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  • PHP Code Problem...

    - by aamir Fayyaz
    function check_login($array_val) { $strQury = "Select * from tblsignup where usr_email ='".$array_val[0]."' and usr_password = '".$array_val[1]."'" ; $result = mysql_query($strQury); $row_user = mysql_fetch_array($result); if(mysql_num_rows($result)>0) { $msg = "true"; } else { $msg = "false"; } return $msg ; } The return value is Object id #1true???? what is object id#1?

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  • help me with the following sql query

    - by rupeshmalviya
    could somebody correct my following query, i am novice to software development realm, i am to a string builder object in comma separated form to my query but it's not producing desired result qyery is as follows and string cmd = "SELECT * FROM [placed_student] WHERE passout_year=@passout AND company_id=@companyId AND course_id=@courseId AND branch_id IN('" + sb + "')"; StringBuilder sb = new StringBuilder(); foreach (ListItem li in branch.Items) { if (li.Selected == true) { sb.Append(Convert.ToInt32(li.Value) +", "); } } li is integer value of my check box list which are getting generated may be differne at different time ...please also suggest me some good source to learn sql..

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  • Looping Through Database Query

    - by DrakeNET
    I am creating a very simple script. The purpose of the script is to pull a question from the database and display all answers associated with that particular question. We are dealing with two tables here and there is a foreign key from the question database to the answer database so answers are associated with questions. Hope that is enough explanation. Here is my code. I was wondering if this is the most efficient way to complete this or is there an easier way? <html> <head> <title>Advise Me</title> <head> <body> <h1>Today's Question</h1> <?php //Establish connection to database require_once('config.php'); require_once('open_connection.php'); //Pull the "active" question from the database $todays_question = mysql_query("SELECT name, question FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Variable to hold $todays_question aQID $questionID = mysql_query("SELECT commentID FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Print today's question echo $todays_question; //Print comments associated with today's question $sql = "SELECT commentID FROM approvedQuestions WHERE status = active"; $result_set = mysql_query($sql); $result_num = mysql_numrows($result_set); for ($a = 0; $a < $result_num; $a++) { echo $sql; } ?> </body> </html>

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  • Is it possible to integrate user databases between Drupal and an ASP&SQL Server platform?

    - by hecatomber
    We have a game project designed on ASP&SQL Server, and we need to integrate it's user database with Drupal. This would be easier from Project to Drupal (since there is user_save and user_delete functions available globally by using drupal bootstrap) but I'm not sure if we can execute PHP functions on an ASP platform. Is there any documentation for this kind of problems? What do you suggest?

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  • Zend Framework Multiple Table Query

    - by Jeff
    I am looking to execute this statement via Zend Framework. As I understand it, I can use Zend_Db_Select. Is it possible to use Zend_Db_Table? Three tables: classes, students, and class_students select classes.name, students.student_id, students.fname, students.lname from students, classes, class_students where class_students.student_id=students.student_id AND class_students.class_id=classes.class_id;

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  • Binary string search on one field.

    - by CrazyJoe
    I have 300 boolean fields in one table, and im trying to do somithing like that: One string field: 10000010000100100100100100010001 Ha a simple way to do a simple search os this field like: select * from table where field xor "10000010000100100100000000010001" Im tring this but is to long: select * from teste where mid(info,2,1) and mid(info,3,1) :) Help!!

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  • How to know record has been updated successfully in php

    - by Lisa Ray
    $sql = "UPDATE...."; if(mysql_query($sql)) { $_SESSION['Trans']="COMMIT"; header("location:result.php"); exit; } else { $_SESSION['Trans']="FAIL"; $_SESSION['errors'] = "Error: Sorry! We are unable to update your Profile, Please contact to PNP HelpDesk."; header("location:result.php"); exit; }//end IF data is getting updated then why compiler is not coming inside IF condition.

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  • How to skip an empty LIKE operator in a multiple LIKE query?

    - by alex
    I notice my query doesn't behave correctly if one of the like variables is empty: SELECT name FROM employee WHERE name LIKE '%a%' AND color LIKE '%A%' AND city LIKE '%b%' AND country LIKE '%B%' AND sport LIKE '%c%' AND hobby LIKE '%C%' Now when a and A are not empty it works but when a, A and c are not empty the c part is not excuted so it seems? How can I fix this?

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  • login/logout problem in PHP

    - by user356170
    i have a question. I have a website in which i am giving security like login id and password( as usual). No what i want is that, 1) I don't want to allow a single user to login in different machine at the same time. 2) For this i am using a column in database which is keeping the current status of user(i.e. loging/logout). I am allowing user to login only when has session has not closed and status is login. 3) So my problem is that when i am logging out manually. it is closing the session as well as updating the database with status "logout". 4) but when i am closing the window from Cross buttonat top right corner. it is closing the ssion but table data is still "login". so later on i can't be able to login into the same user. 5) So how could i solve this problem. Please help me!

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  • How do you display a PHPmyadmin table onto a web browser/web page?

    - by user1390754
    Just a simple question, i've been searching around and for some reason i cannot find an answer to this. after creating a database/table in Phpmyadmin using xampp, what command do i need to put into my webpage (PHP) to show the table I made? I know the first step involves connecting to the database and i think i've done that properly. This is the code i found from somewhere about connecting (w3 tutorials I believe) $con = mysql_connect("localhost","xxxxxx,""); if (!$con) { die('Could not connect: ' . mysql_error()); }

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  • num_rows is 0 when it should be >0 for php mysqli code

    - by jpporterVA
    My num_rows is coming back as 0, and I've tried calling it several ways, but I'm stuck. Here is my code: $conn = new mysqli($dbserver, "dbuser", "dbpass", $dbname); // get the data $sql = 'SELECT AT.activityName, AT.createdOn FROM userActivity UA, users U, activityType AT WHERE U.userId = UA.userId and AT.activityType = UA.activityType and U.username = ? order by AT.createdOn'; $stmt = $conn->stmt_init(); $stmt->prepare($sql); $stmt->bind_param('s', $requestedUsername); $stmt->bind_result($activityName, $createdOn); $stmt->execute(); // display the data $numrows = $stmt->num_rows; $result=array("user activity report for: " . $requestedUsername . " with " . $numrows . " rows:"); $result[]="Created On --- Activity Name"; while ($stmt->fetch()) { $msg = " " . $createdOn . " --- " . $activityName . " "; $result[] = $msg; } $stmt->close(); There are multiple rows found, and the fetch loop process them just fine. Any suggestions on what will enable me to get the number of rows returned in the query? Suggestions are much appreciated. Thanks in advance.

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  • Creating dynamic icons based on data entered into database from django forms

    - by John Hoke
    So I'm using Django to create a projects page with multiple forms for each project. Let's call them form 1, 2, 3, and 4. Once you create a project you can fill out any of these forms. I want to create "buttons" or links for each one of the forms that would show up on the main page. Now this is the part I need help with: Step 1. I want it so that if you click on a button for a form (say form 1) and none exists for that project yet a pop up would come up saying "This form does not exist yet, are you sure you want to create one?". And if you'd answer yes you would be directed to the form page. Step 2. But if that form does exist, I don't want any pop up to open and I want the link to take the user directly to that page. Step 3. My next problem is this. These forms are in order, so if you didn't create form 1 but created form 2, I don't want to give the user access to form 1. So in this scenario, if you click on form 1 I want a pop up to open and say "This form can no longer be created", and the link wouldn't function anymore. Basically the button will have 3 function. First it should look at the database and if data for that specific form exists it should do "Step 2", if data for that form and the proceeding forms don't exist it should do "Step 1", and if data for that form doesn't exist but data for proceeding form's does exist is should do "Step 3". Is this possible? Please help as I need to find a solution to this soon. Any help would be highly appreciated. Thank you

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  • Stop Submit With empty and error Input Values Using PHP

    - by user3615781
    I am using the following code for sending data to database, but it sends the data even the values of the fields are incorrect or empty. So can anyone help me solve this by using php? Here is my code: <?php //Connecting to sql db $connect = mysqli_connect("localhost","root","","form"); /* check connection */ if (!$connect) { die('Connect Error: ' . mysqli_connect_error()); } //Sending data to sql db $result = mysqli_query($connect,"INSERT INTO students(name,email,website,comment,gender) VALUES('$name','$email','$website','$comment','$gender')"); if (!$result) { die('Query Error:'. mysqli_error($connect)); } ?>

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  • check if a table exsist in where

    - by Luca Romagnoli
    This query generates an error because table2 doesn't exist: Select * FROM table WHERE table2.id IS NOT NULL Is there anything like this for check the table2 before apply the check on the id? Select * FROM table WHERE (EXIST(table2) AND table2.id IS NOT NULL) or not EXIST(table2) Thanks

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  • How can I fake sql data while preserving statements without commenting my server-side code?

    - by Fedor
    I have to use hardcoded values for certain fields because at this moment we don't have access to the real data. When we do get access, I don't want to go through a lot of work uncommenting. Is it possible to keep this statement the way it is, except use '25' as the alias for ratecode? IF(special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode, I have about 8 or so IF statements similar to this and I'm just too lazy ( even with vim ) to re-append while commenting out each if statement line by line. I would have to do this: $sql = 'SELECT u.*,'; // IF ( special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode $sql.= '25 AS ratecode';

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  • MULTIPLE CRITERIA TABLE JOIN

    - by user1447203
    I have a table listing clothing items (shirt, trousers, etc) named . Each item is identified with a unique CLOTHING.CLOTHING_ID. So a blue shirt is 01, a flowery shirt is 12 and jeans are 07 say. I have a second table identifying outfits with a column for shirts, for trousers, shoes etc. For example Outfit 1: shirt 01, trousers 07 (i.e. blue shirt with jeans) Outfit 2: shirt 12, trousers 07 (so flowery shirt with jeans). This table is named and each outfit is unique with OUTFIT_LIST.OUTFIT_ID. I want to produce a select statement that will list each outfit's contents, i.e. find the clothing specified in Outfit 1. Any help would be very much appreciated, and apologies in advance if I am missing a very simple solution. I have been playing with JOINS of all descriptions and CONCATS and so on with now luck - I am very new to this. Thanks.

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  • TV Guide script - getting current date programmes to show

    - by whitstone86
    This is part of my TV guide script: //Connect to the database mysql_connect("localhost","root","PASSWORD"); //Select DB mysql_select_db("mytvguide"); //Select only results for today and future $result = mysql_query("SELECT programme, channel, episode, airdate, expiration, setreminder FROM mediumonair where airdate >= now()"); The episodes show up, so there are no issues there. However, it's getting the database to find data that's the issue. If I add a record for a programme that airs today this should show: Medium showing on TV4 8:30pm "Episode" Set Reminder Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder but this shows instead: Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder I almost have the SQL working; just not sure what the right code is here, to avoid the second mistake showing - as the record (which indicates a show currently airing) does not seem to work at present. Please can anyone help me with this? Thanks

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  • Allow users to pull temporary data then delete table?

    - by JM4
    I don't know the best way to title this question but am trying to accomplish the following goal: When a client logs into their profile, they are presented with a link to download data from an existing database in CSV format. The process works, however, I would like for this data to be 'fresh' each time they click the link so my plan was - once a user has clicked the link and downloaded the CSV file, the database table would 'erase' all of its data and start fresh (be empty) until the next set of data populated it. My EXISTING CSV creation code: <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'database'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= '"'.$rowr[$j].'",'; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> any ideas?

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  • Help constructing query - Compare columns and replace numbers

    - by Tommy
    I have a feeling that this query is pretty easy to construct, I just can't figure it out. I want to replace all numbers in table X column C, with numbers in table Z column A, where numbers from table X column C matches numbers in table Z column B. I hope that makes sense. Perhaps a little background information will make it clearer. I've converted from one CMS to another, and the module I used to convert mapped the ids to the new database. Table X column A is the new id's. Table X column B is the old id's. Table Z is the table for an image gallery that I migrated, and column C contains the id's of the images owners. Can anyone crack this nut?

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  • Php random row help...

    - by Skillman
    I've created some code that will return a random row, (well, all the rows in a random order) But i'm assuming its VERY uneffiecent and is gonna be a problem in a big database... Anyone know of a better way? Here is my current code: $count3 = 1; $count4 = 1; //Civilian stuff... $query = ("SELECT * FROM `*Table Name*` ORDER BY `Id` ASC"); $result = mysql_query($query); while($row = mysql_fetch_array($result)) { $count = $count + 1; $civilianid = $row['Id']; $arrayofids[$count] = $civilianid; //echo $arrayofids[$count]; } while($alldone != true) { $randomnum = (rand()%$count) + 1; //echo $randomnum . "<br>"; //echo $arrayofids[$randomnum] . "<br>"; $currentuserid = $arrayofids[$randomnum]; $count3 += 1; while($count4 < $count3) { $count4 += 1; $currentarrayid = $listdone[$count4]; //echo "<b>" . $currentarrayid . ":" . $currentuserid . "</b> "; if ($currentarrayid == $currentuserid){ $found = true; //echo " '" .$found. "' "; } } if ($found == true) { //Reset array/variables... $count4 = 1; $found = false; } else { $listdone[$count3] = $currentuserid; //echo "<u>" . $count3 .";". $listdone[$count3] . "</u> "; $query = ("SELECT * FROM `*Tablesname*` WHERE Id = '$currentuserid'"); $result = mysql_query($query); $row = mysql_fetch_array($result); $username = $row['Username']; echo $username . "<br>"; $count4 = 1; $amountdone += 1; if ($amountdone == $count) { //$count $alldone = true; } } } Basically it will loop until its gets an id (randomly) that hasnt been chosen yet. -So the last username could take hours :P Is this 'bad' code? :P :(

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  • Select past date from database x days from now

    - by Pr0no
    Consider the following table daterange _date trading_day ------------------------ 2011-08-01 1 2011-07-31 0 2011-07-30 0 2011-07-29 1 2011-07-28 1 2011-07-27 1 2011-07-26 1 2011-07-25 1 2011-07-24 0 2011-07-23 0 2011-07-22 1 2011-07-21 1 2011-07-20 1 2011-07-19 1 2011-07-18 1 2011-07-17 0 I'm in need of a query that returns a _date, x days before a given _date. When counting back, _days with trading_day = 0 should be ignored. A few examples: input | output -------------------------+------------ 1 day before 2011-07-19 | 2011-07-18 2 days before 2011-08-01 | 2011-07-28 (trading_day = 0 don't count) 3 days before 2011-07-29 | 2001-07-26 The first one is easy: SELECT _date FROM daterange WHERE trading_day = 0 AND _date < '2011-07-19' LIMIT 1 But I don't know how to query for the other examples. Do you?

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