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  • How can I prevent a double submit with jQuery or Javascript?

    - by kielie
    Hi guys, I keep getting duplicate entries in my database because of impatient users clicking the submit button multiple times. I googled and googled and found a few scripts, but none of them seem to be sufficient. How can I prevent these duplicate entries from occurring using javascript or preferably jQuery? Thanx in advance!

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  • How to select DISTINCT rows without having the ORDER BY field selected

    - by JannieT
    So I have two tables students (PK sID) and mentors (PK pID). This query SELECT s.pID FROM students s JOIN mentors m ON s.pID = m.pID WHERE m.tags LIKE '%a%' ORDER BY s.sID DESC; delivers this result pID ------------- 9 9 3 9 3 9 9 9 10 9 3 10 etc... I am trying to get a list of distinct mentor ID's with this ordering so I am looking for the SQL to produce pID ------------- 9 3 10 If I simply insert a DISTINCT in the SELECT clause I get an unexpected result of 10, 9, 3 (wrong order). Any help much appreciated.

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  • Display a ranking grid for game : optimization of left outer join and find a player

    - by Jerome C.
    Hello, I want to do a ranking grid. I have a table with different values indexed by a key: Table SimpleValue : key varchar, value int, playerId int I have a player which have several SimpleValue. Table Player: id int, nickname varchar Now imagine these records: SimpleValue: Key value playerId for 1 1 int 2 1 agi 2 1 lvl 5 1 for 6 2 int 3 2 agi 1 2 lvl 4 2 Player: id nickname 1 Bob 2 John I want to display a rank of these players on various SimpleValue. Something like: nickname for lvl Bob 1 5 John 6 4 For the moment I generate an sql query based on which SimpleValue key you want to display and on which SimpleValue key you want to order players. eg: I want to display 'lvl' and 'for' of each player and order them on the 'lvl' The generated query is: SELECT p.nickname as nickname, v1.value as lvl, v2.value as for FROM Player p LEFT OUTER JOIN SimpleValue v1 ON p.id=v1.playerId and v1.key = 'lvl' LEFT OUTER JOIN SimpleValue v2 ON p.id=v2.playerId and v2.key = 'for' ORDER BY v1.value This query runs perfectly. BUT if I want to display 10 different values, it generates 10 'left outer join'. Is there a way to simplify this query ? I've got a second question: Is there a way to display a portion of this ranking. Imagine I've 1000 players and I want to display TOP 10, I use the LIMIT keyword. Now I want to display the rank of the player Bob which is 326/1000 and I want to display 5 rank player above and below (so from 321 to 331 position). How can I achieve it ? thanks.

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  • PHP mysqli error return time

    - by Dori
    Hello. Can i ask a fundamental question. Why when I try to create a new mysqli object in php with invalid database infomation (say an incorrect database name) does it not return an error intstantly? I usually program server stuff in Java and something like this would throw back an error strait away, not after 20 seconds or so. For example $conn = new mysqli($host, $username, $password, $database); Thanks!

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  • Preventing spam bots on site?

    - by Mike
    We're having an issue on one of our fairly large websites with spam bots. It appears the bots are creating user accounts and then posting journal entries which lead to various spam links. It appears they are bypassing our captcha somehow -- either it's been cracked or they're using another method to create accounts. We're looking to do email activation for the accounts, but we're about a week away from implementing such changes (due to busy schedules). However, I don't feel like this will be enough if they're using an SQL exploit somewhere on the site and doing the whole cross site scripting thing. So my question to you: If they are using some kind of XSS exploit, how can I find it? I'm securing statements where I can but, again, its a fairly large site and it'd take me awhile to actively clean up SQL statements to prevent XSS. Can you recommend anything to help our situation?

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  • Can you automatically create a mysqldump file that doesn't enforce foreign key constraints?

    - by Tai Squared
    When I run a mysqldump command on my database and then try to import it, it fails as it attempts to create the tables alphabetically, even though they may have a foreign key that references a table later in the file. There doesn't appear to be anything in the documentation and I've found answers like this that say to update the file after it's created to include: set FOREIGN_KEY_CHECKS = 0; ...original mysqldump file contents... set FOREIGN_KEY_CHECKS = 1; Is there no way to automatically set those lines or export the tables in the necessary order (without having to manually specify all table names as that can be tedious and error prone)? I could wrap those lines in a script, but was wondering if there is an easy way to ensure I can dump a file and then import it without manually updating it.

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  • User Getting Logged Out After Making First Comment

    - by John
    Hello, I am using a login system that works well. I am also using a comment system. The comment function does not show up unless the user is logged in (as shown in commentformonoff.php below). When a user makes a comment, the info is passed from the function "show_commentbox" to the file comments2a.php. Then, the info is passed to the file comments2.php. When the site is first pulled up on a browser, after logging in and making a comment, the user is logged out. After logging in a second time during the same browser session, the user is no longer logged out after making a comment. How can I keep the user logged in after making the first comment? Thanks in advance, John Commentformonoff.php: <?php if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } ?> Function "show_commentbox": function show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl) { echo '<form action="http://www...com/.../comments/comments2a.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <input type="hidden" value="'.$_SESSION['username'].'" name="u"> <input type="hidden" value="'.$submissionid.'" name="submissionid"> <input type="hidden" value="'.stripslashes($submission).'" name="submission"> <input type="hidden" value="'.$url.'" name="url"> <input type="hidden" value="'.$submittor.'" name="submittor"> <input type="hidden" value="'.$submissiondate.'" name="submissiondate"> <input type="hidden" value="'.$countcomments.'" name="countcomments"> <input type="hidden" value="'.$dispurl.'" name="dispurl"> <label class="addacomment" for="title">Add a comment:</label> <textarea class="checkMax" name="comment" type="comment" id="comment" maxlength="1000"></textarea> <div class="commentsubbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; } Included in comments2a.php: $uid = mysql_real_escape_string($_POST['uid']); $u = mysql_real_escape_string($_POST['u']); $query = sprintf("INSERT INTO comment VALUES (NULL, %d, %d, '%s', NULL)", $uid, $subid, $comment); mysql_query($query) or die(mysql_error()); $lastcommentid = mysql_insert_id(); header("Location: comments2.php?submission=".$submission."&submissionid=".$submissionid."&url=".$url."&submissiondate=".$submissiondate."&comment=".$comment."&subid=".$subid."&uid=".$uid."&u=".$u."&submittor=".$submittor."&countcomments=".$countcomments."&dispurl=".$dispurl."#comment-$lastcommentid"); exit(); Included in comments2.php: if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../comments/comments2.php?submission='.$submission.'&submissionid='.$submissionid.'&url='.$url.'&submissiondate='.$submissiondate.'&submittor='.$submittor.'&countcomments='.$countcomments.'&dispurl='.$dispurl.'');} $uid = mysql_real_escape_string($_GET['uid']); $u = mysql_real_escape_string($_GET['u']);

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  • Minimizing SQL queries using join with one-to-many relationship

    - by Brian
    So let me preface this by saying that I'm not an SQL wizard by any means. What I want to do is simple as a concept, but has presented me with a small challenge when trying to minimize the amount of database queries I'm performing. Let's say I have a table of departments. Within each department is a list of employees. What is the most efficient way of listing all the departments and which employees are in each department. So for example if I have a department table with: id name 1 sales 2 marketing And a people table with: id department_id name 1 1 Tom 2 1 Bill 3 2 Jessica 4 1 Rachel 5 2 John What is the best way list all departments and all employees for each department like so: Sales Tom Bill Rachel Marketing Jessica John Pretend both tables are actually massive. (I want to avoid getting a list of departments, and then looping through the result and doing an individual query for each department). Think similarly of selecting the statuses/comments in a Facebook-like system, when statuses and comments are stored in separate tables.

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  • VBA How to find last insert id?

    - by Muiter
    I have this code: With shtControleblad Dim strsql_basis As String strsql_basis = "INSERT INTO is_calculatie (offerte_id) VALUES ('" & Sheets("controleblad").Range("D1").Value & "')" rs.Open strsql_basis, oConn, adOpenDynamic, adLockOptimistic Dim last_id As String last_id = "select last_insert_id()" End With The string last_id is not filled. What is wrong? I need to find te last_insert_id so I can use it in an other query.

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  • Adding to a multidimensional array in PHP

    - by b. e. hollenbeck
    I have an array being returned from the database that looks like so: $data = array(201 => array('description' => blah, 'hours' => 0), 222 => array('description' => feh, 'hours' => 0); In the next bit of code, I'm using a foreach and checking the for the key in another table. If the next query returns data, I want to update the 'hours' value in that key's array with a new hours value: foreach ($data as $row => $value){ $query = $db->query($sql); if ($result){ $value['hours'] = $result['hours']; } I've tried just about every combination of declarations for the foreach loop, but I keep getting the error that it's a non-object. Surely this is easier than my brain is perceiving it.

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  • How to display thumbnails with category by category

    - by shin
    http://pastie.org/962429 I have the above SQL. There are three categories, webdesign, logos and prints. I want to display them as following html. http://pastie.org/962432 Basically pulling thumbnail image and name from database category by category. I am not really sure how to do it. I will appreciate any inputs or help. Thanks in advance.

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  • CakePHP - get last query run

    - by Phantz
    I want to get the last query CakePHP ran. I can't turn debug on in core.php and I can't run the code locally. I need a way to get the last sql query and log it to the error log without effecting the live site. This query is failing but is being run. something like this would be great: $this->log($this->ModelName->lastQuery); Thanks in advance.

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  • 'Attempt to call private method' error when trying to change change case of db entires in migration

    - by Senthil
    class AddTitleToPosts < ActiveRecord::Migration def self.up add_column :posts, :title, :string Post.find(:all).each do |post| post.update(:title => post.name.upcase) end end def self.down end end Like you can nothing particularly complicated, just trying to add new column title by changing case of name column already in db. But I get attempt to call private method error. I'm guessing it has something to do with 'self'? Thanks for your help.

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  • Reset ID autoincrement ? phpmyadmin

    - by Marcelo
    Hi, I was testing some data in my tables of my database, to see if there was any error, now I cleaned all the testing data, but my id (auto increment) does not start from 1 anymore, can (how do) I reset it ? Sorry for any mistake in English, and thanks for the attention.

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  • to take values of checkbox in table attributes

    - by mwj
    i have a database patient with 3-4 tables n each table has about 8 attributes.... i have a table medical history which has attribute additional info ... under which i have 5 checkboxes.... all the values entered are taken up except the chekbox values..... plz help

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  • can you make an sql query for this situation?

    - by saurav
    i have a table as below. name and 10 cities in which he lived during his lifetime. name , city1 , city2 , city3 ,city4 , city5 ,city6 , city7 , city8 , city9 city10 suppose for a particular name i want to fetch other names in table matching with maximum number of cities. for example if i want to fetch other people who have lived in three or more cities lived by this person.

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  • Unique Alpha numeric generator

    - by AAA
    Hi, I want to give our users in the database a unique alpha-numeric id. I am using the code below, will this always generate a unique id? Below is the updated version of the code: old php: // Generate Guid function NewGuid() { $s = strtoupper(md5(uniqid(rand(),true))); $guidText = substr($s,0,8) . '-' . substr($s,8,4) . '-' . substr($s,12,4). '-' . substr($s,16,4). '-' . substr($s,20); return $guidText; } // End Generate Guid $Guid = NewGuid(); echo $Guid; echo "<br><br><br>"; New PHP: // Generate Guid function NewGuid() { $s = strtoupper(uniqid("something",true)); $guidText = substr($s,0,8) . '-' . substr($s,8,4) . '-' . substr($s,12,4). '-' . substr($s,16,4). '-' . substr($s,20); return $guidText; } // End Generate Guid $Guid = NewGuid(); echo $Guid; echo "<br><br><br>"; Will the second (new php) code guarantee 100% uniqueness. Final code: PHP // Generate Guid function NewGuid() { $s = strtoupper(uniqid(rand(),true)); $guidText = substr($s,0,8) . '-' . substr($s,8,4) . '-' . substr($s,12,4). '-' . substr($s,16,4). '-' . substr($s,20); return $guidText; } // End Generate Guid $Guid = NewGuid(); echo $Guid; $alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; function base_encode($num, $alphabet) { $base_count = strlen($alphabet); $encoded = ''; while ($num >= $base_count) { $div = $num/$base_count; $mod = ($num-($base_count*intval($div))); $encoded = $alphabet[$mod] . $encoded; $num = intval($div); } if ($num) $encoded = $alphabet[$num] . $encoded; return $encoded; } function base_decode($num, $alphabet) { $decoded = 0; $multi = 1; while (strlen($num) > 0) { $digit = $num[strlen($num)-1]; $decoded += $multi * strpos($alphabet, $digit); $multi = $multi * strlen($alphabet); $num = substr($num, 0, -1); } return $decoded; } echo base_encode($Guid, $alphabet); } So for more stronger uniqueness, i am using the $Guid as the key generator. That should be ok right?

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  • WordPress on other parts of my site

    - by SHiNKiROU
    I have a WordPress installation on my site, and I want to display WP posts on other parts of my site (that is outside the WP installation). How do I do that with PHP? I tried to search this type of question on Stack Overflow, Google and WP official site but I didn't find anything.

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  • Unknown column even thoug it exits

    - by george
    I have SELECT servisler.geo_location, servisler.ADRES_MERKEZ, servisler.ADRES_ILCE, servisler.ADRES_IL, servisler.FIRMA_UNVANI, servisler.ADRES_ISTEL, servisler.YETKILI_ADISOYADI, urun_gruplari.GRUP_ADI FROM servisler INNER JOIN urun_gruplari ON kullanici_cihaz.URUN_GRUP_NO= urun_gruplari.RECNO INNER JOIN kullanici ON kullanici.SERVIS_RECNO = servisler.RECNO INNER JOIN kullanici_cihaz ON kullanici.RECNO = kullanici_cihaz.KUL_RECNO AND kullanici_cihaz.URUN_GRUP_NO = urun_gruplari.RECNO where kullanici.kullanici = 'MAR.EDI.003' but it says [Err] 1054 - Unknown column 'kullanici_cihaz.URUN_GRUP_NO' in 'on clause' enen though the column exits. What is its problem? schema Server version: 5.1.33-community-log

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  • Why is the ( ) mandatory in the SQL statement select * from gifts INNER JOIN sentgifts using (giftID

    - by Jian Lin
    Why is the ( ) mandatory in the SQL statement select * from gifts INNER JOIN sentgifts using (giftID); ? The ( ) usually is for specifying grouping of something. But in this case, are we supposed to be able to use 2 or more field names... in the example above, it can be all clear that it is 1 field, is it just that the parser is not made to bypass the ( ) when it is all clear? (such as in the language Ruby).

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  • Lost with hibernate - OneToMany resulting in the one being pulled back many times..

    - by Andy
    I have this DB design: CREATE TABLE report ( ID MEDIUMINT PRIMARY KEY NOT NULL AUTO_INCREMENT, user MEDIUMINT NOT NULL, created TIMESTAMP NOT NULL, state INT NOT NULL, FOREIGN KEY (user) REFERENCES user(ID) ON UPDATE CASCADE ON DELETE CASCADE ); CREATE TABLE reportProperties ( ID MEDIUMINT NOT NULL, k VARCHAR(128) NOT NULL, v TEXT NOT NULL, PRIMARY KEY( ID, k ), FOREIGN KEY (ID) REFERENCES report(ID) ON UPDATE CASCADE ON DELETE CASCADE ); and this Hibernate Markup: @Table(name="report") @Entity(name="ReportEntity") public class ReportEntity extends Report{ @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name="ID") private Integer ID; @Column(name="user") private Integer user; @Column(name="created") private Timestamp created; @Column(name="state") private Integer state = ReportState.RUNNING.getLevel(); @OneToMany(mappedBy="pk.ID", fetch=FetchType.EAGER) @JoinColumns( @JoinColumn(name="ID", referencedColumnName="ID") ) @MapKey(name="pk.key") private Map<String, ReportPropertyEntity> reportProperties = new HashMap<String, ReportPropertyEntity>(); } and @Table(name="reportProperties") @Entity(name="ReportPropertyEntity") public class ReportPropertyEntity extends ReportProperty{ @Embeddable public static class ReportPropertyEntityPk implements Serializable{ /** * long#serialVersionUID */ private static final long serialVersionUID = 2545373078182672152L; @Column(name="ID") protected int ID; @Column(name="k") protected String key; } @EmbeddedId protected ReportPropertyEntityPk pk = new ReportPropertyEntityPk(); @Column(name="v") protected String value; } And i have inserted on Report and 4 Properties for that report. Now when i execute this: this.findByCriteria( Order.asc("created"), Restrictions.eq("user", user.getObject(UserField.ID)) ) ); I get back the report 4 times, instead of just the once with a Map with the 4 properties in. I'm not great at Hibernate to be honest, prefer straight SQL but I must learn, but i can't see what it is that is wrong.....? Any suggestions?

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  • SQL to CodeIgniter Array Missing Data Issue

    - by SamD
    $query = $this->db->query("SELECT t1.numberofbets, t1.profit, t2.seven_profit, t3.28profit, user.user_id, username, password, email, balance, user.date_added, activation_code, activated FROM user LEFT JOIN (SELECT user_id, SUM(amount_won) AS profit, count(tip_id) AS numberofbets FROM tip GROUP BY user_id) as t1 ON user.user_id = t1.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS seven_profit FROM tip WHERE date_settled > '$seven_daystime' GROUP BY user_id) as t2 ON user.user_id = t2.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS 28profit FROM tip WHERE date_settled > '$twoeight_daystime' GROUP BY user_id) as t3 ON user.user_id = t3.user_id where activated = 1 GROUP BY user.user_id ORDER BY user.date_added DESC"); return $query->result_array(); The query works fine running it in phpMyAdmin and returns complete results (in image attached). However, printing the array in CodeIgniter, it has no value for one field ,seven_profit, where it is there in the SQL query ran in phpMyAdmin, just the discrepancy in this one field, from sql to php array... I just can’t see why, when printing the array, that one field, which should have value of 26, contains nothing? Any ideas? I changed the field name from starting with a number in attempt to fix it, but no difference. I know this is complex and looks horrible, any help or just people coming across something similar would be great to know about, thanks. Sam

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