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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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  • Rails: Generated tokens missing occasionally

    - by Vincent Chan
    We generate an unique token for each user and store it on database. Everything is working fine in the local environment. However, after we upload the codes to the production server on Engine Yard, things become weird. We tried to register an account right after the deploy. It is working fine and we can see the token in the db. But after that, when we register new accounts, we cannot see any tokens. We only have NULL in the db. Not sure what caused this problem because we can't re-produce this in the local machine. Thanks for your help.

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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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  • checking if records exists in DB, in single step or 2 steps?

    - by Sinan
    Suppose you want to get a record from database which returns a large data and requires multiple joins. So my question would be is it better to use a single query to check if data exists and get the result if it exists. Or do a more simple query to check if data exists then id record exists, query once again to get the result knowing that it exists. Example: 3 tables a, b and ab(junction table) select * from from a, b, ab where condition and condition and condition and condition etc... or select id from a, b ab where condition then if exists do the query above. So I don't know if there is any reason to do the second. Any ideas how this affects DB performance or does it matter at all?

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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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  • (My)SQL performance: updating one field vs many unneccesary fields

    - by changokun
    i'm processing a form that has a lot of fields for a user who is editing an existing record. the user may have only changed one field, and i would typically do an update query that sets the values of all the fields, even though most of them don't change. i could do some sort of tracking to see which fields have actually changed, and only update the few that did. is there a performance difference between updating all fields in a record vs only the one that changed? are there other reasons to go with either method? the shotgun method is pretty easy...

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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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  • 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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  • Create fulltext index on a VIEW

    - by kylex
    Is it possible to create a full text index on a VIEW? If so, given two columns column1 and column2 on a VIEW, what is the SQL to get this done? The reason I'd like to do this is I have two very large tables, where I need to do a FULLTEXT search of a single column on each table and combine the results. The results need to be ordered as a single unit. Suggestions? EDIT: This was my attempt at creating a UNION and ordering by each statements scoring. (SELECT a_name AS name, MATCH(a_name) AGAINST('$keyword') as ascore FROM a WHERE MATCH a_name AGAINST('$keyword')) UNION (SELECT s_name AS name,MATCH(s_name) AGAINST('$keyword') as sscore FROM s WHERE MATCH s_name AGAINST('$keyword')) ORDER BY (ascore + sscore) ASC sscore was not recognized.

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  • 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?

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  • How to add SQL elements to an array in PHP

    - by DanLeaningphp
    So this question is probably pretty basic. I am wanting to create an array from selected elements from a SQL table. I am currently using: $rcount = mysql_num_rows($result); for ($j = 0; $j <= $rcount; $j++) { $row = mysql_fetch_row($result); $patients = array($row[0] => $row[2]); } I would like this to return an array like this: $patients = (bob=>1, sam=>2, john=>3, etc...) Unfortunately, in its current form, this code is either copying nothing to the array or only copying the last element.

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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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  • 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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  • 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 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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  • 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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  • 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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  • 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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  • Database time data retrieval, time based queries

    - by Raphael Pineda
    I am new to time manipulation or time arithmetic operations and am currently developing a navigation system with Web server based information and currently I have this Database that contains a table peek hours whose columns are id, start_time, end_time , edge_id, day_of_the_week, edge_weight ------------------------------------------------------------------------ | Peek Hours | ------------------------------------------------------------------------ | | | | | | | | id | start_time | end_time | edge_id | day_of_the_week | edge_weight | | | | | | | | ------------------------------------------------------------------------ I am using PHP as a webservice and so based on the current time i want to get all the records that would fit this equation start_time< current_time < end_time

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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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  • 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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