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  • how store date in mysql database?

    - by Syom
    i have date in dd/mm/yyyy format. how can i store it in databse, if i fant to do some operations on them after? for example i must find out the rows, where date > something what type i must set to date field? thanks

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  • Adding up row number and displaying total using COUNT (PHP MySQL)

    - by Yvonne
    I'm attempting to run a query that adds up the total number of subjects in a class. A class has many subjects. There is a 'teachersclasses' table between teachers (the user table) and classes. The principles sounds pretty simple but I'm having some trouble in getting my page to display the number of subjects for each class (directly associated with the teacher) This is what I have so far, trying to make use of the COUNT with a nested SELECT: SELECT (SELECT count(*) FROM subjects WHERE subjects.classid = class.classid) AS total_subjects, class.classname, class.classid FROM class Then I am calling up 'num_subjects' to present the total within a while loop: <?php echo $row['total_subjects']?> From the above, I am receiving the total subjects for a class, but within the same table row (for one class) and my other while loop doesnt run anymore, which returns all of the classes associated with a teacher :( ... Bit of a mess now! I know to return the classes for a particular teacher, I can do an additional WHERE clause on the session of 'teacherid' but I think my query is getting too complicated for me that errors are popping up everywhere. Anyone have a quick fix for this! Thanks very much

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  • mysql query trouble

    - by Bharanikumar
    Hi , in my database i have phone numbers with country code , which look somthing like 0044-123456 0044-123456 0014-123456 0014-123456 0024-123456 0024-123456 0034-123456 0044-123456 0044-123456 0024-123456 0034-123456 084-123456 084-123456 i want to total up the numbers by country, something like this output 0044 (2) 0024 (2) 0034 (1) 084 (2) 064 (5) Is it possible to do this with a SQL query?

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  • Creating CFArray from MySQL Result Array

    - by Andrew
    Is there an easy way to dump an array returned from mysql_fetch_row into a CFArray? (part of the PHP implementation of CFPropertyList) I'm bummed by the lack of documentation on CFPropertyList for PHP. Iterating through each item in the array seems inefficient. I'm open to using a different mysql_fetch_... command. I'd like to just say: $NewArray = new CFArray( $ResultArray ) But that deosn't seem to work. This is my current code: $plist = new CFPropertyList(); $ResultRow = mysqli_fetch_row( $result ); $plist-add( $TableRow = new CFArray() ); foreach ( $ResultRow as $Item ){ $TableRow-add( new CFString( $Item ) ); }

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  • limiting mysql results by range of a specific key INCLUDING DUPLICATES

    - by aVC
    I have a query SELECT p.*, m.*, (SELECT COUNT(*) FROM newPhotoonAlert n WHERE n.userIDfor='$id' AND n.threadID=p.threadID and n.seen='0') AS unReadCount FROM posts p JOIN myMembers m ON m.id = p.user_id LEFT JOIN following f ON (p.user_id = f.user_id AND f.follower_id='$id' AND f.request='0' AND f.status='1') JOIN myMembers searcher ON searcher.id = '$id' WHERE ((f.follower_id = searcher.id) OR m.id='$id') AND p.flagged <'5' ORDER BY p.threadID DESC,p.positionID It brings result as expected but I want to add Another CLAUSE to limit the results. Say a sample (minimal shown) set of data looks like this with the above query. threadID postID positionID url 564 1254 2 a.com 564 1245 1 a1.com 541 1215 3 b1.com 541 1212 2 b2.com 541 1210 1 b3.com 523 745 1 c1.com 435 689 2 d2.com 435 688 1 a4.com 256 345 1 s3.com 164 316 1 f1.com . . I want to get ROWS corresponding to 2 DISTINCT threadIDs starting from MAX, but I want to include duplicates as well. Something like AND p.threadID IN (Select just Two of all threadIDs currently selected, but include duplicate rows) So my result should be threadID postID positionID url 564 1254 2 a.com 564 1245 1 a1.com 541 1215 3 b1.com 541 1212 2 b2.com 541 1210 1 b3.com

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  • how to remove mysql root password

    - by nectar
    I want to remove password for user root in localhost how can I do that?by mistake I have set the password of root user thats why phpmyadmin is giving error- #1045 - Access denied for user 'root'@'localhost' (using password: NO)

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  • onUpdate in MySQL means?

    - by ajsie
    i know that if you create a foreign key on a field (parent_id) in a child table that refer to a parent table's primary key (id), then if this parent table is deleted the child class will be deleted as well if you set onDelete to cascade when creating the foreign key in the child class. but what happens if i set it to onUpdate = cascade?

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  • mysql_num_rows(): supplied argument is not a valid MySQL result resource

    - by php-b-grader
    I am getting this error when I pass an invalid SQL string... I spent the last hour trying to find the problem assuming - It's not my SQL it must be the db handle... ANyway, I've now figured out that it was bad SQL... What I want to do is test the result of the mysql_query() for a valid resultset. I am simply using empty($result)... Is this the most effective test? Is there a more widely accepted method of testing a resultset for a valid result?

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  • MYSQL UPDATE with IN and Subquery

    - by Johal
    Hi i have tables like this : table entry : id | total_comments ___________ 1 | 0 2 | 0 3 | 0 4 | 0 table comments : id | eid | comment ___________ 1 | 1 | comment sdfd 2 | 1 | testing testing 3 | 1 | comment text 4 | 2 | dummy comment 5 | 2 | sample comment 6 | 1 | fg fgh dfh Query i write : UPDATE entry SET total_comments = total_comments + 1 WHERE id IN ( SELECT eid FROM comments WHERE id IN (1,2,3,4,5,6)) Results i get is : table entry : id | total_comments ___________ 1 | 1 2 | 1 3 | 0 4 | 0 Expected results : table entry : id | total_comments ___________ 1 | 4 2 | 2 3 | 0 4 | 0 Any help will be appreciated.

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  • MySQL Query : Advise Needed.

    - by Rachel
    "SELECT id as Id from dbTable WHERE code = ? AND CURDATE() BETWEEN start_date AND end_date AND offerId IN ('12321', '12124')"; //Passing arguments for the query $args = array_merge(array(51342),$offerid); //Execute the prepared query $statement->execute($args); Now array(51342) represents combination of code+value, aside my database has value, code columns and so I want a query which would look logically like "SELECT id as Id from dbTable WHERE code and value (Note here I do not know the syntax, what am looking at is (code+value = ?), please advise on query) = ? AND CURDATE() BETWEEN start_date AND end_date AND offerId IN ('12321', '12124')";

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  • Problem with MySQL query

    - by Psyche
    This time my setup looks like this: one table with galleries names (gallery_id, gallery_name) and another table with galleries photos (photo_id, photo_gallery_id, photo_name). What I need is to get all the galleries with one random picture for each gallery. Is it possible to do this with a single query?

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  • MySQL: group by and IF statement

    - by notset
    By default, parent_id = 0. I want to select all records with parent_id = 0 and only the last ones with parent_id 0. I tried this, but it didn't work: SELECT * FROM `articles` IF `parent_id` > 0 THEN GROUP BY `parent_id` HAVING COUNT(`parent_id`) >= 1 END; ORDER BY `time` DESC What could be the solution?

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  • MySQL - generate numbers for groups of a result

    - by FreeIX
    I need a query to return this result: +---------+-----+-------+ | ref_nid | nid | delta | +---------+-----+-------+ | AA | 97 | 1 | | BB | 97 | 2 | | CC | 97 | 3 | | DD | 98 | 1 | | EE | 98 | 2 | | FF | 98 | 3 | +---------+-----+-------+ However, I do not have the delta column. I need to generate it for each nid group. In other words, I need an auto incremented number for each group of the result.

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  • How to optimize this MYSQL table?

    - by Lost_in_code
    This is for an upcoming project. I have two tables - first one keeps tracks of photos, and the second one keeps track of the photo's rank Photos: +-------+-----------+------------------+ | id | photo | current_rank | +-------+-----------+------------------+ | 1 | apple | 5 | | 2 | orange | 9 | +-------+-----------+------------------+ The photo rank keeps changing on a regular basis and this is the table that tracks it: Ranks: +-------+-----------+----------+-------------+ | id | photo_id | ranks | timestamp | +-------+-----------+----------+-------------+ | 1 | 1 | 8 | * | | 2 | 2 | 2 | * | | 3 | 1 | 3 | * | | 4 | 1 | 7 | * | | 5 | 1 | 5 | * | | 6 | 2 | 9 | * | +-------+-----------+----------+-------------+ * = current timestamp Every rank is tracked for reporting/analysis purpose. I talked to someone who has experience in this field and he told me that storing ranks like above is the way to go. But I'm not so sure yet. The problem here is data redundancy. There are going to be tens of thousands of photos. The photo rank changes on a hourly basis (many time within minutes) for recent photos but less frequently for older photos. At this rate the table will have millions of records within months. And since I do not have experience in working with large databases, this makes me a little nervous. I thought of this: Ranks: +-------+-----------+--------------------+ | id | photo_id | ranks | +-------+-----------+--------------------+ | 1 | 1 | 8:*,3:*,7:*,5:* | | 2 | 2 | 2:*,9:* | +-------+-----------+--------------------+ * = current timestamp That means some extra code in PHP to split the rank/time (and sorting) but that looks OK to me. Is this a correct way to optimize the table for performance? What would you recommend? Any suggestions would be great.

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  • MySQL charset issue.

    - by Shagymoe
    I'm not sure exactly why this happened, but I'm assuming it was an dump and import. The db is full of characters like — for commas and such. I've tried various solutions on the web, but nothing seems to work. I've verified that the html header specifies utf8. Any ideas on how I can get the entire db back to normal characters?

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  • PHP & MySQL - Undefined variable problem?

    - by TaG
    I keep getting the following error Undefined variable: password on line 33 how do I correct this problem? So this error will stop showing. Here is the php code. $first_name = mysqli_real_escape_string($mysqli, $purifier->purify(htmlentities(strip_tags($_POST['first_name'])))); $password1 = mysqli_real_escape_string($mysqli, $purifier->purify(strip_tags($_POST['password1']))); $password2 = mysqli_real_escape_string($mysqli, $purifier->purify(strip_tags($_POST['password2']))); // Check for a password and match against the confirmed password: if ($password1 == $password2) { $sha512 = hash('sha512', $password1); $password = mysqli_real_escape_string($mysqli, $sha512); } else { echo '<p class="error">Your password did not match the confirmed password!</p>'; } //If the table is not found add it to the database if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, first_name, password) VALUES ('$user_id', '$first_name', '$password')"); } //If the table is in the database update each field when needed if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE users SET first_name = '$first_name', password = '$password' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; }

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