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  • Input questions mysql php html

    - by Marcelo
    (Q1)Hi I'm using textbox in my project and I can't receive the values that are typed <textarea rows="5" cols="60"> Type your suggestion </textarea> <br> <input type="submit" name="sugestao" value="Submit" /> Sorry I don't know how to 'kill' html code, that's why < is missing. All I'm getting in the column of the database from this text box is "Submit", I'd like to receive whatever is written in the text area. How can I make the value equal whaterever is typed? (Q2) How can I make sure that I'll only store the same type(int,varchar,text) that I setted,declared in the database. For example: age(int), but if someone types "abc" in the input it will be stored in my database as the value 0 . How can I forbid this, and only save the age when it's just int and all the other fields(like name, email) are filled ?. And if is still possible warn the user that he is typing something wrong, don't need to say where. Sorry for any mistake in English and Thanks for the attention.

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  • How to handle large table in MySQL ?

    - by Frantz Miccoli
    I've a database used to store items and properties about these items. The number of properties is extensible, thus there is a join table to store each property associated to an item value. CREATE TABLE `item_property` ( `property_id` int(11) NOT NULL, `item_id` int(11) NOT NULL, `value` double NOT NULL, PRIMARY KEY (`property_id`,`item_id`), KEY `item_id` (`item_id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; This database has two goals : storing (which has first priority and has to be very quick, I would like to perform many inserts (hundreds) in few seconds), retrieving data (selects using item_id and property_id) (this is a second priority, it can be slower but not too much because this would ruin my usage of the DB). Currently this table hosts 1.6 billions entries and a simple count can take up to 2 minutes... Inserting isn't fast enough to be usable. I'm using Zend_Db to access my data and would really be happy if you don't suggest me to develop any php side part. Thanks for your advices !

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  • mysql boolean joins

    - by user280381
    I want to use a JOIN to return a boolean result. Here's an example of the data... t1 id | data | 1 | abcd | 2 | 2425 | 3 | xyz | t2 id | data | t1_id | 1 | 75 | 2 | 2 | 79 | 2 | 3 | 45 | 3 | So with these two tables I want to select all the data from t1, and also whether a given variable appears in t2.data for each id. So say the variable is 79, the results should be id | data | t2_boolean 1 | abcd | 0 2 | abcd | 1 3 | xyz | 0 So I'm thinking some sort of join is needed, but without a WHERE clause. I've been banging my head about this one. Is it possible? I really need it inside the same statement as I want to sort results by the boolean field. As the boolean needs to be a field, can I put a join inside of a field? Thanks...

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  • many to many relationship mysql select

    - by zeina
    Let's consider 2 tables "schools" and "students". Now a student may belong to different schools in his life, and a school have many students. So this is a many to many example. A third table "links" specify the relation between student and school. Now to query this I do the following: Select sc.sid , -- stands for school id st.uid, -- stands for student id sc.sname, -- stands for school name st.uname, -- stands for student name -- select more data about the student joining other tables for that from students s left join links l on l.uid=st.uid -- l.uid stands for the student id on the links table left join schools sc on sc.sid=l.sid -- l.sid is the id of the school in the links table where st.uid=3 -- 3 is an example this query will return duplicate data for the user id if he has more than one school, so to fix this I added group by st.uid, yet I also need the list of school name related to the same user. Is there a way to do it with fixing the query I wrote instead of having 2 queries? So as example I want to have Luci of schools ( X, Y, Z, R, ...) etc

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  • mySQL: Joining three tables - how?

    - by Industrial
    Hi everybody, I have the following query in my application. It works well, but I need it to also contain the number of products that are associated with each manufacturer. The current query: SELECT * FROM (`manufacturers`) JOIN `languages` ON `manufacturers`.`lang` = `languages`.`id` ORDER BY `languages`.`id` asc, `id` asc My products table looks like this: id | name | manufacturerid 0 | Product1 | 0

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  • Help with MySQL date operation

    - by jartaud
    Hello i have two columns: duration (int) and dateCol (date) in a table. How should i proceed if i want to know when CURDATE() - dateCol will iqual to duration. ie: CURDATE() - dateCol (2010-05-08) = duration (14 days), and then send an e-amil?

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  • Add results from MySQL?

    - by NardCake
    Not quite sure how descriptive that title was but this is what I want to do. I scripted a URL shortener today and it's working fine, I just want to add some stats on the bottom of it saying how many links there are and how many clicks there are. Now everytime a user clicks one of the links it +1 the column in the database then it redirects the user. I want to query that and add all of those numbers together from each row. I attempted a while loop which im not surprised didn't work: while($rows = mysql_fetch_assoc($check_count)){ $clicks = $rows['clicks']; $clicks = $clicks+$clicks; } If you don't understand please ask and if you do understand it means alot for any response!

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  • Does introducing foreign keys to MySQL reduce performance

    - by Tam
    I'm building Ruby on Rails 2.3.5 app. By default, Ruby on Rails doesn't provide foreign key contraints so I have to do it manually. I was wondering if introducing foreign keys reduces query performance on the database side enough to make it not worth doing. Performance in this case is my first priority as I can check for data consistency with code. What is your recommendation in general? do you recommend using foreign keys? and how do you suggest I should measure this?

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  • sql jdbc getgeneratedkeys with mysql returns column "id" not found

    - by iamrohitbanga
    I want to retrieve the most recently updated value in the table using an insert query. these are the datatypes in my sql table. int(11) // primary key auto increment, not being assigned by sqlQuery varchar(30) timestamp // has a default value. but i am explicit assigning it using CURRENT_TIMESTAMP varchar(300) varchar(300) varchar(300) int(11) varchar(300) // java code statement.executeUpdate(sqlQuery, Statement.RETURN_GENERATED_KEYS); ResultSet rs = statement.getGeneratedKeys(); System.out.println("here: " + rs.getMetaData().getColumnCount()); System.out.println("here1: " + rs.getMetaData().getColumnName(1)); // none of the following 3 works System.out.println("id: " + rs.getInt(1)); System.out.println("id: " + rs.getInt("GENERATED_KEY")); System.out.println("id: " + rs.getInt("id")); for a bit of background see this

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  • csv file upload and update mysql db

    - by Indra
    I am very to new to php. i am using the following code to open a csv file and update my database. i need to check the value of first row-first column of the csv file. if it is matching "some text 1", then i need to run code1, if it is "some text 2", run code2, else code3. I can use if else condition but since i am using while loop It fails. Can anyone help me. $handle = fopen($file_tmp,"r"); while(($fileop = fgetcsv($handle,",")) !== false) { // I need to check here $companycode = mysql_real_escape_string($fileop[0]); $Item = mysql_real_escape_string($fileop[3]); $pack = preg_replace('/[^A-Za-z0-9\. -]/', '', $fileop[4]); $lastmonth = mysql_real_escape_string($fileop[5]); $ltlmonth = mysql_real_escape_string($fileop[6]); $op = mysql_real_escape_string($fileop[9]); $pur = mysql_real_escape_string($fileop[10]); $sale = mysql_real_escape_string($fileop[12]); $bal = mysql_real_escape_string($fileop[17]); $bval = mysql_real_escape_string($fileop[18]); $sval = mysql_real_escape_string($fileop[19]); $sq1 = mysql_query("INSERT INTO `sas` (companycode,Item,pack,lastmonth,ltlmonth,op,pur,sale,bal,bval,sval) VALUES ('$companycode','$Item','$pack','$lastmonth','$ltlmonth','$op','$pur','$sale','$bal','$bval','$sval')"); }

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  • How to refactor this MySQL code?

    - by Jader Dias
    SELECT * ( SELECT * FROM `table1` WHERE `id` NOT IN ( SELECT `id` FROM `table2` WHERE `col4` = 5 ) group by `col2` having sum(`col3`) > 0 UNION SELECT * FROM `table1` WHERE `id` NOT IN ( SELECT `id` FROM `table2` WHERE `col4` = 5 ) group by `col2` having sum(`col3`) = 0 ) t1; For readability and performance reasons, I think this code could be refactored. But how?

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  • MySQL - Join as zero if record Not IN

    - by Zurahn
    To explain by example, take two tables, A and B Table A id foo 1 x 2 y 3 z Table B id aid bar 1 3 50 2 1 100 An example join SELECT foo, bar FROM a, b WHERE a.id = b.aid; Garners a result of foo bar z 50 x 100 What I would like to do is get all values of foo and for any instances where there isn't a corresponding bar value, return 0 for that column. My best guess was something along the lines of SELECT foo, bar AS br FROM a, b WHERE a.id = b.aid OR a.id NOT IN (SELECT aid FROM b); But that returns duplicates and non-zero values for bar. Possible?

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  • Mysql / Django - Begin auto increment to 1

    - by ark
    Hi, I have a django script which loads data, the beginning of the script deletes all datas in database. So when I execute 1st time this script, the auto increment primary keys begin to 1 to 15 (if 15 objects) and if I want to reload data, I reexecute the script. My issue is when I execute it again, pks numbers begin to 16 (for 2nd launch), I would like each time auto_increment begins to 1, is it possible whitout regenerating tables structure each time ? Thanks

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  • how to update mysql with inputs and checkboxes

    - by Ronnie Chester Lynwood
    hi. with this code: <? if (isset($_POST['onay'])) { foreach ($_POST['secilen'] as $zuha) { $olay = mysql_query("update mp3 SET aktif = '1' WHERE id = '$zuha'"); } if ($olay) { echo "islem tamam"; exit; } } ?> and with this form: (select * from mp3 aktif = '0') <form name="form" method="post"> <input readonly type="text" name="id" value="<?=$haciosman['id']?>" /> <input type="text" name="baslik" value="<?=$haciosman['baslik']?> <input type="checkbox" name="secilen[]" value="<?=$haciosman['id']?>"> <input type="submit" name="onay" value="Onayla" /> I can set "aktif to 1" for each row that i checked. but i want to update "baslik" too! how can i update each "baslik" with that php code?

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  • MySQL join headaches, please help!

    - by Andrew Heath
    Ok, I've hit the wall here and need some help. Sample tables are as follows: SCENARIO_NATIONS [scenID] [side] [nation] scen001 1 Germany scen001 2 Britain scen001 2 Canada SCENARIO_NEEDUNITS [scenID] [unitID] scen001 0001 scen001 0003 scen001 0107 scen001 0258 scen001 0759 UNIT_BASIC_DATA [unitID] [nation] [name] 0001 Germany Mortars 0003 Germany Infantry 0107 Britain Lt 0258 Britain Infantry 0759 Canada Kilted Yaksmen Goal: given a scenID, pull a list of units from the database sorted by side, nation, name. I can do everything except for the side inclusion with: SELECT scenario_needunits.scenID, unit_basic_data.nation, unit_basic_data.name FROM scenario_needunits LEFT OUTER JOIN unit_basic_data ON scenario_needunits.unitID=unit_basic_data.unitID WHERE scenario_needunits.scenID='scen001' ORDER BY unit_basic_data.nation ASC, unit_basic_data.name ASC I've tried just dropping the SCENARIO_NATIONS table in as a LEFT OUTER JOIN on scenID but what ends up happening is that ALL units come back with a side of 1 because that's always the first side listed for the scenID in the SCENARIO_NATIONS table. Conceptually, what I think needs to happen is SCENARIO_NATIONS must be joined to both the scenID (to restrict it to just that scenario) and to each unit's nation but I don't have any idea how to do that and my Google-fu is inadequate. :-/

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  • mysql update unique index

    - by atno
    Is there a way to say UPDATE items SET qty=10 WHERE **unique key** instead of saying UPDATE items SET qty=10 WHERE userID=1 AND listID=10 and itemID=100 on the following table? CREATE TABLE IF NOT EXISTS `items` ( `userID` int(20) NOT NULL, `listID` int(20) NOT NULL, `itemID` int(20) NOT NULL, `qty` int(10) NOT NULL, UNIQUE KEY `unique` (`userID`,`listID`,`itemID`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

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  • php and mysql user tracking and reporting

    - by inertiahz
    Hi, I currently have a table which consists of user information and lesson id; the table layout looks like: ---------------------------------------------------- |employeeID|numVisits|lessonID1|lessonID2|lessonID3| ---------------------------------------------------- |33388 |2 |1 |0 |3 | and a lessons table which contains the information about the lesson: ------------------------------------------------------ |lessonID |cateogry |title |filepath |numberviews| ------------------------------------------------------ |1 |beginner |lesson |file:// |10 | Within the lessonID fields in the user table is an integer which tracks how many times someone has clicked on a lesson. Now what I am trying to do is in a report I have the top 5 people who have visited the site and would like to then be able to drill down into what lessons they have clicked on. Can anyone help with this? Or would restructuring the way the database is be an easier task? Thanks

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  • Need help optimizing MYSQL query with join

    - by makeee
    I'm doing a join between the "favorites" table (3 million rows) the "items" table (600k rows). The query is taking anywhere from .3 seconds to 2 seconds, and I'm hoping I can optimize it some. Favorites.faver_profile_id and Items.id are indexed. Instead of using the faver_profile_id index I created a new index on (faver_profile_id,id), which eliminated the filesort needed when sorting by id. Unfortunately this index doesn't help at all and I'll probably remove it (yay, 3 more hours of downtime to drop the index..) Any ideas on how I can optimize this query? In case it helps: Favorite.removed and Item.removed are "0" 98% of the time. Favorite.collection_id is NULL about 80% of the time. SELECT `Item`.`id`, `Item`.`source_image`, `Item`.`cached_image`, `Item`.`source_title`, `Item`.`source_url`, `Item`.`width`, `Item`.`height`, `Item`.`fave_count`, `Item`.`created` FROM `favorites` AS `Favorite` LEFT JOIN `items` AS `Item` ON (`Item`.`removed` = 0 AND `Favorite`.`notice_id` = `Item`.`id`) WHERE ((`faver_profile_id` = 1) AND (`collection_id` IS NULL) AND (`Favorite`.`removed` = 0) AND (`Item`.`removed` = '0')) ORDER BY `Favorite`.`id` desc LIMIT 50;

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  • mysql get table based on common column between two tables

    - by Zentdayn
    while trying to learn sql i came across "Learn SQL The Hard Way" and i started reading it. Everything was going fine then i thought ,as a way to practice, to make something like given example in the book (example consists in 3 tables pet,person,person_pet and the person_pet table 'links' pets to their owners). I made this: report table +----+-------------+ | id | content | +----+-------------+ | 1 | bank robbery| | 2 | invalid | | 3 | cat on tree | +----+-------------+ notes table +-----------+--------------------+ | report_id | content | +-----------+--------------------+ | 1 | they had guns | | 3 | cat was saved | +-----------+--------------------+ wanted result +-----------+--------------------+---------------+ | report_id | report_content | report_notes | +-----------+--------------------+---------------+ | 1 | bank robbery | they had guns | | 2 | invalid | null or '' | | 3 | cat on tree | cat was saved | +-----------+--------------------+---------------+ I tried a few combinations but no success. My first thought was SELECT report.id,report.content AS report_content,note.content AS note_content FROM report,note WHERE report.id = note.report_id but this only returns the ones that have a match (would not return the invalid report). after this i tried adding IF conditions but i just made it worse. My question is, is this something i will figure out after getting past basic sql or can this be done in simple way? Anyway i would appreciate any help, i pretty much lost with this. Thank you. EDIT: i have looked into related questions but havent yet found one that solves my problem. I probably need to look into other statements such as join or something to sort this out.

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  • MySQL query against pseudo-key-value pair data in WordPress custom query

    - by andrevr
    I'm writing a custom WordPress query to use some of the data which the Woothemes Diarise theme creates. Diarise is an event planner theme with calendar blah, blah... and uses custom fields to store the event start and end dates in WP custom fields in the *wp_postmeta* table, which implements a key-value store. So for each post in the "event" category, there are 2 records in *wp_postmeta*, named *event_start_date* and *event_end_date* that I'm interested in. The task is to compare a tourist's arrival and departure dates with the start and end dates of events, yielding a what's on list of events available. We thought we'd killed it with a grand flash of logic, that goes like this: Disregard any event that ends before the tourist arrives, and any that begin after the departure date. I wrote this query: SELECT wposts.* FROM wp_posts wposts LEFT JOIN wp_postmeta wpostmeta ON wposts.ID = wpostmeta.post_id LEFT JOIN wp_term_relationships ON (wposts.ID = wp_term_relationships.object_id) LEFT JOIN wp_term_taxonomy ON (wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id) WHERE wp_term_taxonomy.taxonomy = 'category' AND wp_term_taxonomy.term_id IN(3,4) AND ( wpostmeta.meta_key = 'event_start_date' AND NOT ( concat(subst(wpostmeta.meta_value,7,4),'-',subst(wpostmeta.meta_value,4,2),'-',subst(wpostmeta.meta_value,1,2) > '2010-07-31' ) ) AND ( wpostmeta.meta_key = 'event_end_date' AND NOT ( concat(subst(wpostmeta.meta_value,7,4),'-',subst(wpostmeta.meta_value,4,2),'-',subst(wpostmeta.meta_value,1,2) < '2010-05-01' ) ) ) ORDER BY wpostmeta.meta_value ASC And, of course it returns no records. The problem I believe is in the dual reference to wpostmeta.meta_key, but how to get around that?

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