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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • Error during data INSERT in php

    - by nectar
    here my code- $sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES "; for($i=0; $i0) { $sql .= ", "; } $sql .= "('$pin[$i]', '$ownerid', 'Free', '1')"; } $sql .= ";"; echo $sql; mysql_query($sql); if(mysql_affected_rows() 0) { echo "done"; } else { echo "Fail"; } output: ** INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES ('13837927', 'admin', 'Free', '1'), ('59576082', 'admin', 'Free', '1'); Fail why it is not inserting values when $sql query is right?

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  • sql count function

    - by suryll
    Hi I have three tables and I want to know how much jobs with the wage of 1000 an employee has had The first SQL query gives me the names of all the employees that has recieved 1000 for a job SELECT distinct first_name FROM employee, job, link WHERE job.wage = 1000 AND job.job_id = link.job_id and employee.employee_id = link.employee_id; The second SQL query gives me the total number for all employees of how much jobs they have made for 1000 SELECT count(wage) FROM employee, job, link WHERE job.wage = 1000 AND job.job_id = link.job_id and employee.employee_id = link.employee_id; I was wondering if there was a way of joining both queries and also making the second for each specific employee???

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  • how store date in myqsl database?

    - by Syom
    i have date in dd/mm/yyyy format. how can i store it in databse, if i fant to do some operations on them after? for example i must find out the rows, where date > something what type i must set to date field? thanks

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  • Problem while redirecting user after registration

    - by Eternal Learner
    I am creating a simple website . My situation is like this. After registering an user, I want to redirect the user after say 3 seconds to a main page(if the registration succeeds) . The code I have now is as below $query = "INSERT INTO Privileges VALUES('$user','$password1','$role')"; $result = mysql_query($query, $dbcon) or die('Registration Failed: ' . mysql_error()); print 'Thanks for Registering , You will be redirected shortly'; ob_start(); echo "Test"; header("Location: http://www.php.net"); ob_flush() I get the error message Warning: Cannot modify header information - headers already sent by (output started at/home/srinivasa/public_html/ThanksForRegistering.php:27) in /home/srinivasa /public_html/ThanksForRegistering.php on line 35. What do I need to do now ?

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  • problem during data fetch

    - by nectar
    here is my code $sql="SELECT * FROM $tbl_name WHERE ownerId='$UserId'"; $result=mysql_query($sql,$link)or die(mysql_error()); $row = mysql_fetch_array($result, MYSQL_ASSOC); <?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<tr>"; echo "<td>".$row['pinId']."</td>"; echo "<td>".$row['usedby']."</td>"; echo "<td>".$row['status']."</td>"; echo "</tr>"; } ?> it is ignoring the first record means if 4 rows are in $row its ignoring the 1st one rest three are coming on page. ownerId is not primary key.

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  • How to properly design a simple favorites and blocked table?

    - by Nils Riedemann
    Hey, i am currently writing a webapp in rails where users can mark items as favorites and also block them. I came up two ways and wondered which one is more common/better way. 1. Separate join tables Would it be wise to have 2 tables for this? Like: users_favorites - user_id - item_id users_blocked - user_id - item_id 2. single table users_marks (or so) - users_id - item_id - type (["fav", "blk"]) Both ways seem to have advantages. Which one would you use and why?

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  • Why does this properly escaped SQL query fail?

    - by Jason Rhodes
    Here's the query: INSERT INTO jobemails (jobid, to, subject, message, headers, datesent) VALUES ('340', '[email protected]', 'We\'ve received your request for a photo shoot called \'another\'.', 'message', 'headers', '2010-04-22 15:55:06') The datatypes are all correct, it always fails at the subject, so it must be how I'm escaping the values, I assume. I'm sure one of you will see my idiot mistake right away. A little help?

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  • where id = multiple artists

    - by pixel
    Any time there is an update within my music community (song comment, artist update, new song added, yadda yadda yadda), a new row is inserted in my "updates" table. The row houses the artist id involved along with other information (what type of change, time and date, etc). My users have a "favorite artists" section where they can do just that -- mark artists as their favorites. As such, I'd like to create a new feature that shows the user the changes made to their various favorite artists. How should I be doing this efficiently? SELECT * FROM table_updates WHERE artist_id = 1 and artist_id = 500 and artist_id = 60032 Keep in mind, a user could have 43,000 of our artists marked as a favorite. Thoughts?

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  • PHP - How to display other values, when a query is limited by 3?

    - by Dodi300
    Hello. Can anyone tell me how to display the other values, when a query is limited my 3. In this question I asked how to order and limit values, but now I want to show the others in another query. How would I go about doing this? Here's the code I used before: $query = "SELECT gmd FROM account ORDER BY gmd DESC LIMIT 3"; $result = mysql_query($query); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { } Thanks!

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