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  • LINQ expression precedence with Skip(), Take() and OrderBy()

    - by Robert Koritnik
    I'm using LINQ to Entities and display paged results. But I'm having issues with the combination of Skip(), Take() and OrderBy() calls. Everything works fine, except that OrderBy() is assigned too late. It's executed after result set has been cut down by Skip() and Take(). So each page of results has items in order. But ordering is done on a page handful of data instead of ordering of the whole set and then limiting those records with Skip() and Take(). How do I set precedence with these statements? My example (simplified) var query = ctx.EntitySet.Where(/* filter */).OrderBy(/* expression */); int total = query.Count(); var result = query.Skip(n).Take(x).ToList();

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  • Can a binary tree or tree be always represented in a Database as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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  • please help me in this query

    - by testkhan
    I have three tables (user, friends, posts) and two users (user1 and user2). When user1 adds user2 as friend then user1 can see the posts of user2 just like on Facebook. But only the posts after the date when user1 added user2 as friend. My query is like this: $sql = mysql_query("SELECT * FROM posts p JOIN friends f ON p.currentuserid = f.friendid AND p.time >= f.friend_since OR p.currentuserid='user1id' WHERE f.myid='user1id' ORDER BY p.postid DESC LIMIT 20"); it is working all the way fine but with a little problem.....!! it displays user2, user3 (all the users as friends of user1) posts for single time but shows user1 posts multiple.......i.e user2. hi user1. userssfsfsfsfsdf user1. userssfsfsfsfsdf user3. dddddddd user1. sdfsdsdfsdsfsf user1. sdfsdsdfsdsfsf but i in database it is single entry/post why it is happening........!! How can I fix it?

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  • 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.

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  • 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?

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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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  • PHP error problem.

    - by TaG
    I get the following error on line 8: Undefined index: real_name which is $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); I was wondering how can I fix this problem? Here is the PHP. if (isset($_POST['submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.* FROM users WHERE user_id=3"); $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, privacy_policy) VALUES ('$user_id', '$privacy_policy')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE users SET privacy_policy = '$privacy_policy' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { print mysqli_error($mysqli); return; } } Here is the HTML. <form method="post" action="index.php"> <fieldset> <ul> <li><input type="checkbox" name="privacy_policy" id="privacy_policy" value="yes" <?php if (isset($_POST['privacy_policy'])) { echo 'checked="checked"'; } else if($privacy_policy == "yes") { echo 'checked="checked"'; } ?> /></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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  • 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!

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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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  • 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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  • 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.

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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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  • 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".

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  • 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?

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  • [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

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  • 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.

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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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  • 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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  • 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

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  • 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.

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  • Wordpress - Total User Count who only have posts

    - by knightrider
    I want to display total number of user who only have posts at Wordpress. I can get all users by this query <?php $user_count = $wpdb->get_var("SELECT COUNT(*) FROM $wpdb->users;"); echo $user_count ?> But for the user count only with posts, i think i might need to join another table, does anyone have snippets ? Thanks.

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  • Problem in Fetching Table contents when adding rows in same table

    - by jasmine
    Im trying to write a function for adding category: function addCategory() { $cname = mysql_fix_string($_POST['cname']); $kabst = mysql_fix_string($_POST['kabst']); $kselect = $_POST['kselect']; $kradio = $_POST['kradio']; $ksubmit = $_POST['ksubmit']; $id = $_POST['id']; if($ksubmit){ $query = "INSERT INTO category VALUES (' ', '{$cname}', '{$kabst}', {$kselect}, {$kradio}, ' ') "; $result = mysql_query($query); if ($result) { echo "ok"; } else{ echo $query ; } } $text .= '<div class="form"> <h2>ADD new category</h2> <form action="?page=addCategory" method="post"> <ul> <li><label>Category</label></li> <li><input name="cname" type="text" class="inp" /></li> <li><label>Description</label></li> <li><textarea name="kabst" cols="40" rows="10" class="inx"></textarea></li> <li>Published:</li> <li> <select name="kselect" class="ins"> <option value="1">Active</option> <option value="0">Passive</option> </select> </li> <li>Show in home page:</li> <li> <input type="radio" name="kradio" value="1" /> yes <input type="radio" name="kradio" value="0" /> no </li> <li>Subcategory of</li> <li> <select>'; while ($row = mysql_fetch_assoc(mysql_query("SELECT * FROM category"))){ $text .= '<option>'.$row['name'].'</option>'; } $text .= '</select> </li> <li><input name="ksubmit" type="submit" value="ekle" class="int"/></li> </ul> </form> '; return $text;} And the error: Fatal error: Maximum execution time of 30 seconds exceeded What is wrong in my function?

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  • uploading image & getting back from database

    - by Anup Prakash
    Putting a set of code which is pushing image to database and fetching back from database: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } if ($resultbytes!='') { echo $resultbytes; } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> Above set of code has one problem. The problem is whenever i pressing the "submit" button. It is just displaying the image on a page. But it is leaving all the html codes. even any new line message after the // Printing image on browser echo $resultbytes; //************************// So, for this i put this set of code in html tag: This is other sample code: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <?php if ($resultbytes!='') { // Printing image on browser echo $resultbytes; } ?> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> ** But in this It is showing the image in format of special charaters and digits. 1) So, Please help me to print the image with some HTML code. So that i can print it in my form to display the image. 2) Is there any way to convert the database image into real image, so that i can store it into my hard-disk and call it from tag? Please help me.

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