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  • How to construct this query? (Ordering by COUNT() and joining with users table)

    - by Andrew
    users table: id-user-other columns scores table: id-user_id-score-other columns They're are more than one rows for each user, but there's only two scores you can have. (0 or 1, == win or loss). So I want to output all the users ordered by the number of wins, and all the users ordered by the numbers of losses. I know how to do this by looping through each user, but I was wondering how to do it with one query. Any help is appreciated!

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  • What does MySqlDataAdapter.Fill return when the results are empty?

    - by Brian
    I have a 'worker' function that will be processing any and all sql queries in my program. I will need to execute queries that return result sets and ones that just execute stored procedures without any results. Is this possible with MySqlDataAdapter.Fill or do I need to use the MySqlCommand.ExecuteNonQuery() method? Here is my 'worker' function for reference: private DataSet RunQuery(string SQL) { MySqlConnection connection; MySqlCommand command; MySqlDataAdapter adapter; DataSet dataset = new DataSet(); lock(locker) { connection = new MySqlConnection(MyConString); command = new MySqlCommand(); command = connection.CreateCommand(); command.CommandText = SQL; adapter = new MySqlDataAdapter(); adapter.Fill(dataset); } return dataset; }

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  • Find the % character in a LIKE query

    - by Jensen
    Hi, I've an SQL database and I would like to do a query who show all the datas containing the sign "%". Normally, to find a character (for example: "z") in a database I use a query like this : mysql_query("SELECT * FROM mytable WHERE tag LIKE '%z%'"); But here, I want to found the % character, but in SQL it's a joker so when I write: mysql_query("SELECT * FROM mytable WHERE tag LIKE '%%%'"); It show me all my datas. So how to found the % character in my SQL datas ? Thanks

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  • change password code error.....

    - by shimaTun
    I've created a code to change a password. Now it seem contain an error.before i fill the form. the page display the error message: Parse error: parse error, unexpected $end in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 222 this the code: <?php # userChangePass.php //this page allows logged in user to change their password. $page_title='Change Your Password'; //if no first_name variable exists, redirect the user if(!isset($_SESSION['nameuser'])){ header("Location: http://" .$_SERVER['HTTP_HOST']. dirname($_SERVER['PHP_SELF'])."/index.php"); ob_end_clean(); exit(); }else{ if(isset($_POST['submit'])) {//handle form. require_once('connectioncomplaint.php'); //connec to the database //check for a new password and match againts the confirmed password. if(eregi ("^[[:alnum:]]{4,20}$", stripslashes(trim($_POST['password1'])))){ if($_POST['password1'] == $_POST['password2']){ $p =escape_data($_POST['password1']); }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Your password did not match the confirmed password!</font></p>'; } }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Please Enter a valid password!</font></p>'; } if($p){ //if everything OK. //make the query $query="UPDATE access SET password=PASSWORD('$p') WHERE userid={$_SESSION['userid']}"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; //include('templates/footer.inc');//include the HTML footer. exit(); }else{//if it did not run ok $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } mysql_close();//close the database connection. }else{//failed the validation test. echo '<p><font color="red" size="+1"> Please try again.</font></p>'; } }//end of the main Submit conditional. ?> And code for form: <h1>Change Your Password</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <fieldset> <p><b>New Password:</b><input type="password" name="password1" size="20" maxlength="20" /> <small>Use only letters and numbers.Must be between 4 and 20 characters long.</small></p> <p><b>Confirm New Password:</b><input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form-->

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  • change password code error

    - by ejah85
    I've created a code to change a password. Now it seem contain an error. When I fill in the form to change password, and click save the error message: Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 I really don’t know what the error message means. Please guys. Help me fix it. Here's is the code: <?php session_start(); ?> <?php # change password.php //set the page title and include the html header. $page_title = 'Change Your Password'; //include('templates/header.inc'); if(isset($_POST['submit'])){//handle the form require_once('connectioncomplaint.php');//connect to the db. //include "connectioncomplaint.php"; //create a function for escaping the data. function escape_data($data){ global $dbc;//need the connection. if(ini_get('magic_quotes_gpc')){ $data=stripslashes($data); } return mysql_real_escape_string($data, $dbc); }//end function $message=NULL;//create the empty new variable. //check for a username if(empty($_POST['userid'])){ $u=FALSE; $message .='<p> You forgot enter your userid!</p>'; }else{ $u=escape_data($_POST['userid']); } //check for existing password if(empty($_POST['password'])){ $p=FALSE; $message .='<p>You forgot to enter your existing password!</p>'; }else{ $p=escape_data($_POST['password']); } //check for a password and match againts the comfirmed password. if(empty($_POST['password1'])) { $np=FALSE; $message .='<p> you forgot to enter your new password!</p>'; }else{ if($_POST['password1'] == $_POST['password2']){ $np=escape_data($_POST['password1']); }else{ $np=FALSE; $message .='<p> your new password did not match the confirmed new password!</p>'; } } if($u && $p && $np){//if everything's ok. $query="SELECT userid FROM access WHERE (userid='$u' AND password=PASSWORD('$p'))"; $result=@mysql_query($query); $num=mysql_num_rows($result); if($num == 1){ $row=mysql_fetch_array($result, MYSQL_NUM); //make the query $query="UPDATE access SET password=PASSWORD('$np') WHERE userid=$row[0]"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; include('templates/footer.inc');//include the HTML footer. exit();//quit the script. }else{//if it did not run OK. $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } }else{ $message= '<p> Your username and password do not match our records.</p>'; } mysql_close();//close the database connection. }else{ $message .='<p>Please try again.</p>'; } }//end oh=f the submit conditional. //print the error message if there is one. if(isset($message)){ echo'<font color="red">' , $message, '</font>'; } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <body> <script language="JavaScript1.2">mmLoadMenus();</script> <table width="604" height="599" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td height="130" colspan="7"><img src="images/banner(E-Complaint)-.jpg" width="759" height="130" /></td> </tr> <tr> <td width="100" height="30" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="160" bgcolor="#ABD519"> <?php include "header.php"; ?>&nbsp;</td> </tr> <tr> <td colspan="7" bgcolor="#FFFFFF"> <fieldset><legend> Enter your information in the form below:</legend> <p><b>User ID:</b> <input type="text" name="username" size="10" maxlength="20" value="<?php if(isset($_POST['userid'])) echo $_POST['userid']; ?>" /></p> <p><b>Current Password:</b> <input type="password" name="password" size="20" maxlength="20" /></p> <p><b>New Password:</b> <input type="password" name="password1" size="20" maxlength="20" /></p> <p><b>Confirm New Password:</b> <input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form--> </td> </tr> </table> </body> </html>

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  • group by query issue

    - by user319088
    gorup by query issue i have one table, which has three fields and data. Name , Top , total cat , 1 ,10 dog , 2, 7 cat , 3 ,20 hourse 4, 4 cat, 5,10 Dog 6 9 i want to select record which has highest value of "total" for each Name so my result should be like this. Name , Top , total cat , 3 , 20 hourse , 4 , 4 Dog , 6 , 9 i tried group by name order by total, but it give top most record of group by result. any one can guide me , please!!!!

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  • How to add condition on multiple-join table

    - by Jean-Philippe
    Hi, I have those two tables: client: id (int) #PK name (varchar) client_category: id (int) #PK client_id (int) category (int) Let's say I have those datas: client: {(1, "JP"), (2, "Simon")} client_category: {(1, 1, 1), (2, 1, 2), (3, 1, 3), (4,2,2)} tl;dr client #1 has category 1, 2, 3 and client #2 has only category 2 I am trying to build a query that would allow me to search multiple categories. For example, I would like to search every clients that has at least category 1 and 2 (would return client #1). How can I achieve that? Thanks!

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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • Overriding unique indexed values

    - by Yeti
    This is what I'm doing right now (name is UNIQUE): SELECT * FROM fruits WHERE name='apple'; Check if the query returned any result. If yes, don't do anything. If no, a new value has to be inserted: INSERT INTO fruits (name) VALUES ('apple'); Instead of the above is it ok to insert the value into the table without checking if it already exists? If the name already exists in the table, an error will be thrown and if it doesn't, a new record will be inserted. Right now I am having to insert 500 records in a for loop, which results in 1000 queries. Will it be ok to skip the "already-exists" check?

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  • What database field should this be?

    - by alex
    I have a table called "Event". A column has "duration"--which is...how long that event lasts. What database type should this be? Integer? (in seconds)? In the future, I will do statements such as: If now() > event.duration THEN don't display the event.

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  • PHP drop down which each are dependable

    - by user147685
    Hi all, I have this problems. using html and php. May I know how to do this. I have 2 drop down, eg A and B. Drop down B is depend to the drop down A. Example, A have these options which will be called from dbase(no prob with this, tq) (Jack, Carol), and B wil have options depend on A: if select Jack(T1, T2, T3), if select carol(T1,T2,T3,T4,T5). Here are the sample interface. Can someone help me with this? thank you.

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  • question with its query

    - by user329820
    Hi this is my homework and the question is this: List the average balance of customers by city and short zip code (the first five digits of thezip code). Only include customers residing in Washington State (‘WA’). also the Customer table has 5 columns(Name,Family,CustZip,CustCity,CustAVGBal) I wrote the query like below is this correct? SELECT CustCity,LEFT(CustZip,5) AS NewCustZip,CustAVGBal FROM Customer WHERE CustCity = 'WA' THANKS!!

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  • 500 internal server error at form connection

    - by klox
    hi..all..i've a problem i can't connect to database what's wrong with my code?this is my code: $("#mod").change(function() { var barcode; barCode=$("#mod").val(); var data=barCode.split(" "); $("#mod").val(data[0]); $("#seri").val(data[1]); var str=data[0]; var matches=str.match(/(EE|[EJU]).*(D)/i); $.ajax({ type:"post", url:"process1.php", data:"value="+matches+"action=tunermatches", cache:false, async:false, success: function(res){ $('#rslt').replaceWith( "<div id='value'><h6>Tuner range is" + res + " .</h6></div>" ); } }); }); and this is my process file: switch(postVar('action')) { case 'tunermatches' : tunermatches(postVar('tuner')); break; function tunermatches($tuner)){ $Tuner=mysql_real_escape_string($tuner); $sql= "SELECT remark FROM settingdata WHERE itemname="Tuner_range" AND itemdata="$Tunermatches"; $res=mysql_query($sql); $dat=mysql_fetch_array($res,MYSQL_NUM); if($dat[0]>0) { echo $dat[0]; } mysql_close($dbc); }

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  • Enhancing an 'ORDER BY' clause to judge condition by more than 1 integer

    - by Yvonne
    Hi folks, I have some PHP code which allows me to sort a column into ascending and descending order (upon click of a table row title), which is good. It works perfectly for my D.O.B colum (with date/time field type), but not for a quantity column. For example, I have quantites of 10, 50, 100, 30 and another 100. The order seems to be only appreciating the 1st integer, so my sorting of the column ends up in this order: 10, 100, 100, 30, 50... and 50, 30, 100, 100, 10. This is obviously incorrect as 100 is bigger than 50, therefore both 100 values should appear at the end surely? It seems to me that 100 is only being taken into account as having the '1' value, then it appears before 10 because the system recognises it has another 0. Is this normal to happen? Is there any way I can solve this problem? Thanks for any help. P.S. I can show code if necessary, but would like to know if this is a common issue by default.

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  • need help on my query.

    - by Dharmendra
    i have one table : nobel(yr, subject, winner) and i have this query : In which years was the Physics prize awarded but no Chemistry prize. this is what i tried : select distinct yr from nobel where subject='physics' and subject!='chemistry' but is not working where i am going wrong. see, i am not here to make my homework from someone. i am here to learn something. so, please give me suggetion.

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  • Need help with many-to-many relationships....

    - by yuudachi
    I have a student and faculty table. The primary key for student is studendID (SID) and faculty's primary key is facultyID, naturally. Student has an advisor column and a requested advisor column, which are foreign key to faculty. That's simple enough, right? However, now I have to throw in dates. I want to be able to view who their advisor was for a certain quarter (such as 2009 Winter) and who they had requested. The result will be a table like this: Year | Term | SID | Current | Requested ------------------------------------------------ 2009 | Winter | 860123456 | 1 | NULL 2009 | Winter | 860445566 | 3 | NULL 2009 | Winter | 860369147 | 5 | 1 And then if I feel like it, I could also go ahead and view a different year and a different term. I am not sure how these new table(s) will look like. Will there be a year table with three columns that are Fall, Spring and Winter? And what will the Fall, Spring, Winter table have? I am new to the art of tables, so this is baffling me... Also, I feel I should clarify how the site works so far now. Admin can approve student requests, and what happens is that the student's current advisor gets overwritten with their request. However, I think I should not do that anymore, right?

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  • Wordpress inserting comments via wp_insert_comment()

    - by Cyber Junkie
    Hello all happy holidays! :) I'm trying to insert comments in my wordpress blog via the wp_insert_comment() function. It's for a plugin I'm trying to make. I have this code in my header for testing. It works every time I refresh the page. $agent = $_SERVER['HTTP_USER_AGENT']; $data = array( 'comment_post_ID' => 256, 'comment_author' => 'Dave', 'comment_author_email' => '[email protected]', 'comment_author_url' => 'http://www.someiste.com', 'comment_content' => 'Lorem ipsum dolor sit amet...', 'comment_author_IP' => '127.3.1.1', 'comment_agent' => $agent, 'comment_date' => date('Y-m-d H:i:s'), 'comment_date_gmt' => date('Y-m-d H:i:s'), 'comment_approved' => 1, ); $comment_id = wp_insert_comment($data); It successfully inserts comments into the database. The problem: Comments don't show via the Disqus comment system. I compared table rows and I noticed that user_agent differs. Normal comments use for example, Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv... and Disqus comments use Disqus/1.1(2.61):119598902 numbers are different for each comment. Does anyone know how to insert comments with wp_insert_comment() when Disqus is enabled?

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  • Accessing data entered into multiple Django forms and generating them onto a new URL

    - by pedjk
    I have a projects page where users can start up new projects. Each project has two forms. The two forms are: class ProjectForm(forms.Form): Title = forms.CharField(max_length=100, widget=_hfill) class SsdForm(forms.Form): Status = forms.ModelChoiceField(queryset=P.ProjectStatus.objects.all()) With their respective models as follows: class Project(DeleteFlagModel): Title = models.CharField(max_length=100) class Ssd(models.Model): Status = models.ForeignKey(ProjectStatus) Now when a user fills out these two forms, the data is saved into the database. What I want to do is access this data and generate it onto a new URL. So I want to get the "Title" and the "Status" from these two forms and then show them on a new page for that one project. I don't want the "Title" and "Status" from all the projects to show up, just for one project at a time. If this makes sense, how would I do this? I'm very new to Django and Python (though I've read the Django tutorials) so I need as much help as possible. Thanks in advance Edit: The ProjectStatus code is (under models): class ProjectStatus(models.Model): Name = models.CharField(max_length=30) def __unicode__(self): return self.Name

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  • getting sql records

    - by droidus
    when i run this code, it returns the topic fine... $query = mysql_query("SELECT topic FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; } but when I change it to this, it doesn't run at all. why is this happening? $query = mysql_query("SELECT topic, lock FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; $lockedThread = $row['lock']; echo "here: " . $lockedThread; }

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