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  • Database Design for One to One relationships

    - by Greelmo
    I'm trying to finalize my design of the data model for my project, and am having difficulty figuring out which way to go with it. I have a table of users, and an undetermined number of attributes that apply to that user. The attributes are in almost every case optional, so null values are allowed. Each of these attributes are one to one for the user. Should I put them on the same table, and keep adding columns when attributes are added (making the user table quite wide), or should I put each attribute on a separate table with a foreign key to the user table. I have decided against using the EAV model. Thanks!

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  • Trouble creating a SQL query

    - by JoBu1324
    I've been thinking about how to compose this SQL query for a while now, but after thinking about it for a few hours I thought I'd ask the SO community to see if they have any ideas. Here is a mock up of the relevant portion of the tables: contracts id date ar (yes/no) term payments contract_id payment_date The object of the query is to determine, per month, how many payments we expect, vs how many payments we received. conditions for expecting a payment Expected payments begin on contracts.term months after contracts.date, if contracts.ar is "yes". Payments continue to be expected until the month after the first missed payment. There is one other complication to this: payments might be late, but they need to show up as if they were paid on the date expected. The data is all there, but I've been having trouble wrapping my head around the SQL query. I am not an SQL guru - I merely have a decent amount of experience handling simpler queries. I'd like to avoid filtering the results in code, if possible - but without your help that may be what I have to do. Expected Output Month Expected Payments Received Payments January 500 450 February 498 478 March 234 211 April 987 789 ... SQL Fiddle I've created an SQL Fiddle: http://sqlfiddle.com/#!2/a2c3f/2

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  • storing image_id after uploading image in article table

    - by Bader
    According to this question i successed to create upload image , but now i need to store the image_id to another table called articles , i do not know if this is correct , but i tried to select the image_id from table image like this $select_image=mysql_query("select image_id from image where image_name = $fileName") or die(mysql_error()); and fetch the result to my article insert query like this $fetch=mysql_fetch_array($select_image); $qeuery=mysql_query("insert into articles (article_name,article_category,article_subcategory,article_body,article_summary,article_tags,article_photo,article_timedate) values ('$article_title','$CategoryID','$ProductID','$article_body','$article_summary','$fetch[image_id]','$time')") or die ('Error, Query Faild'.mysql_error()); is this correct ? the mysql_error keeps saying " Unknown column 'Penguins.jpg' in 'where clause'"

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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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  • Form validation

    - by kielie
    Hi guys, I need to create a form that has many of the same fields, that have to be inserted into a database, but the problem I have is that if a user only fills in one or two of the rows, the form will still submit the blank data of the empty fields along with the one or two fields the user has filled in. How can I check for the rows that have not been filled in and leave them out of the query? or check for those that have been filled in and add them to the query. . . The thank_you.php file will capture the $_POST variables and add them to the database. <form method="post" action="thank_you.php"> Name: <input type="text" size="28" name="name1" /> E-mail: <input type="text" size="28" name="email1" /> <br /> Name: <input type="text" size="28" name="name2" /> E-mail: <input type="text" size="28" name="email2" /> <br /> Name: <input type="text" size="28" name="name3" /> E-mail: <input type="text" size="28" name="email3" /> <br /> Name: <input type="text" size="28" name="name4" /> E-mail: <input type="text" size="28" name="email4" /> <input type="image" src="images/btn_s.jpg" /> </form> I am assuming that I could use javascript or jQuery to accomplish this, how would I go about doing this? Thanx in advance for the help.

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  • How to retrieve column total when rows are paginated?

    - by Rick
    Hey guys I have a column "price" in a table and I used a pagination script to generate the data displayed. Now the pagination is working perfectly however I am trying to have a final row in my HTML table to show the total of all the price. So I wrote a script to do just that with a foreach loop and it sort of works where it does give me the total of all the price summed up together however it is the sum of all the rows, even the ones that are on following pages. How can I retrieve just the sum of the rows displayed within the pagination? Thank you! Here is the query.. SELECT purchase_log.id, purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM purchase_log LEFT JOIN cart_contents ON purchase_log.id = cart_contents.purchase_id LEFT JOIN members ON purchase_log.member_id = members.id GROUP BY id ORDER BY id DESC LIMIT 0,30";

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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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  • PHP MySQLi isn't letting me alter a table (adding a new column)

    - by asdasd
    Well thats pretty much it. This is my query: $query = 'ALTER TABLE permissions ADD '.$name.' INT NOT NULL DEFAULT \'0\''; Where $name is already checked to exist with only lower case alpha letters, and not more than 20 length. Im just starting this out with very simple names. The next 4 lines of code after that one are: if($stmt = $db -> prepare($query)) { $success = $stmt -> execute(); $stmt -> close(); if(!$success) echo 'ERROR: Unsuccessful query: ',$db->error,PHP_EOL; } And I get back, every time ERROR: Unsuccessful query: And no error message. Is there a way to get more error messages so I can see what is failing? I can add new columns through phpmyadmin, but that really doesnt help me at all. The $db is fine, i do lots of stuff before and after this one section. It is only adding new column to the table that fails. side question: prepare() rejected my query every time when i tried to make those 2 variables, the $name and the 0 value as ? ? prepared statement values. Thats why they are in the real query and not bound later. If i could change that too I would like that.

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  • Resetting AUTO_INCREMENT on myISAM without rebuilding the table

    - by Artem
    Please help I am in major trouble with our production database. I had accidentally inserted a key with a very large value into an autoincrement column, and now I can't seem to change this value without a huge rebuild time. "ALTER TABLE tracks_copy AUTO_INCREMENT = 661482981" Is super-slow. How can I fix this in production? I can't get this to work either (has no effect): myisamchk tracks.MYI --set-auto-increment=661482982 Any ideas? Basically, no matter what I do I get an overflow: SHOW CREATE TABLE tracks CREATE TABLE tracks ( ... ) ENGINE=MYISAM AUTO_INCREMENT=2147483648 DEFAULT CHARSET=latin1

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  • JOIN two tables to show already purchased items

    - by Norbert
    I have a table where I keep all my templates: templates template_id template_name template_price These templates can be purchased by a registered user and then are inserted in the payments table: payments payment_id template_id user_id Is there a way to join these two tables and get not just a list of templates that have been purchased by a certain user, but all the templates? And then figure out from there which ones have already been purchased? I used this SELECT, but only the ones that the user bought showed up. I would like to have all the rows from templates, but empty in case the user_id doesn't match. SELECT * FROM templates LEFT JOIN payments ON templates.template_id = payments.template_id WHERE user_id = 2 GROUP BY templates.template_id

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  • Does a SELECT happen all at once, or progressively

    - by AmbroseChapel
    I have a process which finds a list of files to be deleted using a SELECT wheredelete= 'Y'. I set this process running the other day but it takes a while because it actually does the file deletions too. And in the middle of its long operation, I was using the application and deleted one more file. At this point I realised I didn't know if that file would be deleted, because I didn't know if the SELECT would have found all the files at the start, or if it was finding them progressively and would get to my newly-deleted file eventually.

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  • How do I get the median/mode/range of a column in SQL using Java?

    - by Derek
    I have to get the median, mode and range of test scores from one column in a table but I am unsure how to go about doing that. When you connect to the database using java, you are normally returned a ResultSet that you can make a table or something out of but how do you get particular numbers or digits? Is there an SQL command to get the median/mode/range or will I have to calculate this myself, and how do you pull out numbers from the table in order to be able to calculate the mode/median/range? Thanks.

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  • Having a Link Only Appear If a Logged-In User Appears on a Dynamic List

    - by John
    Hello, For the function below, I would like the link <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> to only appear if the logged in user currently appears on editorlist.php. (I. e. if the loginid in the function corresponds to any of the usernames that currently appear in editorlist.php.) Appearing on editorlist.php is something that is dynamic. How can I do this? Thanks in advance, John function show_userbox() { // retrieve the session information $u = $_SESSION['username']; $uid = $_SESSION['loginid']; // display the user box echo '<div id="userbox"> <div class="username">'.$u.'</div> <div class="submit"><a href="http://www...com/.../submit.php">Submit an item.</a></div> <div class="changepassword"><a href="http://www...com/.../changepassword.php">Change Password</a></div> <div class="logout"><a href="http://www...com/.../logout.php">Logout</a></div> <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> </div>'; } On editorlist.php: $sqlStr = "SELECT l.loginid, l.username, l.created, DATEDIFF(NOW(), l.created) AS days, COALESCE(s.total, 0) AS countSubmissions, COALESCE(c.total, 0) AS countComments, COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore, DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2 FROM login l LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM submission GROUP BY loginid ) s ON l.loginid = s.loginid LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM comment GROUP BY loginid ) c ON l.loginid = c.loginid GROUP BY l.loginid ORDER BY totalScore2 DESC LIMIT 10"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec1edit\">"; while ($row = mysql_fetch_array($result)) { echo '<tr>'; echo '<td class="sitename1edit1"><a href="http://www...com/.../members/index.php?profile='.$row["username"].'">'.stripslashes($row["username"]).'</a></td>'; echo '<td class="sitename1edit2">'.($row["countSubmissions"]).'</td>'; echo '<td class="sitename1edit2">'.($row["countComments"]).'</td>'; echo '<td class="sitename1edit2">'.($row["days"]).'</td>'; echo '<td class="sitename1edit2">'.($row["totalScore2"]).'</td>'; echo '</tr>'; } echo "</table>";

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  • How add column order in careers table???

    - by Mahran Elneel
    this table i want to create and how assign job to first position??? job_id dynamic Jobs Title text Job Description text Order combo box to choose after what job or at first position in the website i create this table and cannot choose first job to view in my website

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  • Rails find :conditions

    - by Sam
    I have a Reservation model that I'm searching for with three fields. The container_id must always be self.id but as confirmed and auto_confirmed only one needs to be true. I have the following but it doesn't preform what I need: Reservation.find(:all, :conditions => ['container_id = ? AND confirmed = ? OR auto_confirm = ?', self.id, true, true,]) How should I change this?

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  • PHP form validation submit problem

    - by TaG
    Every time I try to submit the form and I have not entered nothing in the year field I get Incorrect year! how can I still submit the form without having to enter a year. In other words leaving the year field blank and not getting a warning? Here is the PHP code. if(preg_match('/^\d{4,}$/', $_POST['year'])) { $year = mysqli_real_escape_string($mysqli, $_POST['year']); } else { $year = NULL; } if($year == NULL) { echo '<p class="error">Incorrect year!</p>'; } else { //do something }

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  • Passing dynamic string in hyperlink as parameter in jsp

    - by user3660263
    I am trying to pass a dynamic string builder variable in jsp I am generating a string through code. String Builder variable has some value but i am not able to pass it in at run time.It doesn't get the value. CODE FOR VARIABLE <% StringBuilder sb=new StringBuilder(""); if(request.getAttribute("Brand")!=null) { String Brand[]=(String[])request.getAttribute("Brand"); for(String brand:Brand) { sb.append("Brand="); sb.append(brand); sb.append("&&"); } } if(request.getAttribute("Flavour")!=null) { String Flavour[]=(String[])request.getAttribute("Flavour"); for(String flavour:Flavour) { sb.append(flavour); sb.append("&&"); } sb.trimToSize(); pageContext.setAttribute("sb", sb); } out.print("this is string"+sb); %> CODE FOR HYPERLINK <a href="Filter_Products?${sb}page=${currentPage + 1}" style="color: white;text-decoration: none;">Next</a></td>

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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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  • 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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  • Product Name Print Several times, How to fix.?

    - by mans
    i had added the following Opencart module for my order report list... http://www.opencart.com/index.php?route=extension/extension/info&extension_id=3597&filter_search=order%20list%20filter%20model&page=4 I have problems with the column "Products". If there are more than one option the products name prints several times. So if I got a product with three options the product name prints three times. Is there any way to fix this problem? i want print product name and model number only once, any idea.? i will attach the results what i got now... this is my sql query... public function getOrders($data = array()) { $sql = "select o.order_id,o.email,o.telephone,CONCAT(o.shipping_address_1, ' ', o.shipping_address_2) AS address,CONCAT(o.firstname, ' ', o.lastname) AS customer,o.payment_zone AS state,o.payment_address_2 AS block, o.payment_address_1 AS address,o.payment_postcode AS postcode,(SELECT os.name FROM " . DB_PREFIX . "order_status os WHERE os.order_status_id = o.order_status_id AND os.language_id = '" . (int)$this->config->get('config_language_id') . "') AS status,o.payment_city AS city,GROUP_CONCAT(pd.name) AS pdtname,GROUP_CONCAT(op.model) AS model,o.date_added,sum(op.quantity) AS quantity,GROUP_CONCAT(opt.value ) AS options, GROUP_CONCAT(opt.order_product_id ) AS ordprdid,GROUP_CONCAT(op.order_product_id ) AS optprdid, GROUP_CONCAT(op.quantity) AS opquantity from `" . DB_PREFIX . "order` o LEFT JOIN " . DB_PREFIX . "order_product op ON (op.order_id = o.order_id) LEFT JOIN " . DB_PREFIX . "product_description pd ON (pd.product_id = op.product_id and pd.language_id = '" . (int)$this->config->get('config_language_id') . "') LEFT JOIN " . DB_PREFIX . "order_option opt ON (opt.order_product_id = op.order_product_id) "; Product Name = GROUP_CONCAT(pd.name) AS pdtname,

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  • How to make an add friend/defriend function in PHP?

    - by user300371
    I have created a site where people can create a profile. But I am trying to figure out how to start on making an add friend button so users can have friends. In my user table, i have user_id, first_name, last_name, email, etc. Should I somehow relate the user_id of the user and the friend in a friend table? I am a novice to programming, so these things are still new to me. Thanks!

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