Search Results

Search found 63877 results on 2556 pages for 'mysql error 1452'.

Page 539/2556 | < Previous Page | 535 536 537 538 539 540 541 542 543 544 545 546  | Next Page >

  • In SQL, in what situation do we want to Index a field in a table, or 2 fields in a table at the same

    - by Jian Lin
    In SQL, it is obvious that whenever we want to do a search on millions of record, say CustomerID in a Transactios table, then we want to add an index for CustomerID. Is another situation we want to add an index to a field when we need to do inner join or outer join using that field as a criteria? Such as Inner join on t1.custumerID = t2.customerID. Then if we don't have an index on customerID on both tables, we are looking at O(n^2) because we need to loop through the 2 tables sequentially. If we have index on customerID on both tables, then it becomes O( (log n) ^ 2 ) and it is much faster. Any other situation where we want to add an index to a field in a table? What about adding index for 2 fields combined in a table. That is, one index, for 2 fields together?

    Read the article

  • tiny mce sql error when adding links

    - by Anders Kitson
    I am using tiny mce with a script I built for uploading some content to a blog like system. Whenever I add a link via tiny mce I get this error. The field type in mysql for $content which is the one carrying the link is longblob if that helps. here is the link error first and then my code You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'google test" href="http://www.google.ca" target="_blank"google est laborum' at line 4 /* GRAB FORM DATA */ $title = $_POST['title']; $date = $_POST['date']; $content = $_POST['content']; $imageName1 = $_FILES["file"]["name"]; $date = date("Y-m-d"); $sql = "INSERT INTO blog (title,date,content,image)VALUES( \"$title\", \"$date\", \"$content\", \"$imageName1\" )"; $results = mysql_query($sql)or die(mysql_error());

    Read the article

  • 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.

    Read the article

  • run-time error '429' activex component can't create object

    - by kojof
    I've created a simple application .Net Class that converts an excel spreadsheet into a a pdf file. I then get a Excel 2007 application to call this dll which works fine on my development machine. However when i deploy it on to a machine that has both the .net framework and excel 2007, i get this error - run-time error '429' activex component can't create object when i try and register the tlb or the dll , i get error below. the module c:\temp\test\Printlibarary.dll2 was loaded but the entry-point dllregisterserver was not found. Make sure that ...... is a valid dll or ocx file and then try again. Can someone please help me to resolve this.

    Read the article

  • 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?

    Read the article

  • PHP / Zend Framework: Force prepend table name to column name in result array?

    - by Brian Lacy
    I am using Zend_Db_Select currently to retrieve hierarchical data from several joined tables. I need to be able to convert this easily into an array. Short of using a switch statement and listing out all the columns individually in order to sort the data, my thought was that if I could get the table names auto-prepended to the keys in the result array, that would solve my problem. So considering the following (assembled) SQL: SELECT user.*, contact.* FROM user INNER JOIN contact ON contact.user_id = user.user_id I would normally get a result array like this: [username] => 'bob', [contact_id] => 5, [user_id] => 2, [firstname] => 'bob', [lastname] => 'larsen' But instead I want this: [user.user_id] => 2, [user.username] => 'bob', [contact.contact_id] => 5, [contact.firstname] => 'bob', [contact.lastname] => 'larsen' Does anyone have an idea how to achieve this? Thanks!

    Read the article

  • openssl error wehn compiling Ruby from source

    - by Florian Salihovic
    Prelude: I don't want to use rvm. I installed ruby 2 with the following configuration on Mac OS X 10.8.5 ./configure --prefix=/usr/local \ --enable-pthread \ --with-readline-dir=/usr/local \ --enable-shared It is installed, version is printed correctly ... Now, when invoking gem install jekyll I get the following error: ERROR: Loading command: install (LoadError) cannot load such file -- openssl ERROR: While executing gem ... (NoMethodError) undefined method `invoke_with_build_args' for nil:NilClass I installed openssl into /usr/local but i'm really banging my head against the wall on how installing gems. It can't be that big of a deal right?

    Read the article

  • ob_start() is partially capturing data

    - by AAA
    I am using the following 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(); $alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; function base_encode($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 $decoded; } ob_start(); echo base_encode($Guid, $alphabet); //should output: bUKpk $theid = ob_get_contents(); ob_get_clean(); The problem: When i echo $theid, it shows the complete entry, but as it is being inserted into the database, only the first entry in the sequence gets inserted, for example for the entry buKPK, only 'b' is being inserted not the rest.

    Read the article

  • Check for a unique value within a count, but get all results

    - by pedalpete
    I'm trying to create a single query which, similar to stack overflow, will give me the number of votes, but also make sure that the currently viewing user can't upvote again if they've already upvoted. my query currently looks like SELECT cid, text, COUNT(votes.parentid) FROM comments LEFT JOIN votes ON comments.cid=votes.parentid AND votes.type=3 WHERE comments.type=0 AND comments.parentid='$commentParentid' GROUP BY comments.cid But I'm completely stumpted on how to add the check to see if the userid is in the votes table. The other option is to add a seperate query where SELECT COUNT(*) FROM votes WHERE userid='$userid' AND parentid='$commentParentid' AND type=3 I'm just realizing I'm so lost with this that I don't even really know what tags to provide.

    Read the article

  • Can anybody help me out with this error.?

    - by kumar
    Error during serialization or deserialization using the JSON JavaScriptSerializer. The length of the string exceeds the value set on the maxJsonLength property. Description: An unhandled exception occurred during the execution of the current web request. Please review the stack trace for more information about the error and where it originated in the code. Exception Details: System.InvalidOperationException: Error during serialization or deserialization using the JSON JavaScriptSerializer. The length of the string exceeds the value set on the maxJsonLength property. in jquery gird on button click i am displaying something like 28000 rows? I know some of them are sujjested to define the JsonmaxLength in web config file.. but its not working for me? can anybody tell me about this? thanks

    Read the article

  • Managing Foreign Keys

    - by jwzk
    So I have a database with a few tables. The first table contains the user ID, first name and last name. The second table contains the user ID, interest ID, and interest rating. There is another table that has all of the interest ID's. For every interest ID (even when new ones are added), I need to make sure that each user has an entry for that interest ID (even if its blank, or has defaults). Will foreign keys help with this scenario? or will I need to use PHP to update each and every record when I add a new key?

    Read the article

  • Trouble making login page?

    - by Ken
    Okay, so I want to make a simple login page. I've created a register page successfully, but i can't get the login thing down. login.php: <?php session_start(); include("mainmenu.php"); $usrname = mysql_real_escape_string($_POST['usrname']); $password = md5($_POST['password']); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } mysql_select_db("users", $con) or die(mysql_error()); $login = "SELECT * FROM `users` WHERE (usrname = '$usrname' AND password = '$password')"; $result = mysql_query($login); if(mysql_num_rows($result) == 1 { $_SESSION = true; header('Location: indexlogin.php'); } else { echo = "Wrong username or password." ; } ?> indexlogin.php just echoes "Login successful." What am I doing wrong? Oh, and just FYI- my database is "users" and my table is "data".

    Read the article

  • atk4 advanced crud?

    - by thindery
    I have the following tables: -- ----------------------------------------------------- -- Table `product` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `product` ( `id` INT NOT NULL AUTO_INCREMENT , `productName` VARCHAR(255) NULL , `s7location` VARCHAR(255) NULL , PRIMARY KEY (`id`) ) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `pages` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `pages` ( `id` INT NOT NULL AUTO_INCREMENT , `productID` INT NULL , `pageName` VARCHAR(255) NOT NULL , `isBlank` TINYINT(1) NULL , `pageOrder` INT(11) NULL , `s7page` INT(11) NULL , PRIMARY KEY (`id`) , INDEX `productID` (`productID` ASC) , CONSTRAINT `productID` FOREIGN KEY (`productID` ) REFERENCES `product` (`id` ) ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `field` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `field` ( `id` INT NOT NULL AUTO_INCREMENT , `pagesID` INT NULL , `fieldName` VARCHAR(255) NOT NULL , `fieldType` VARCHAR(255) NOT NULL , `fieldDefaultValue` VARCHAR(255) NULL , PRIMARY KEY (`id`) , INDEX `id` (`pagesID` ASC) , CONSTRAINT `pagesID` FOREIGN KEY (`pagesID` ) REFERENCES `pages` (`id` ) ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; I have gotten CRUD to work on the 'product' table. //addproduct.php class page_addproduct extends Page { function init(){ parent::init(); $crud=$this->add('CRUD')->setModel('Product'); } } This works. but I need to get it so that when a new product is created it basically allows me to add new rows into the pages and field tables. For example, the products in the tables are a print product(like a greeting card) that has multiple pages to render. Page 1 may have 2 text fields that can be customized, page 2 may have 3 text fields, a slider to define text size, and a drop down list to pick a color, and page 3 may have five text fields that can all be customized. All three pages (and all form elements, 12 in this example) are associated with 1 product. So when I create the product, could i add a button to create a page for that product, then within the page i can add a button to add a new form element field? I'm still somewhat new to this, so my db structure may not be ideal. i'd appreciate any suggestions and feedback! Could someone point me toward some information, tutorials, documentation, ideas, suggestions, on how I can implement this?

    Read the article

  • [WEB] Local/Dev/Live deployment - best workflow

    - by Adam Kiss
    Hello, situation We our little company with 3 people, each has a localhost webserver and most projects (previous and current) are on one PC network shared disk. We have virtual server, where some of our clients' sites and our site. Our standard workflow is: Coder PC ? Programmer localhost ? dev domain (client.company.com) ? live version (client.com) It often happens, that there are two or three guys working on same projects at the same time - one is on dev version, two are on localhost. When finished, we try to synchronize the files on dev version and ideally not to mess (thanks ILMV:]) up any files, which *knock knock * doesn't happen often. And then one of us deploys dev version on live webserver. question we are looking for a way to simplify this workflow while updating websites - ideally some sort of diff uploader or VCS probably (Git/SVN/VCS/...), but we are not completely sure where to begin or what way would be ideal, therefore I ask you, fellow stackoverflowers for your experience with website / application deployment and recommended workflow. We probably will also need to use Mac in process, so if it won't be a problem, that would be even better. Thank you

    Read the article

  • ASP.net SessionState Error in Design Mode

    - by stringo0
    I'm getting a weird error in the design view for a user creation page for 2 controls: Error Creating Control - wCreateUser Session state can only be used when enableSessionState is set to true, either in a configuration file or in the Page directive. (There's some more) I've done both of these, but I'm still getting the error in design mode. The controls work fine when compiled, and on the live site - this is just in the Visual Web Developer 2010 Design view for the page. Any ideas as to how I can resolve this? Thanks!

    Read the article

  • How to use where condition for the for a selected column using subquery?

    - by Holicreature
    I have two columns as company and product. I use the following query to get the products matching particular string... select id,(select name from company where product.cid=company.id) as company,name,selling_price,mrp from product where name like '$qry_string%' But when i need to list products of specific company how can i do? i tried the following but in vein select id,(select name from company where product.cid=company.id) as company,name,selling_price,mrp from product where company like '$qry_string%' Help me

    Read the article

  • PHP4 No Error When Missing Method is Called

    - by INTPnerd
    I have come across an annoying problem while writing some PHP4 code. I renamed the method of a class but I forgot to rename it where it was being called from. The annoying part is it was hard to track down where the problem was because no error was triggered. The script simply aborted leaving the web page partially rendered. Is it normal for this not to trigger an error or is there something wacky going on here? If this is normal is there a way to force this kind of thing to cause an error?

    Read the article

  • PHP auto refresh page without losing user input

    - by Tony
    I'm working on a PHP collaboration software project. I have a page that shows the latest updates from other users who are adding content to the database, but also has a form input to allow the user to enter text. I am currently using this code to refresh the page automatically every 30 seconds: header('Refresh: 30'); The problem is that the header code refreshes the entire page, and not just what is pulled from the database. Is there any PHP code that will just pull any new data from the database without refreshing the entire page? If someone could point me in the right direction I'd appreciate it.

    Read the article

  • Need help in Select query..

    - by Parth
    If I have four rows for a field with different values with other fields similar and then other four rows with same condition, as given below here u can see there different rows for insert with only difference in the "newvalue" and "field" excluding "id" for the table_name=jos_menu, operation=INSERT and live=0 now here what select query should be used to get only single row from the table on every change of table_name...??

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • Get number of posts in a topic PHP

    - by Wayne
    How do I get to display the number of posts on a topic like a forum. I used this... (how very noobish): function numberofposts($n) { $sql = "SELECT * FROM posts WHERE topic_id = '" . $n . "'"; $result = mysql_query($sql) or die(mysql_error()); $count = mysql_num_rows($result); echo number_format($count); } The while loop of listing topics: <div class="topics"> <div class="topic-name"> <p><?php echo $row['topic_title']; ?></p> </div> <div class="topic-posts"> <p><?php echo numberofposts($row['topic_id']); ?></p> </div> </div> Although it is a bad method of doing this... All I need is to know what would be the best method, don't just point me out to a website, do it here, because I'm trying to learn much. Okay? :D Thanks.

    Read the article

  • ASP.net error message when using REST starter kit

    - by jonhobbs
    Hi all, I've written some code using the REST starter kit and it works fine on my development machine. However, when I upload it to our server the page gives me the following error message... CS1684: Warning as Error: Reference to type 'System.Runtime.Serialization.Json.DataContractJsonSerializer' claims it is defined in 'c:\WINNT\assembly\GAC_MSIL\System.ServiceModel.Web\3.5.0.0__31bf3856ad364e35\System.ServiceModel.Web.dll', but it could not be found I've removed code line by line and it appears that the following line of code is triggering the error... HttpContent newOrganizationContent = HttpContentExtensions.CreateXmlSerializable(newOrganizationXml); Really haven't got a clue how to fix it. I assumed it might be because it needs a newer version of the framework to run, but looking in IIS it says it's running version 2.0.50727 which I think is the lates version because it says that even when we're using framework 3.5 Very confused, any ideas? Jon

    Read the article

< Previous Page | 535 536 537 538 539 540 541 542 543 544 545 546  | Next Page >