr id="row_0" class="datarow"> <td>Injuries</td> <td><input class="tabcell" value="01"></td> Here is the first part of the function: function setRowLabels() { var row = []; $('.dataRow').each(function(i) { row.push($('td input:eq(0)', this).val() + ' - '); $('td input:eq(0)', this).replaceWith($('td input:eq(0)', this).val()); $('td input:gt(0)', this).each(function(e) { etcetera But when the page reloads, the first column is not an input type, so it changes the second column to text too! Can I tell it to only change the first column, no matter what the type is? I tried $('td:eq(0)', this).replaceWith($('td:eq(0)', this).val()); but it does not work. Any suggestions appreciated! Thanks

    Read the article

  • Export CSV from Mysql

    - by ss888
    Hi, I'm having a bit of trouble exporting a csv file that is created from one of my mysql tables using php. The code I'm using prints the correct data, but I can't see how to download this data in a csv file, providing a download link to the created file. I thought the browser was supposed to automatically provide the file for download, but it doesn't. (Could it be because the below code is called using ajax?) Any help greatly appreciated - code below, S. include('../config/config.php'); //db connection settings $query = "SELECT * FROM isregistered"; $export = mysql_query ($query ) or die ( "Sql error : " . mysql_error( ) ); $fields = mysql_num_fields ( $export ); for ( $i = 0; $i < $fields; $i++ ) { $header .= mysql_field_name( $export , $i ) . "\t"; } while( $row = mysql_fetch_row( $export ) ) { $line = ''; foreach( $row as $value ) { if ( ( !isset( $value ) ) || ( $value == "" ) ) { $value = "\t"; } else { $value = str_replace( '"' , '""' , $value ); $value = '"' . $value . '"' . "\t"; } $line .= $value; } $data .= trim( $line ) . "\n"; } $data = str_replace( "\r" , "" , $data ); if ( $data == "" ) { $data = "\n(0) Records Found!\n"; } //header("Content-type: application/octet-stream"); //have tried all of these at sometime //header("Content-type: text/x-csv"); header("Content-type: text/csv"); //header("Content-type: application/csv"); header("Content-Disposition: attachment; filename=export.csv"); //header("Content-Disposition: attachment; filename=export.xls"); header("Pragma: no-cache"); header("Expires: 0"); echo '<a href="">Download Exported Data</a>'; //want my link to go in here... print "$header\n$data";

    Read the article

  • MSYQL ~ Why does this query only select a single row?

    - by Joe
    SELECT * FROM tbl_houses WHERE (SELECT HousesList FROM tbl_lists WHERE tbl_lists.ID = '123') LIKE CONCAT('% ', tbl_houses.ID, '#') ^ It only selects the row from tbl_houses of the last occuring "tbl_houses.ID" inside tbl_lists.HousesList I need it to select ALL the rows where any ID from tbl_houses exists within tbl_lists.HousesList

    Read the article

  • php pconnect vs connect

    - by user192344
    if i have a script which insert a data then exit the script will be opened by 100 user at same time or within 2 mins actually im doing email tracking so pconnect is bettwe or connect is better to reduce the resource i have close when after insert

    Read the article

  • Is this a bad indexing strategy for a table?

    - by llamaoo7
    The table in question is part of a database that a vendor's software uses on our network. The table contains metadata about files. The schema of the table is as follows Metadata ResultID (PK, int, not null) MappedFieldname (char(50), not null) Fieldname (PK, char(50), not null) Fieldvalue (text, null) There is a clustered index on ResultID and Fieldname. This table typically contains millions of rows (in one case, it contains 500 million). The table is populated by 24 workers running 4 threads each when data is being "processed". This results in many non-sequential inserts. Later after processing, more data is inserted into this table by some of our in-house software. The fragmentation for a given table is at least 50%. In the case of the largest table, it is at 90%. We do not have a DBA. I am aware we desperately need a DB maintenance strategy. As far as my background, I'm a college student working part time at this company. My question is this, is a clustered index the best way to go about this? Should another index be considered? Are there any good references for this type and similar ad-hoc DBA tasks?

    Read the article

  • PHP/MySQL Interview - How would you have answered?

    - by martincarlin87
    I was asked this interview question so thought I would post it here to see how other users would answer: Please write some code which connects to a MySQL database (any host/user/pass), retrieves the current date & time from the database, compares it to the current date & time on the local server (i.e. where the application is running), and reports on the difference. The reporting aspect should be a simple HTML page, so that in theory this script can be put on a web server, set to point to a particular database server, and it would tell us whether the two servers’ times are in sync (or close to being in sync). This is what I put: // Connect to database server $dbhost = 'localhost'; $dbuser = 'xxx'; $dbpass = 'xxx'; $dbname = 'xxx'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die (mysql_error()); // Select database mysql_select_db($dbname) or die(mysql_error()); // Retrieve the current time from the database server $sql = 'SELECT NOW() AS db_server_time'; // Execute the query $result = mysql_query($sql) or die(mysql_error()); // Since query has now completed, get the time of the web server $php_server_time = date("Y-m-d h:m:s"); // Store query results in an array $row = mysql_fetch_array($result); // Retrieve time result from the array $db_server_time = $row['db_server_time']; echo $db_server_time . '<br />'; echo $php_server_time; if ($php_server_time != $db_server_time) { // Server times are not identical echo '<p>Database server and web server are not in sync!</p>'; // Convert the time stamps into seconds since 01/01/1970 $php_seconds = strtotime($php_server_time); $sql_seconds = strtotime($db_server_time); // Subtract smaller number from biggest number to avoid getting a negative result if ($php_seconds > $sql_seconds) { $time_difference = $php_seconds - $sql_seconds; } else { $time_difference = $sql_seconds - $php_seconds; } // convert the time difference in seconds to a formatted string displaying hours, minutes and seconds $nice_time_difference = gmdate("H:i:s", $time_difference); echo '<p>Time difference between the servers is ' . $nice_time_difference; } else { // Timestamps are exactly the same echo '<p>Database server and web server are in sync with each other!</p>'; } Yes, I know that I have used the deprecated mysql_* functions but that aside, how would you have answered, i.e. what changes would you make and why? Are there any factors I have omitted which I should take into consideration? The interesting thing is that my results always seem to be an exact number of minutes apart when executed on my hosting account: 2012-12-06 11:47:07 2012-12-06 11:12:07

    Read the article

  • login not working when changing from mysql to mysqli

    - by user1438647
    I have a code below where it logs a teacher in by matching it's username and password in the database, if correct, then log in, if incorrect, then display a message. <?php session_start(); $username="xxx"; $password="xxx"; $database="mobile_app"; $link = mysqli_connect('localhost',$username,$password); mysqli_select_db($link, $database) or die( "Unable to select database"); foreach (array('teacherusername','teacherpassword') as $varname) { $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : ''; } ?> <form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm"> <p>Username</p><p><input type="text" name="teacherusername" /></p> <!-- Enter Teacher Username--> <p>Password</p><p><input type="password" name="teacherpassword" /></p> <!-- Enter Teacher Password--> <p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p> </form> <?php if (isset($_POST['submit'])) { $query = " SELECT * FROM Teacher t WHERE (t.TeacherUsername = '".mysqli_real_escape_string($teacherusername)."') AND (t.TeacherPassword = '".mysqli_real_escape_string($teacherpassword)."') "; $result = mysqli_query($link, $query); $num = mysqli_num_rows($result); $loged = false; while($row = mysqli_fetch_array($result)) { if ($_POST['teacherusername'] == ($row['TeacherUsername']) && $_POST['teacherpassword'] == ($row['TeacherPassword'])) { $loged = true; } $_SESSION['teacherforename'] = $row['TeacherForename']; $_SESSION['teachersurname'] = $row['TeacherSurname']; $_SESSION['teacherusername'] = $row['TeacherUsername']; } if ($loged == true){ header( 'Location: menu.php' ) ; }else{ echo "The Username or Password that you Entered is not Valid. Try Entering it Again."; } mysqli_close($link); } ?> Now the problem is that even if the teacher has entered in the correct username and password, it still doesn't let the teacher log in. When the code above was the old mysql() code, it worked fine as teacher was able to login when username and password match, but when trying to change the code into mysqli then it causes login to not work even though username and password match. What am I doing wrong?

    Read the article

  • The [2] table entry '[3]' has no associated entry in the Media table. (error 2602)

    - by derekf
    Coworker started getting the above message in the event log and as dialog during install.  Argument [2] was File and argument [3] was a specific file. Error dialog read   Product: (app name) -- The installer has encountered an unexpected error installing this package. This may indicate a problem with this package. The error code is 2602. Package was a vendor-provided MSI that had been installed administratively, and then a patch (.msp) applied to the administrative install point. With some digging we found that the MSI still had the entries in the media table pointing at the CAB files, and that there were several files at the end of the sequence that did not have corresponding entries in the Media table (last sequence 990 in Media table, last entry in File table had sequence 994).  Attributes on files in the File table all had the msidbFileAttributesCompressed (&16384) attribute set, so they were all expecting to be within the CAB files, but since this was an admin install there were no CAB files. Resolved by clearing the Media table (replace with a single entry: Disk ID 1, LastSequence 994) and going through the file table and subtracting 8192 from each entry to mark files as not compressed.  Tested and worked.

    Read the article

  • Sort MySQL query result by a alphanumeric field

    - by Jason Shultz
    I'm querying a table in a db using php. one of the fields is a column called "rank" and has data like the following: none 1-bronze 2-silver 3-gold ... 10-ambassador 11-president I want to be able to sort the results based on that "rank" column. any results where the field is "none" get excluded, so those don't factor in. As you can already guess, right now the results are coming back like this: 1-bronze 10-ambassador 11-president 2-silver 3-gold Of course, I would like for it to be sorted so it is like the following: 1-bronze 2-silver 3-gold ... 10-ambassador 11-president Right now the query is being returned as an object. I've tried different sort options like natsort, sort, array_multisort but haven't got it to work the way I'm sure it can. I would prefer keeping the results in an object form if possible. I'm passing the data on to a view in the next step. although, it's perfectly acceptable to pass the object to the view and then do the work there. so it's not an issue after all. :) thank you for your help. i'm hoping I'm making sense.

    Read the article

  • What does `?` stand for in SQL?

    - by user295189
    I have this SQL by a programmer: $sql = " INSERT INTO `{$database}`.`table` ( `my_id`, `xType`, `subType`, `recordID`, `textarea` ) VALUES ( {$my_id}, ?xType, ?subType, {$recordID}, ?areaText ) "; My question is why is he using ? before values? How do I see what values are coming in? I did echo and it shows ?xType as ?xType. No values. What does ? stand for in SQL?

    Read the article

< Previous Page | 202 203 204 205 206 207 208 209 210 211 212 213  | Next Page >