Search Results

Search found 20931 results on 838 pages for 'mysql insert'.

Page 410/838 | < Previous Page | 406 407 408 409 410 411 412 413 414 415 416 417  | Next Page >

  • SQL Server Mapping a user to a login and adding roles programmatically

    - by user163457
    In my SQL Server 2005 server I create databases and logins using Management Studio. My application requires that I give a newly created user read and write permissions to another database. To do this I right-click the newly created login, select properties and go to User Mapping. I put a check beside the database to map this login to the db and select db_datareader and db_datawriter as the roles to map. Can this be done programmatically? I've read about using Alter User and sp_change_users_login but I'm having problems getting these to work, since sp_change_users_login has been deprecated so I'd prefer to use Alter User. Please note my understanding of SQL Server database users/logins/roles is basic

    Read the article

  • add array element to row returned from sql query

    - by bert
    I want to add an additional value into an array before passing it to json_encode function, but I can't get the syntax right. $result = db_query($query); // $row is a database query result resource while ($row = db_fetch_object($result)) { $stack[] = $row; // I am trying to 'inject' array element here $stack[]['x'] = "test"; } echo json_encode($stack);

    Read the article

  • PHP looping through an array to fetch a value for each key from database (third normal form)

    - by zomboble
    I am building a system, mostly for consolidating learning but will be used in practice. I will try and verbally explain the part of the E-R diagram I am focusing on: Each cadet can have many uniformID's Each Uniform ID is a new entry in table uniform, so cadets (table) may look like: id | name | ... | uniformID 1 | Example | ... | 1,2,3 uniform table: id | notes | cadet 1 | Need new blahh | 1 2 | Some stuff needed | 1 3 | Whatever you like | 1 On second thought, looks like I wont need that third column in the db. I am trying to iterate through each id in uniformID, code: <?php $cadet = $_GET['id']; // set from URL $query = mysql_query("SELECT `uniformID` FROM `cadets` WHERE id = '$cadet' LIMIT 1") or die(mysql_error()); // get uniform needed as string // store it while ($row = mysql_fetch_array($query)) { $uniformArray = $row['uniformID']; } echo $uniformArray . " "; $exploded = explode(",", $uniformArray); // convert into an array // for each key in the array perform a new query foreach ($exploded as $key => $value) { $query(count($exploded)); $query[$key] = mysql_query("SELECT * FROM `uniform` WHERE `id` = '$value'"); } ? As I say, this is mainly for consolidation purposes but I have come up with a error, sql is saying: Fatal error: Function name must be a string in C:\wamp\www\intranet\uniform.php on line 82 line 82 is: $query[$key] = mysql_query("SELECT * FROM `uniform` WHERE `id` = '$value'"); I wasn't sure it would work so I tried it and now i'm stuck! EDIT: Thanks to everyone who has contributed to this! This is now the working code: foreach ($exploded as $key => $value) { //$query(count($exploded)); $query = mysql_query("SELECT * FROM `uniform` WHERE `id` = '$value'"); while ($row = mysql_fetch_array($query)) { echo "<tr> <td>" . $row['id'] . "</td> <td>" . $row['note'] . "</td> </tr>"; } } Added the while and did the iteration by nesting it in the foreach

    Read the article

  • How many indexes will actually get used?

    - by Ender
    I'm writing a page that does very simple search queries, resulting in something like: SELECT * FROM my_table WHERE A in (a1, a2, a3) AND B in (b1, b2) AND C in (c1, c2, c3, c4) AND And so on for a variable number of columns, usually ~5. If I create a separate index for each column (one for A, one for B, one for C, not (A,B,C)), will all of them be used in the above query?

    Read the article

  • PHP Foreach statement issue. Multiple rows are returned

    - by Daniel Patilea
    I'm a PHP beginner and lately i've been having a problem with my source code. Here it is: <html> <head> <title> Bot </title> <link type="text/css" rel="stylesheet" href="main.css" /> </head> <body> <form action="bot.php "method="post"> <lable>You:<input type="text" name="intrebare"></lable> <input type="submit" name="introdu" value="Send"> </form> </body> </html> <?php //error_reporting(E_ALL & ~E_NOTICE); mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("robo") or die(mysql_error()); $intrebare=$_POST['intrebare']; $query = "SELECT * FROM dialog where intrebare like '%$intrebare%'"; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result) or die(mysql_error()); ?> <div id="history"> <?php foreach($row as $rows){ echo "<b>The robot says: </b><br />"; echo $row['raspuns']; } ?> </div> It returns me the result x6 times. This problem appeared when I've made that foreach because I wanted the results to stuck on the page one by one after every sql querry. Can you please tell me what seems to be the problem? Thanks!

    Read the article

  • Time diff calculations where date and time are in seperate columns

    - by pedalpete
    I've got a query where I'm trying to get the hours in duration (eg 6.5 hours) between two different times. In my database, time and date are held in different fields so I can efficiently query on just a startDate, or endDate as I never query specifically on time. My query looks like this SELECT COUNT(*), IFNULL(SUM(TIMEDIFF(endTime,startTime)),0) FROM events WHERE user=18 Sometimes an event will go overnight, so the difference between times needs to take into account the differences between the dates as well. I've been trying SELECT COUNT(*), IFNULL(SUM(TIMEDIFF(CONCAT(endDate,' ',endTime),CONCAT(startDate,' ',startTime))),0) FROM events WHERE user=18 Unfortunately I only get errors when I do this, and I can't seem to combine the two fields into a single timestamp.

    Read the article

  • Where binary in SQL

    - by fire
    I have an SQL statement: SELECT * FROM customers WHERE BINARY login='xxx' AND password='yyyy' There are no blob/binary fields in the table, do I need the BINARY after the WHERE what else does it do?

    Read the article

  • PHP Ajax not working

    - by Kostis
    I have 3 buttons on my page and depending on which one the user is clickingi want to run through ajax call a delete query in my database. When the user clicks on a button the javascript function seems to work but it doesn't run the query in php script. The html page: <?php session_start(); ?> <!DOCTYPE HTML> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-7"> <script> function myFunction(name) { var r=confirm("Are you sure? This action cannot be undone!"); if (r==true) { alert(name); // check if is getting in if statement and confirm the parameter's value var xmlhttp; if (str.length==0) { document.getElementById("clearMessage").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("clearMessage").innerHTML= responseText; } } xmlhttp.open("GET","clearDatabase.php?q="+name,true); xmlhttp.send(); } else alert('pff'); } </script> </head> <body> <div id="wrapper"> <div id="header"></div> <div id="main"> <?php if (session_is_registered("username")){ ?> <!--<a href="#">???a????s? pa?a??? µ???µ?t??</a><br /> <a href="#">???a????s? pa?a??? s??ed????</a><br /> <a href="#">???a????s? push notifications</a><br />--> <input type="button" value="???a????s? pa?a??? µ???µ?t??" onclick="myFunction('messages')" /> <input type="button" value="???a????s? pa?a??? s??ed????" onclick="myFunction('conferences')" /> <input type="button" value="???a????s? push notifications" onclick="myFunction('notifications')" /> <div id="clearMessage"></div> <?php } else echo "Login first."; ?> </div> <div id="footer"></div> </div> </body> </html> and the php script: <?php if (isset($_GET["q"])) $q=$_GET["q"]; $host = "localhost"; $database = "dbname"; $user = "dbuser"; $pass = "dbpass"; $con = mysql_connect($host,$user,$pass) or die(mysql_error()); mysql_select_db($database,$con) or die(mysql_error()); if ($q=="messages") $query = "DELETE FROM push_message WHERE time_sent IS NOT NULL"; else if ($q=="conferences") $query = "DELETE FROM push_message WHERE time_sent IS NOT NULL"; else if ($q=="notifications") { $query = "DELETE FROM push_friend WHERE time_sent IS NOT NULL"; } $res = mysql_query($query,$con) or die(mysql_error()); if ($res) echo "success"; else echo "failed"; mysql_close($con); ?>

    Read the article

  • PHP - How to retrieve session in php

    - by Klaus Jasper
    I created a table that contains id - names - jobs and page that shows the names only and beside each name there is button Job and session that contains the id. this is my code $query = mysql_query("SELECT * FROM table"); while($fetch = mysql_fetch_array("$query")){ $name = $fetch['names']; $id = $fetch['id']; echo '</br>'; echo $name; $_SESSION['name'] = $id; echo "<button>Job</button>"; } I want when the user click on button Job redirect to a page that contains the job of that session. so how can I do it?

    Read the article

  • how can we change the value by using radio buttons

    - by magna
    I am making a website in Adobe Dreamweaver with php. In the site there’s a 3 buttons for selecting payment method that will act as the continue button. What I want is when the user checks a radio buttons (I agree button), it will be add with that amount and display with previous amount.. there is three buttons which has the corresponding values(amount in pounds).. plz check my website http://www.spsmobile.co.uk in this linkgo to mobile phone unlocking and after add the cart click make payment it will go to next page there is a delivery mail details.. for that delivery mail details only am asking.. here i mentioned code: <input id="radio-1" type="radio" name="rmr" value="1"> <label for="radio-1">£3</label> <input id="radio-2" type="radio" name="rmr" value="2"> <label for="radio-2">£5.5</label> <input id="radio-3" type="radio" name="rmr" value="4"> <label for="radio-3">£10</label> <div class="total-text" style="font-size:36px">£10</div> var total = parseInt($("div.total-text").text().substring(1), 10); $("input[name='rmr']").bind('change', function() { var amount = 0; switch (this.value) { case "1": amount = 3; break; case "2": amount = 5.5; break; case "4": amount = 10; break; } $("div.total-text").text("£" + (total + amount)); }); but there is no change , my previous amount did not add with that. while am clicking previous amount only displayed on browser.. i need when i cliks radio button the value should change correspondingly.. where i did that mistake...plz give me some idea and what should i do..is there any need for storing db.. thanks in adv

    Read the article

  • Unique constraint with nullable column

    - by Álvaro G. Vicario
    I have a table that holds nested categories. I want to avoid duplicate names on same-level items (i.e., categories with same parent). I've come with this: CREATE TABLE `category` ( `category_id` int(10) unsigned NOT NULL AUTO_INCREMENT, `category_name` varchar(100) NOT NULL, `parent_id` int(10) unsigned DEFAULT NULL, PRIMARY KEY (`category_id`), UNIQUE KEY `category_name_UNIQUE` (`category_name`,`parent_id`), KEY `fk_category_category1` (`parent_id`,`category_id`), CONSTRAINT `fk_category_category1` FOREIGN KEY (`parent_id`) REFERENCES `category` (`category_id`) ON DELETE SET NULL ON UPDATE CASCADE ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_spanish_ci Unluckily, category_name_UNIQUE does not enforce my rule for root level categories (those where parent_id is NULL). Is there a reasonable workaround?

    Read the article

  • How to create a better tables Structure.

    - by user160820
    For my website i have tables Category :: id | name Product :: id | name | categoryid Now each category may have different sizes, for that I have also created a table Size :: id | name | categoryid | price Now the problem is that each category has also different ingredients that customer can choose to add to his purchased product. And these ingredients have different prices for different sizes. For that I also have a table like Ingredient :: id | name | sizeid | categoryid | price I am not sure if this Structure really normalized is. Can someone please help me to optimize this structure and which indexed do i need for this Structure?

    Read the article

  • group by with 3 diffrent

    - by NN
    I have 2 table and I wanna a query with 3 column result in on of them 2 column with view count and title name and in the other 1 column with type_ and i wanna to grouping type_ with max(view count) and show the them title but i didn't have any idea about grouping expression. i think we can solve in by using sub query but i don't know which column use in group by. 2 table join with this expression class pk=resource key i exam this query: SELECT t.title,j.type_ FROM tags asset t,journal article j where type_ in (select type_ from journal article,tags asset where class pk=resource key group by type_) but the answer was wrong

    Read the article

  • What is the best way to reduce code and loop through a hierarchial commission script?

    - by JM4
    I have a script which currently "works" but is nearly 3600 lines of code and makes well over 50 database calls within a single script. From my experience, there is no way to really "loop" the script and minimize it because each call to the database is a subquery of the ones before based on referral ids. Perhaps I can give a very simple example of what I am trying to accomplish and see if anybody has experience with something similar. In my example, there are three tables: Table 1 - Sellers ID | Comm_level | Parent ----------------------------------- 1 | 4 | NULL 2 | 3 | 1 3 | 2 | 1 4 | 2 | 2 5 | 2 | 2 6 | 1 | 3 Where ID is the id of one of our sales agents, comm_level will determine what his commission percentage is for each product he sells, parent indicates the ID for whom recruited that particular agent. In the example above, 1 is the top agent, he recruited two agents, 2 and 3. 2 recruited two agents, 4 and 5. 3 recruited one agent, 6. NOTE: An agent can NEVER recruit anybody equal to or higher than their own level. Table 2 - Commissions Level | Item 1 | Item 2 | Item 3 ----------------------------------------------------- 4 | .5 | .4 | .3 3 | .45 | .35 | .25 2 | .4 | .3 | .2 1 | .35 | .25 | .15 This table lays out the commission percentages for each agent based on their actual comm_level (if an agent is at a level 4, he will receive 50% on every item 1 sold, 40% on every item 2, 30% on every item 3 and so on. Table 3 - Items Sold ID | Item --------------------- 4 | item_1 4 | item_2 1 | item_1 2 | item_3 6 | item_2 1 | item_3 This table pairs the actual item sold with the seller who sold the item. When generating the commission report, calculating individual values is very simple. Calculating their commission based on their sub_sellers however is very difficult. In this example, Seller ID 1 gets a piece of every single item sold. The commission percentages indicate individual sales or the height of their commission. For example: When seller ID 6 sold one of item_2 above, the tree for commissions will look like the following: -ID 6 - 25% of cost(item_1) -ID 3 - 5% of cost(item_1) - (30% is his comm - 25% comm of seller id 6) -ID 1 - 10% of cost(item_1) - (40% is his comm - 30% of seller id 3) This must be calculated for every agent in the system from the top down (hence the DB calls within while loops throughout my enormous script). Anybody have a good suggestion or samples they may have used in the past?

    Read the article

  • Symfony 1.4: Deleting a sfGuardUser

    - by Tom
    Hi, I'm having some trouble with the following... I have a sfGuardUser table set up normally, and it has a one-to-one relationship with a Profile table, which contains some additional user info. When a user wants to delete themselves from the site, I'd like to retain their info in the Profile table for various purposes BUT delete the sfGuardUser in order to keep that table cleaner/shorter (not just set it to inactive). I was under the impression that I could set the FK in the Profile table to NULL and then delete the sfGuardUser, but it seems the FK-constraint fails. Indeed, it seems I can't delete either and the queries fail: If I try to delete the sfGuardUser, the Profile table will have an invalid FK If I try to delete a Profile, the sfGuardUser will have an invalid FK Other than leaving outdated sfGuardUsers and Profiles in these tables, or having to use a cascaded delete to get rid of both, can anyone tell me if there's any other way around this? Thank you.

    Read the article

  • Fill in missing values in a SELECT statement

    - by benjamin button
    If i have a table with two fields.customer id and order. let's say i have in total order ID 1,2,3,4 all the customer can have all the four orders.like below 1234 1 1234 2 1234 3 1234 4 3245 3 3245 4 5436 2 5436 4 you can see above that 3245 customer doesnt have order id 1 and 2. how could i print in the query output like 3245 1 3245 2 5436 1 5436 3 EDIT: i dont have order table but i have list of order's like we can hard code it in the query(1,2,3,4) i dont have an orders table.

    Read the article

  • Store database, good pattern for simultaneous access

    - by dygi
    I am kinda new to database designing so i ask for some advices or some kind of a good pattern. The situation is that, there is one database, few tables and many users. How should i design the database, or / and which types of queries should i use, to make it work, if users can interact with the database simultaneously? I mean, they have access to and can change the same set of data. I was thinking about transactions, but I am not sure, if that is the right / good / the only solution. I will appreciate some google keywords too.

    Read the article

  • Help needed to construct a SQL query

    - by song202y
    Need your help to get the list of suggested friends (who aren't friends of the current user but are friends of 2 or more of the current user's friends). The primary ordering should put people at the same school at the top, and the secondary ordering should put people with more common friends (that is, the number of people who are friends of that person and the current user) near the top. Users: user_id PK, user_name Profiles: user_id PK, school_name, ... Friendships: id PK, user_id FK, friend_id FK Thank you in advance. Joe

    Read the article

  • paging php error - undefined index

    - by fusion
    i've a search form with paging. when the user enters the keyword initially, it works fine and displays the required output; however, it also shows this error: Notice: Undefined index: page in C:\Users\user\Documents\Projects\Pro\search.php on line 21 Call Stack: 0.0011 372344 1. {main}() C:\Users\user\Documents\Projects\Pro\search.php:0 . .and if the user clicks on the 'next' page, it shows no output with this error thrown: Notice: Undefined index: q in C:\Users\user\Documents\Projects\Pro\search.php on line 19 Call Stack: 0.0016 372048 1. {main}() C:\Users\user\Documents\Projects\Pro\search.php:0 this is my code: <?php ini_set('display_errors',1); error_reporting(E_ALL|E_STRICT); include 'config.php'; $search_result = ""; $search_result = trim($_GET["q"]); $page= $_GET["page"]; //Get the page number to show if($page == "") $page=1; $search_result = mysql_real_escape_string($search_result); //Check if the string is empty if ($search_result == "") { echo "<p class='error'>Search Error. Please Enter Your Search Query.</p>" ; exit(); } if ($search_result == "%" || $search_result == "_" || $search_result == "+" ) { echo "<p class='error1'>Search Error. Please Enter a Valid Search Query.</p>" ; exit(); } if(!empty($search_result)) { // the table to search $table = "thquotes"; // explode search words into an array $arraySearch = explode(" ", $search_result); // table fields to search $arrayFields = array(0 => "cQuotes"); $countSearch = count($arraySearch); $a = 0; $b = 0; $query = "SELECT cQuotes, vAuthor, cArabic, vReference FROM ".$table." WHERE ("; $countFields = count($arrayFields); while ($a < $countFields) { while ($b < $countSearch) { $query = $query."$arrayFields[$a] LIKE '%$arraySearch[$b]%'"; $b++; if ($b < $countSearch) { $query = $query." AND "; } } $b = 0; $a++; if ($a < $countFields) { $query = $query.") OR ("; } } $query = $query.")"; $result = mysql_query($query, $conn) or die ('Error: '.mysql_error()); $totalrows = mysql_num_rows($result); if($totalrows < 1) { echo '<span class="error2">No matches found for "'.$search_result.'"</span>'; } else { $limit = 3; //Number of results per page $numpages=ceil($totalrows/$limit); $query = $query." ORDER BY idQuotes LIMIT " . ($page-1)*$limit . ",$limit"; $result = mysql_query($query, $conn) or die('Error:' .mysql_error()); ?> <div class="caption">Search Results</div> <div class="center_div"> <table> <?php while ($row= mysql_fetch_array($result, MYSQL_ASSOC)) { $cQuote = highlightWords(htmlspecialchars($row['cQuotes']), $search_result); ?> <tr> <td style="text-align:right; font-size:15px;"><?php h($row['cArabic']); ?></td> <td style="font-size:16px;"><?php echo $cQuote; ?></td> <td style="font-size:12px;"><?php h($row['vAuthor']); ?></td> <td style="font-size:12px; font-style:italic; text-align:right;"><?php h($row['vReference']); ?></td> </tr> <?php } ?> </table> </div> <?php //Create and print the Navigation bar $nav=""; if($page > 1) { $nav .= "<a href=\"search.php?page=" . ($page-1) . "&string=" .urlencode($search_result) . "\"><< Prev</a>"; } for($i = 1 ; $i <= $numpages ; $i++) { if($i == $page) { $nav .= "<B>$i</B>"; }else{ $nav .= "<a href=\"search.php?page=" . $i . "&string=" .urlencode($search_result) . "\">$i</a>"; } } if($page < $numpages) { $nav .= "<a href=\"search.php?page=" . ($page+1) . "&searchstring=" .urlencode($search_result) . "\">Next >></a>"; } echo "<br /><br />" . $nav; } } else { exit(); } ?>

    Read the article

< Previous Page | 406 407 408 409 410 411 412 413 414 415 416 417  | Next Page >