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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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  • Difficulty with sql query

    - by João Madureira Pires
    I have the following tables: TableA (id, tableB_id, tableC_id) TableB (id, expirationDate) TableC (id, expirationDate) I want to retrieve all the results from TableA ordered by tableB.expirationDate and tableC.expirationDate. How can I do this?

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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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  • Performance with timestamp conditions

    - by Tim Whitlock
    Which of the following is faster, or are they equivalent? (grabbing recent most records from a TIMESTAMP COLUMN) SELECT UNIX_TIMESTAMP(`modified`) stamp FROM `some_table` HAVING stamp > 127068799 ORDER BY stamp DESC or SELECT UNIX_TIMESTAMP(`modified`) stamp FROM `some_table` WHERE UNIX_TIMESTAMP(`modified`) > 127068799 ORDER BY `modified` DESC or even another combination?

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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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  • 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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  • Determining the popularity of a video with ratings and views

    - by user295825
    I am about to embark on a new project - a video website. Users will be able to register, and vote on videos by clicking "like" or "dislike", or something to that effect. In any event, it will be a 2-option voting system, not a 5-star system. Every X number of days, I will be generating a "chart" of the most popular videos. So my question is: how should I determine the popularity of a given video? If I went the route of tallying up the videos with the most views, this could have the effect of exceptionally bad videos making it to the of the charts (just because they're so bad). If I go the route of a scoring system based on the amount of "like" and "dislike" votes (eg. 100 like votes, and 50 dislike votes equals a score of 2), videos with few views could appear on the top of the charts. So, what I need to do is a combination of the two. Barring, of course, spammy views and votes. What's your guys' thoughts on the subject?

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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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  • 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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  • Table format for Click stats

    - by Francesc
    Hello. I'm currently developing an URL shortening service. I want to allow users to see the stats for their URLs. How has to be the table. First, it has to be the url ID, but then, how I can sort the clicks per day?

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  • making sure "expiration_date - X" falls on a valid "date_of_price" (if not, use the next valid date_

    - by bobbyh
    I have two tables. The first table has two columns: ID and date_of_price. The date_of_price field skips weekend days and bank holidays when markets are closed. table: trading_dates ID date_of_price 1 8/7/2008 2 8/8/2008 3 8/11/2008 4 8/12/2008 The second table also has two columns: ID and expiration_date. The expiration_date field is the one day in each month when options expire. table: expiration_dates ID expiration_date 1 9/20/2008 2 10/18/2008 3 11/22/2008 I would like to do a query that subtracts a certain number of days from the expiration dates, with the caveat that the resulting date must be a valid date_of_price. If not, then the result should be the next valid date_of_price. For instance, say we are subtracting 41 days from the expiration_date. 41 days prior to 9/20/2008 is 8/10/2008. Since 8/10/2008 is not a valid date_of_price, we have to skip 8/10/2008. The query should return 8/11/2008, which is the next valid date_of_price. Any advice would be appreciated! :-)

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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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  • 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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  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

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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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  • Excessive use of Inner Join for more than 3 tables

    - by Archangel08
    Good Day, I have 4 tables on my DB (not the actual name but almost similar) which are the ff: employee,education,employment_history,referrence employee_id is the name of the foreign key from employee table. Here's the example (not actual) data: **Employee** ID Name Birthday Gender Email 1 John Smith 08-15-2014 Male [email protected] 2 Jane Doe 00-00-0000 Female [email protected] 3 John Doe 00-00-0000 Male [email protected] **Education** Employee_ID Primary Secondary Vocation 1 Westside School Westshore H.S SouthernBay College 2 Eastside School Eastshore H.S NorthernBay College 3 Northern School SouthernShore H.S WesternBay College **Employment_History** Employee_ID WorkOne StartDate Enddate 1 StarBean Cafe 12-31-2012 01-01-2013 2 Coffebucks Cafe 11-01-2012 11-02-2012 3 Latte Cafe 01-02-2013 04-05-2013 Referrence Employee_ID ReferrenceOne Address Contact 1 Abraham Lincoln Lincoln Memorial 0000000000 2 Frankie N. Stein Thunder St. 0000000000 3 Peter D. Pan Neverland Ave. 0000000000 NOTE: I've only included few columns though the rest are part of the query. And below are the codes I've been working on for 3 consecutive days: $sql=mysql_query("SELECT emp.id,emp.name,emp.birthday,emp.pob,emp.gender,emp.civil,emp.email,emp.contact,emp.address,emp.paddress,emp.citizenship,educ.employee_id,educ.elementary,educ.egrad,educ.highschool,educ.hgrad,educ.vocational,educ.vgrad,ems.employee_id,ems.workOne,ems.estartDate,ems.eendDate,ems.workTwo,ems.wstartDate,ems.wendDate,ems.workThree,ems.hstartDate,ems.hendDate FROM employee AS emp INNER JOIN education AS educ ON educ.employee_id='emp.id' INNER JOIN employment_history AS ems ON ems.employee_id='emp.id' INNER JOIN referrence AS ref ON ref.employee_id='emp.id' WHERE emp.id='$id'"); Is it okay to use INNER JOIN this way? Or should I modify my query to get the results that I wanted? I've also tried to use LEFT JOIN but still it doesn't return anything .I didn't know where did I go wrong. You see, as I have thought, I've been using the INNER JOIN in correct manner, (since it was placed before the WHILE CLAUSE). So I couldn't think of what could've possible went wrong. Do you guys have a suggestion? Thanks in advance.

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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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  • How do I replace NOT EXISTS with JOIN?

    - by YelizavetaYR
    I've got the following query: select distinct a.id, a.name from Employee a join Dependencies b on a.id = b.eid where not exists ( select * from Dependencies d where b.id = d.id and d.name = 'Apple' ) and exists ( select * from Dependencies c where b.id = c.id and c.name = 'Orange' ); I have two tables, relatively simple. The first Employee has an id column and a name column The second table Dependencies has 3 column, an id, an eid (employee id to link) and names (apple, orange etc). the data looks like this Employee table looks like this id | name ----------- 1 | Pat 2 | Tom 3 | Rob 4 | Sam Dependencies id | eid | Name -------------------- 1 | 1 | Orange 2 | 1 | Apple 3 | 2 | Strawberry 4 | 2 | Apple 5 | 3 | Orange 6 | 3 | Banana As you can see Pat has both Orange and Apple and he needs to be excluded and it has to be via joins and i can't seem to get it to work. Ultimately the data should only return Rob

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  • Printing All Entries in A PHP Table

    - by mgunawan
    I'm trying to insert a php excerpt with SQL (I understand this is outdated, but am trying to grasp the syntax first) into my HTML page, and I've got the following table: ID Name Element1 Element2 0 John John's 1st John's 2nd 1 Bill Bill's 1st Bill's 2nd 2 Steve Steven's 1st Steve's 2nd I'm trying to get the for loop that will essentially print out the following in my html page Name: Name where ID=0 Element1: Element1 where ID=0 Element2: Element2 where ID=0 Name: Name where ID=1 Element1: Element1 where ID=1 Element2: Element2 where ID=1 and so forth. Basically, I am trying to make this process automated so that whenever a new record is added into the table, the HTML page will automatically update with a new "profile". Thank you for your help!

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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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