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  • Using NULL in MySQL

    - by JPro
    The data in my dabatase comes from an external source and where ever there is no data passed, I keep NULL in those places. Can anyone tell me if there are any implications in using NULL to represent empty value? Should I follow some other convention like 'data_not_available'? or something like that? Can anyone suggest?

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  • Fighing system in Php & MYSQL

    - by Gully
    I am working on a game like Mafia Wars and i am trying to get the fighting system working but i keep getting lose trying to work out who is going to win the fight and it still needs to know if the stats are close then there is a random chace of them winning. $strength = $my_strength; $otherplayerinfo = mysql_query("SELECT * FROM accounts WHERE id='$player_id'"); $playerinfo = mysql_fetch_array($otherplayerinfo); $players_strength = $playerinfo['stre']; $players_speed = $playerinfo['speed']; $players_def = $playerinfo['def']; if($players_strength > $strength){ $strength_point_player = 1; $strength_point_your = 0; }else{ $strength_point_your = 1; $strength_point_player = 0; } I was trying a point system but i still could not do it.

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  • PHP & MySQL checked checkbox problem

    - by BeepU
    I'm trying to check if the checkbox has been checked and display the check mark for the user to see when they check there account settings. I want to know how can I fix this problem using PHP so that the check mark is displayed every time the user views their account settings? Here is the HTML. <input type="checkbox" name="privacy_policy" id="privacy_policy" value="yes" />

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  • mysql select update

    - by Tillebeck
    Hi I have read quite a few selcet+update questions in here but cannot understand how to do it. So will have to ask from the beginning. I would like to update a table based on data in another table. Setup is like this: - TABLE a ( int ; string ) ID WORD 1 banana 2 orange 3 apple - TABLE b ( "comma separated" string ; string ) WORDS TEXTAREA 0 banana -> 0,1 0 orange apple apple -> BEST:0,2,3 ELSE 0,2,3,3 0 banana orange apple -> 0,1,2,3 Now I would like to for each word in TABLE a append ",a.ID" to b.WORDS like: SELECT id, word FROM a (for each) -> UPDATE b SET words = CONCAT(words, ',', a.id) WHERE b.textarea like %a.word% Or even better: replace the word found in b.textarea with ",a.id" so it is the b.textarea that ends up beeing a comma separeted string of id's... But I do not know if that is possible.

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  • MySQL whats wrong with my foreign keys?

    - by Skiy
    Hello, what is wrong with the two foreign keys which I have marked with comments? create database db; use db; create table Flug( Flugbez varchar(20), FDatum Date, Ziel varchar(20), Flugzeit int, Entfernung int, Primary Key (Flugbez, FDatum)); create table Flugzeugtyp( Typ varchar(20), Hersteller varchar(20), SitzAnzahl int, Reisegeschw int, primary key (Typ) ); create table flugzeug( Typ varchar(20), SerienNr int, AnschDatum Date, FlugStd int, primary key(Typ,SerienNr), foreign key(Typ) references Flugzeugtyp(Typ)); create table Abflug( Flugbez varchar(20), FDatum Date, Typ varchar(20), Seriennr int, Kaptaen varchar(20), Primary key(Flugbez,FDatum,Typ,SerienNr), Foreign key(Flugbez) references Flug(Flugbez), -- Foreign key(FDatum) references Flug(FDatum), Foreign key(Typ) references Flugzeugtyp(Typ) -- ,Foreign key(SerienNr) references Flugzeug(SerienNr) ); When I uncomment these, I get: ERROR 1005 (HY000): Can't create table 'db.abflug' (errno: 150)

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  • Entering Content Into A MySQL Database Via A Form

    - by ThatMacLad
    I've been working on creating a form that submits content into my database but I decided that rather than using a drop down menu to select the date I'd rather use a textfield. I was wondering what changes I will need to make to my table creation file. <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "CREATE TABLE php_blog ( id int(20) NOT NULL auto_increment, timestamp int(20) NOT NULL, title varchar(255) NOT NULL, entry longtext NOT NULL, PRIMARY KEY (id) )"; $result = mysql_query($sql) or print ("Can't create the table 'php_blog' in the database.<br />" . $sql . "<br />" . mysql_error()); mysql_close(); if ($result != false) { echo "Table 'php_blog' was successfully created."; } ?> It's the timestamp that I need to edit to enter in via a textfield. The Title and Entry are currently being entered via that method anyway.

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  • C#:checking existing record in database Mysql

    - by Meko
    HI.I searched this question inform and I found solution to change column property Unique Index.Now If I try to insert same record cmd.ExecuteNonQuery() gives error that record exist ,but how can use this exception to give user a message that record exist and must enter new one ? I am trying to make some thing like if(cmd.ExecuteNonQuery() !=true ) { MessageBox.Show("User Exists"); } But I dont know what returns cmd.ExecuteNonQuery() ? Or I must get records using reader in table and compare them with text in Textfiel?

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  • JQuery Ajax Updating MySQL Database, But Not Running Success Function

    - by myrmidon16
    I am currently using the JQuery ajax function to call an exterior PHP file, in which I select and add data in a database. Once this is done, I run a success function in JavaScript. What's weird is that the database is updating successfully when ajax is called, however the success function is not running. Here is my code: <!DOCTYPE html> <head> <script type="text/javascript" src="jquery-1.6.4.js"></script> </head> <body> <div onclick="addtask();" style="width:400px; height:200px; background:#000000;"></div> <script> function addtask() { var tid = (Math.floor(Math.random() * 3)) + 1; var tsk = (Math.floor(Math.random() * 10)) + 1; if(tsk !== 1) { $.ajax({ type: "POST", url: "taskcheck.php", dataType: "json", data: {taskid:tid}, success: function(task) {alert(task.name);} }); } } </script> </body> </html> And the PHP file: session_start(); $connect = mysql_connect('x', 'x', 'x') or die('Not Connecting'); mysql_select_db('x') or die ('No Database Selected'); $task = $_REQUEST['taskid']; $uid = $_SESSION['user_id']; $q = "SELECT task_id, taskname FROM tasks WHERE task_id=" .$task. " LIMIT 1"; $gettask = mysql_fetch_assoc(mysql_query($q)); $q = "INSERT INTO user_tasks (ut_id, user_id, task_id, taskstatus, taskactive) VALUES (null, " .$uid. ", '{$gettask['task_id']}', 0, 1)"; $puttask = mysql_fetch_assoc(mysql_query($q)); $json = array( "name" => $gettask['taskname'] ); $output = json_encode($json); echo $output; Let me know if you have any questions or comments, thanks.

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  • Displaying name instead of ID PHP MySQL

    - by Derek
    Hi, I need something simple; I have page where a user clicks an author to see the books associated with that author. On my page displaying the list of books for the author, I want a simple HTML title saying: 'The books for: AUTHORNAME' I can get the page to display author ID but not the name. When the user clicks the link in the previous page of the author, it looks likes this: <a href="viewauthorbooks.php?author_id=<?php echo $row['author_id']?>"><?php echo $row['authorname']?></a> And then on the 'viewauthorbooks.php?author_id=23' I have declared this at the start: $author_id = $_GET['author_id']; $authorname = $_GET['authorname']; And finally, 'The books for: AUTHORNAME, where it says AUTHORNAME, I have this: echo $authorname (With PHP tags, buts its not letting me put them in!) And this doesnt show anything, however if I change it to author_id, it displays the correct author ID that has been clicked, but its not exactly user friendly!! Can anyone help me out!

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  • Best way to limit results in MySQL with user subcategories

    - by JM4
    I am trying to essentially solve for the following: 1) Find all users in the system who ONLY have programID 1. 2) Find all users in the system who have programID 1 AND any other active program. My tables structures (in very simple terms are as follows): users userID | Name ================ 1 | John Smith 2 | Lewis Black 3 | Mickey Mantle 4 | Babe Ruth 5 | Tommy Bahama plans ID | userID | plan | status --------------------------- 1 | 1 | 1 | 1 2 | 1 | 2 | 1 3 | 1 | 3 | 1 4 | 2 | 1 | 1 5 | 2 | 3 | 1 6 | 3 | 1 | 0 7 | 3 | 2 | 1 8 | 3 | 3 | 1 9 | 3 | 4 | 1 10 | 4 | 2 | 1 11 | 4 | 4 | 1 12 | 5 | 1 | 1 I know I can easily find all members with a specific plan with something like the following: SELECT * FROM users a JOIN plans b ON (a.userID = b.userID) WHERE b.plan = 1 AND b.status = 1 but this will only tell me which users have an 'active' plan 1. How can I tell who ONLY has plan 1 (in this case only userID 5) and how to tell who has plan 1 AND any other active plan? Update: This is not to get a count, I will actually need the original member information, including all the plans they have so a COUNT(*) response may not be what I'm trying to achieve.

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  • formatting mysql data for ouptut into a table

    - by bsandrabr
    Following on from a question earlier today this answer was given to read the data into an array and separate it to print vehicle type and then some data for each vehicle. <?php $sql = "SELECT * FROM apparatus ORDER BY vehicleType"; $getSQL = mysql_query($sql); // transform the result set: $data = array(); while ($row = mysql_fetch_assoc($getSQL)) { $data[$row['vehicleType']][] = $row; } ?> <?php foreach ($data as $type => $rows): ?> <h2><?php echo $type?></h2> <ul> <?php foreach ($rows as $vehicleData):?> <li><?php echo $vehicleData['name'];?></li> <?php endforeach ?> </ul> <?php endforeach ?> This is almost perfect for what I want to do but I need to print out two columns from the database ie ford and mondeo before going into the second foreach loop. I've tried print $rows['model'] and all the other combinations I can think of but that doesn't work. Any help much appreciated

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  • incrementing a table column's data by one || mySql

    - by Praveen Prasad
    iam having a table with columns like id || counter if i do something (some event) i want the counter's value(at a particular id) to increase by one , currently iam doing this : //get current value current_value = select counter from myTable where id='someValue' // increase value current_value++ //update table with current value update myTable set counter=current_value where id='someValue'; currently iam running 2 queries for this, please suggest me some way do it in one step.

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  • Help with PHP MySQL join

    - by kester martinez
    Please help me to understand proper join syntax. I have table named inventory which has: trans_id trans_items items -> item_id trans_user employees -> person_id trans_date trans_comment trans_inventory As you can see above, trans_items is a foreign key in items table, and trans_user is a foreign key in employees table. Now what I want to do is to display in HTML the inventory table, but instead of displaying the item id, I want the ITEM NAME to be displayed. Here is what I have done. Please note I'm using CodeIgniter. public function getData(array $inputs) { $this->db->select('trans_items, trans_user, trans_date, trans_inventory, trans_comment'); $this->db->from('inventory'); $this->db->order_by('trans_date desc'); return $this->db->get()->result_array(); }

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  • Help needed to construct a SQL query

    - by song202y
    Need your help to get the list of suggested friends (who aren't friends of the current user but are friends of 2 or more of the current user's friends). The primary ordering should put people at the same school at the top, and the secondary ordering should put people with more common friends (that is, the number of people who are friends of that person and the current user) near the top. Users: user_id PK, user_name Profiles: user_id PK, school_name, ... Friendships: id PK, user_id FK, friend_id FK Thank you in advance. Joe

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  • An array of MySQL results...

    - by Michael Falk
    What am I doing wrong here? I am attempting to return a json object and I can't seem to get past the array... I've built hundreds of regular array and returned them as a json object but I am having a hard time wrapping my head around this one. $rows = array(); $post_array = array(); $i = 0; $result = mysql_query(" SELECT * FROM forum_posts WHERE permalink = '$permalink' AND LOWER(raw_text) LIKE '%$str%' " ); while($row = mysql_fetch_assoc($result)) { $post_array[$i] = $rows[ "id" => htmlentities($row["id"]), "post_content" => htmlentities($row["content"]), "author" => $row["author"], "last_updated" => $row["last_updated"], "author_id" => $row["author_id"], "editing_author" => $row["editing_author"], "date" => $outputQuoteDate ]; $i++; }

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  • MySQL error code:1329 in function

    - by Sharad Sharma
    DELIMITER // CREATE DEFINER=`root`@`localhost` FUNCTION `formatMovieNames`(lID int) RETURNS varchar(1000) CHARSET latin1 BEGIN DECLARE output varchar(1000); DECLARE done INT DEFAULT 0; declare a varchar(200); declare cur1 cursor for select fileName from swlp4_movie where movieID in (select movieID from lesson_movie_map where lessonID = lID order by lm_map_id); DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1; open cur1; read_loop: loop fetch cur1 into a; if done = 1 then leave read_loop; end if; set output = concat(output, 'movie:[',a,']<br/>'); set output = substr(output, 0, length(@output)-5); end loop; close cur1; RETURN output; END// I have create this function and when I run it I do not get any output (select fileName from swlp4_movie where movieID in (select movieID from lesson_movie_map where lessonID = 24 order by lm_map_id)); brings correct result I am trying to get result as movie:['movieName']< br / movie:['movieName1'] and so on (had to change br tag, cause it was adding a line break) cant't figure out what I am doing wrong

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  • MySQL Update Statement + File Upload

    - by Jason Sweet
    Greetings! Been staring at this all day and can't seem to figure out why my update statement fails to update the field 'image_filename': $fileName = $_FILES['image_filename']; if($fileName["name"] <> ""){ $imageFile = $fileName['name']; $destination = "../../../../assets/resources/images/".$fileName['name']; move_uploaded_file($fileName['name'], $destination); } $updateSQL = sprintf("UPDATE content SET image_filename='$imageFile' WHERE id=%s", GetSQLValueString($_POST['resource_id'], "int")); mysql_select_db($database_conn_talent, $conn_talent); $Result1 = mysql_query($updateSQL, $conn_talent) or die(mysql_error()); Can a SQL pro tell me what I"m missing? Much thanks in advance for your feedback!

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  • MySQL Casting in C#

    - by user798080
    Okay, so I'm attempting to print out the contents of a table in a comma-separated file. using (OdbcCommand com = new OdbcCommand("SELECT * FROM pie_data WHERE Pie_ID = ?", con)) { com.Parameters.AddWithValue("", Request.Form["pie_id"]); com.ExecuteNonQuery(); using (OdbcDataReader reader = com.ExecuteReader()) { string finalstring = ""; while (reader.Read()) { finalstring = reader.GetString(9) + ","; for (int i = 0; i <= 8; i = i + 1) { finalstring = finalstring + reader.GetString(i) + ","; } } } Response.Write(finalstring); noredirect = 1; } My table layout is: CREATE TABLE `rent_data` ( `Pies` INT(10) UNSIGNED NOT NULL, `Name` VARCHAR(85) NOT NULL, `Email` VARCHAR(85) NOT NULL, `Pie_Rent` DATE NOT NULL, `Rent_To` DATE NOT NULL, `Returned_Date` DATE NULL DEFAULT NULL, `Place` VARCHAR(100) NOT NULL, `Purpose` MEDIUMTEXT NOT NULL, `Comments` MEDIUMTEXT NULL, `Pie_ID` SMALLINT(5) UNSIGNED ZEROFILL NOT NULL, INDEX `Pie_ID` (`Equipment_ID`) ) The error I'm getting is this: Exception Details: System.InvalidCastException: Unable to cast object of type 'System.Int64' to type 'System.String'. On the line: finalstring = finalstring + reader.GetString(i) + ",";

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