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  • Good article about File- and Folder Permissions on production server?

    - by Camran
    I have a classifieds website, and users may post classifieds, add images, remove classifieds etc etc... I have no idea what to set the permissions to on folders. For instance, a php script which I have uploads a file to a directory. What would you have set the directory permissions to? Nobody need access to the directory, only the php script... Just wonder if anybody has a good (brief) article about setting the "right" permissions? Thanks

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  • Testing a SQL Query for True or False

    - by KickingLettuce
    $sql = "SELECT # FROM users WHERE onduty = 1 AND loc_id = '{$site}';"; $result = mysql_query($sql); I simply want to test if this is true or false. If it returns 0 rows, I want next line to be something like: if (!$result) { //do this; } However, in my test, I am getting false when I know it should be true. Is this sound logic here? (note, yes I know I should be using mysqli_query, that is not what I am asking here)

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  • PhpMyAdmin; Should I disable root login?

    - by Camran
    I have this setup in Phpmyadmin: USER HOST PASSW PRIVILEGES GRANT debian-sys-maint localhost Yes ALL PRIVILEGES YES phpmyadmin localhost Yes USAGE NO root 127.0.0.1 Yes ALL PRIVILEGES YES root localhost Yes ALL PRIVILEGES YES root my_hostname Yes ALL PRIVILEGES YES username localhost Yes ALL PRIVILEGES YES Where "username" is my username and "my_hostname" is my hostname. I am currently only logging in as the last one (username, localhost). Also, I have php which also uses the last ones login details. Should I disable the other ones? And, what other security measures should I take? BTW: My server is Linux and I have root access. Thanks

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  • Lock innoDB table temporarily

    - by Industrial
    Hi everyone, I make bigger inserts consisting of a couple of thousand rows in my current web app and I would like to make sure that no one can do anything but read the table, until the inserts have been done. What is the best way to do this while keeping the read availability open for normal, non-admin users? Thanks!

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  • How to get time from db depending upon conditions

    - by Somebody is in trouble
    I have a table in which the value are Table_hello date col2 2012-01-31 23:01:01 a 2012-06-2 12:01:01 b 2012-06-3 20:01:01 c Now i want to select date in days if it is 3 days before or less in hours if it is 24 hours before or less in minutes if it is 60 minutes before or less in seconds if it is 60 seconds before or less in simple format if it is before 3days or more OUTPUT for row1 2012-01-31 23:01:01 for row2 1 day ago for row3 1 hour ago UPDATE My sql query select case when TIMESTAMPDIFF(SECOND, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(SECOND, `date`,current_timestamp), ' seconds') when TIMESTAMPDIFF(DAY, `date`,current_timestamp) <= 3 then concat(TIMESTAMPDIFF(DAY, `date`,current_timestamp), ' days')end when TIMESTAMPDIFF(HOUR, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(HOUR, `date`,current_timestamp), ' hours') when TIMESTAMPDIFF(MINUTE, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(MINUTE, `date`,current_timestamp), ' minutes') from table_hello Only problem is i am unable to use break and default in sql like switch case in c++

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  • How do I return the numeric value from a database query in PHP?

    - by Luke
    Hello, I am looking to retreive a numerical value from the database function adminLevel() { $q = "SELECT userlevel FROM ".TBL_USERS." WHERE id = '$_SESSION[id]'"; return mysql_query($q, $this->connection); } This is the SQL. I then wrote the following php/html: <?php $q = $database->adminLevel(); if ($q > 7) { ?> <a href="newleague.php">Create a new league</a> <? } ?> The problem I have is that the userlevel returned isn't affecting the if statement. It is always displayed. How do i get it to test the value of userlevel is greater than 7? Thanks

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  • How do you display a PHPmyadmin table onto a web browser/web page?

    - by user1390754
    Just a simple question, i've been searching around and for some reason i cannot find an answer to this. after creating a database/table in Phpmyadmin using xampp, what command do i need to put into my webpage (PHP) to show the table I made? I know the first step involves connecting to the database and i think i've done that properly. This is the code i found from somewhere about connecting (w3 tutorials I believe) $con = mysql_connect("localhost","xxxxxx,""); if (!$con) { die('Could not connect: ' . mysql_error()); }

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  • How to translate this query:

    - by xRobot
    How can I translate this 2 queries in postgresql ? : . CREATE TABLE example ( id int(10) unsigned NOT NULL auto_increment, from varchar(255) NOT NULL default '0', message text NOT NULL, lastactivity timestamp NULL default '0000-00-00 00:00:00', read int(10) unsigned NOT NULL, PRIMARY KEY (id), KEY from (from) ) DEFAULT CHARSET=utf8; . SELECT * FROM table_1 LEFT OUTER JOIN table_2 ON ( table_1.id = table_2.id ) WHERE (table_1.lastactivity > NOW()-100);

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  • Simple PHP query question: LIKE

    - by pg
    When I replace $ordering = "apples, bananas, cranberries, grapes"; with $ordering = "apples, bananas, grapes"; I no longer want cranberries to be returned by my query, which I've written out like this: $query = "SELECT * from dbname where FruitName LIKE '$ordering'"; Of Course this doesn't work, because I used LIKE wrong. I've read through various manuals that describe how to use LIKE and it doesn't quite make sense to me. If I change the end of the db to "LIKE "apples"" that works for limiting it to just apples. Do I have to explode the ordering on the ", " or is there a way to do this in the query?

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  • "slash before every quote" problem

    - by Camran
    I have a php page which contains a form. Sometimes this page is submitted to itself (like when pics are uploaded). I wouldn't want users to have to fill in every field again and again, so I use this as a value of a text-input inside the form: value="<?php echo htmlentities(@$_POST['annonsera_headline'],ENT_COMPAT,'UTF-8');?>"> This works, except it adds a "\" sign before every double-quote... For instance writing 19" wheels gives after page is submitted to itself: 19\" wheels And if I don't even use htmlentities then everything after the quotes dissappears. What is the problem here?

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  • [PHP] 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 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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  • Broken count(*) after adding LEFT JOIN

    - by Iain Urquhart
    Since adding the LEFT JOIN to the query below, the count(*) has been returning some strange values, it seems to have added the total rows returned in the query to the 'level': SELECT `n`.*, exp_channel_titles.*, round((`n`.`rgt` - `n`.`lft` - 1) / 2, 0) AS childs, count(*) - 1 + (`n`.`lft` > 1) + 1 AS level, ((min(`p`.`rgt`) - `n`.`rgt` - (`n`.`lft` > 1)) / 2) > 0 AS lower, (((`n`.`lft` - max(`p`.`lft`) > 1))) AS upper FROM `exp_node_tree_6` `n` LEFT JOIN `exp_channel_titles` ON (`n`.`entry_id`=`exp_channel_titles`.`entry_id`), `exp_node_tree_6` `p`, `exp_node_tree_6` WHERE `n`.`lft` BETWEEN `p`.`lft` AND `p`.`rgt` AND ( `p`.`node_id` != `n`.`node_id` OR `n`.`lft` = 1 ) GROUP BY `n`.`node_id` ORDER BY `n`.`lft` I'm totally stumped... Thank you!

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  • Select multiple records by one query

    - by kofto4ka
    Hello there. Please, give me advice, how to construct select query. I have table table with fields type and obj_id. I want to select all records in concordance with next array: $arr = array( 0 => array('type' => 1, 'obj_id' => 5), 1 => array('type' => 3, 'obj_id' => 15), 2 => array('type' => 4, 'obj_id' => 14), 3 => array('type' => 12, 'obj_id' => 17), ); I want to select needed rows by one query, is it real? Smth like select * from `table` where type in (1,3,4,12) and obj_id in (5,15,14,17) But this query returns also records with type = 3 and obj_id = 14, and for example type = 1 and obj_id = 17. p.s. moderators, please fix my title, I dont know how to describe my question.

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  • Help creating a SQL Query

    - by Wes
    Somewhat new to SQL queries and I need a little help with my join. I am supplied with gid For each of these I need to grab name from table wp_ngg_gallery then join in table wp_ngg_pictures and grab field filename limit 1 order DESC by field imagedate Anyone able to help?

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  • C# SQL SELECT Statement

    - by Feren6
    I have the following code: SqlCommand cmd2 = new SqlCommand("SELECT ClaimId FROM tblPayment WHERE PaymentId = " + PaymentID.ToString(), mvarDBConn); SqlDataReader reader = cmd2.ExecuteReader(); reader.Read(); Int32 ClaimId = reader.GetInt32(0); reader.Close(); If I run the SELECT statement in SQL it returns the number fine, but when I use ExecuteReader all it returns is 0. I've tried multiple methods including ExecuteScalar, ExecuteNonQuery, reader.GetString then casting that to an int, etc. What am I missing? Thanks.

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  • How to do insert one row from one table to another table(what has less columns)?

    - by chobo2
    Hi I am trying to find rows that are not in one table and insert them into another. The table I am trying to insert into has less columns then the other one. These columns are null but it would be cool if I could hardcode a value before I do the insert. But I am having so much trouble with just trying to get it to insert. I have something like this SELECT p.ProductId, p.ProductName INTO SomeTable FROM Product as p WHERE p.ProductName != 'iPad' I will get a error like this though Msg 4104, Level 16, State 1, Line 1 The multi-part identifier "p.ProductId" could not be bound. I am not sure what I am doing wrong. I copied and pasted the names in so I don't think it is a spelling mistake. I am using ms sql 2005 express.

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  • How grouping and totaling are done into three tables using JOIN

    - by text
    Here are my tables respondents: field sample value respondentid : 1 age : 2 gender : male survey_questions: id : 1 question : Q1 answer : sample answer answers: respondentid : 1 question : Q1 answer : 1 --id of survey question I want to display all respondents who answered the certain survey, display all answers and total all the answer and group them according to the age bracket. I tried using this query: $sql = "SELECT res.Age, res.Gender, answer.id, answer.respondentid, SUM(CASE WHEN res.Gender='Male' THEN 1 else 0 END) AS males, SUM(CASE WHEN res.Gender='Female' THEN 1 else 0 END) AS females, CASE WHEN res.Age < 1 THEN 'age1' WHEN res.Age BETWEEN 1 AND 4 THEN 'age2' WHEN res.Age BETWEEN 4 AND 9 THEN 'age3' WHEN res.Age BETWEEN 10 AND 14 THEN 'age4' WHEN res.Age BETWEEN 15 AND 19 THEN 'age5' WHEN res.Age BETWEEN 20 AND 29 THEN 'age6' WHEN res.Age BETWEEN 30 AND 39 THEN 'age7' WHEN res.Age BETWEEN 40 AND 49 THEN 'age8' ELSE 'age9' END AS ageband FROM Respondents AS res INNER JOIN Answers as answer ON answer.respondentid=res.respondentid INNER JOIN Questions as question ON answer.Answer=question.id WHERE answer.Question='Q1' GROUP BY ageband ORDER BY res.Age ASC"; I was able to get the data but the listing of all answers are not present. What's wrong with my query. I want to produce something like this: ex: # of Respondents is 3 ages: 2,3 and 6 Question: what are your favorite subjects? Ages 1-4: subject 1: 1 subject 2: 2 subject 3: 2 total respondents for ages 1-4 : 2 Ages 5-10: subject 1: 1 subject 2: 1 subject 3: 0 total respondents for ages 5-10 : 1

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  • Open Source PHP search engine

    - by Ravi Gupta
    I am looking for an open source search engine plugin written in php for my website(eCommerce). Before anybody answer that I have a doubt regarding the search engine. Usually search engine crawl web pages, create indexes and then use them while looking for contents. But will the same model work for eCommerce websites? Yeah, it can crawl products pages, index them but don't you think it would be better if it crawls the database directly and index the products stored in the database? And when a user search for any product, it will simply give us the rows of the table which matches the user query? May be what I am asking is a stupid question but I am new to web development, so kindly help me to understand the concept. I have looked at a search engine called Sphider but didn't get what all I have to do to make it work with an eCommerce website.

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  • How to find and update the next record in a linking table

    - by someoneinomaha
    I have a classifieds system I'm working on. People are able to add photos to a classified, but I only display one when displaying the list of classifieds. To do that, I have a linking table between classifieds and photos that has a "is_main" boolean field. When someone deletes one of their classified photos, I want to: 1) See if there is more than that photo tied to the classified. 2) If there is, update the next photo and set that "is_main" field to TRUE. Just trying to find out the most efficient way to do this.

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