Search Results

Search found 18661 results on 747 pages for 'linq to mysql'.

Page 491/747 | < Previous Page | 487 488 489 490 491 492 493 494 495 496 497 498  | Next Page >

  • 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

  • How to refresh site if $_SESSION variable has changed

    - by 4ndro1d
    I'm writing in my $_SESSION variable from a database, when i clicked a link. public function getProjectById($id){ $query="SELECT * FROM projects WHERE id=\"$id\""; $result=mysql_query($query); $num=mysql_numrows($result); while ($row = mysql_fetch_object($result)) { $_SESSION['projectid'] = $row->id; $_SESSION['projecttitle'] = $row->title; $_SESSION['projectinfo'] = $row->info; $_SESSION['projecttext'] = $row->text; $_SESSION['projectcategory'] = $row->category; } } Now my variable is overwritten and I want to show these variables in my index.php like this: <div id="textContent"> <?php if(isset($_SESSION['projecttext']) && !empty($_SESSION['projecttext'])) { echo $_SESSION['projecttext']; }else { echo 'No text'; } ?></div> But of course, my page will not refresh automatically. How can I do that?

    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

  • How to get multiple counts with one SQL query?

    - by Crobzilla
    I am wondering how to write this query. I know this actual syntax is bogus, but it will help you understand what I am wanting. I need it in this format, because it is part of a much bigger query. SELECT distributor_id, COUNT(*) AS TOTAL, COUNT(*) WHERE level = 'exec', COUNT(*) WHERE level = 'personal' I need this all returned in one query. Also, it need to be in one row, so the following won't work: 'SELECT distributor_id, COUNT(*) GROUP BY distributor_id'

    Read the article

  • need help to construct query

    - by Learner
    i have the following result and i would like to construct the select query from the following result in java, Please help me how to go about , tablename columnname size order employee name 25 1 employee sex 25 2 employee contactNumber 50 3 employee salary 25 4 address street 25 5 address country 25 6 from this i would like to construct query like select T1.name, T1.sex,T1.contactNumber, T1.salaryT2.street, T2.contry from tablename1[employee] T1, tablename2[address] T2 how to construt the above query in java, here table name can be N also the columname can be also N. Please help me to achieve the above. Thanks and Regards

    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

  • Pagination links do not work properly - incorrect PHP function??

    - by ClarkSKent
    Hi everyone, I am still trying to figure out how to fix my pagination script to work properly. the problem I am having is when I click any of the pagination number links to go the next page, the new content does not load. literally nothing happens and when looking at the console in Firebug, nothing is sent or loaded. I have on the main page 3 links to filter the content and display it. When any of these links are clicked the results are loaded and displayed along with the associated pagination numbers for that specific content. I believe the problem is coming from the function(generate_pagination.php (seen below)). Here is the main page so you can how I am including and starting the function(I'm new to php): <?php include_once('generate_pagination.php'); ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_pagination.js"></script> <div id="loading" ></div> <div id="content" data-page="1"></div> <ul id="pagination"> <?php generate_pagination($sql) ?> </ul> <br /> <br /> <a href="#" class="category" id="marketing">Marketing</a> <a href="#" class="category" id="automotive">Automotive</a> <a href="#" class="category" id="sports">Sports</a> This is as mentioned above, where i think the problem persists since I know nothing of the function formats and how to properly incorporate them: <?php function generate_pagination($sql) { include_once('config.php'); $per_page = 3; //Calculating no of pages $result = mysql_query($sql); $count = mysql_fetch_row($result); $pages = ceil($count[0]/$per_page); //Pagination Numbers for($i=1; $i<=$pages; $i++) { echo '<li class="page_numbers" id="'.$i.'">'.$i.'</li>'; } } $ids=$_GET['ids']; generate_pagination("SELECT COUNT(*) FROM explore WHERE category='$ids'"); ?> I thought I might as well through in the jquery if someone wants to see: $(document).ready(function(){ //Display Loading Image function Display_Load() { $("#loading").fadeIn(900,0); $("#loading").html("<img src='bigLoader.gif' />"); } //Hide Loading Image function Hide_Load() { $("#loading").fadeOut('slow'); }; //Default Starting Page Results $("#pagination li:first").css({'color' : '#FF0084'}).css({'border' : 'none'}); Display_Load(); $("#content").load("pagination_data.php?page=1", Hide_Load()); //Pagination Click $("#pagination li").click(function(){ Display_Load(); //CSS Styles $("#pagination li") .css({'border' : 'solid #dddddd 1px'}) .css({'color' : '#0063DC'}); $(this) .css({'color' : '#FF0084'}) .css({'border' : 'none'}); //Loading Data var pageNum = this.id; $("#content").load("pagination_data.php?page=" + pageNum, function(){ $(this).attr('data-page', pageNum); Hide_Load(); }); }); // Editing below. // Sort content Marketing $("a.category").click(function() { Display_Load(); var this_id = $(this).attr('id'); $.get("pagination.php", { category: this.id }, function(data){ //Load your results into the page var pageNum = $('#content').attr('data-page'); $("#pagination").load('generate_pagination.php?category=' + pageNum +'&ids='+ this_id ); $("#content").load("filter_marketing.php?page=" + pageNum +'&id='+ this_id, Hide_Load()); }); }); }); Any help would be appreciated on getting the function to work properly. Thank you.

    Read the article

  • 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

  • How to verify if two tables have exactly the same data?

    - by SiLent SoNG
    Basically we have one table (original table) and it is backed up into another table (backup table); thus the two tables have exactly the same schema. At the beginning both tables (original table and backup table) contains exactly the same set of data. After sometime for some reason I need to verify whether dataset in the original table has changed or not. In order to do this I have to compare the dataset in the original table against the backup table. Let's say the original table has the following schema: `create table LemmasMapping ( lemma1 int, lemma2 int, index ix_lemma1 using btree (lemma1), index ix_lemma2 using btree (lemma2) )` How could I achieve the dataset comparision? Update: the table does not have a primary key. It simply stores mappings between two ids.

    Read the article

  • get me the latest Change from Select Query in below given condition

    - by OM The Eternity
    I have a Table structure as id, trackid, table_name, operation, oldvalue, newvalue, field, changedonetime Now if I have 3 rows for the same "trackid" same "field", then how can i select the latest out of the three? i.e. for e.g.: id = 100 trackid = 152 table_name = jos_menu operation= UPDATE oldvalue = IPL newvalue = IPLcccc field = name live = 0 changedonetime = 2010-04-30 17:54:39 and id = 101 trackid = 152 table_name = jos_menu operation= UPDATE oldvalue = IPLcccc newvalue = IPL2222 field = name live = 0 changedonetime = 2010-04-30 18:54:39 As u can see above the secind entry is the latest change, Now what query I should use to get the only one and Latest row out of many such rows... $distupdqry = "select DISTINCT trackid,table_name from jos_audittrail where live = 0 AND operation = 'UPDATE'"; $disupdsel = mysql_query($distupdqry); $t_ids = array(); $t_table = array(); while($row3 = mysql_fetch_array($disupdsel)) { $t_ids[] = $row3['trackid']; $t_table[] = $row3['table_name']; //$t_table[] = $row3['table_name']; } //echo "<pre>";print_r($t_table);echo "<pre>"; //exit; for($n=0;$n<count($t_ids);$n++) { $qupd = "SELECT * FROM jos_audittrail WHERE operation = 'UPDATE' AND trackid=$t_ids[$n] order by changedone DESC "; $seletupdaudit = mysql_query($qupd); $row4 = array(); $audit3 = array(); while($row4 = mysql_fetch_array($seletupdaudit)) { $audit3[] = $row4; } $updatefield = ''; for($j=0;$j<count($audit3);$j++) { if($j == 0) { if($audit3[$j]['operation'] == "UPDATE") { //$insqry .= $audit2[$i]['operation']." "; //echo "<br>"; $updatefield .= "UPDATE `".$audit3[$j]['table_name']."` SET "; } } if($audit3[$j]['operation'] == "UPDATE") { $updatefield .= $audit3[$j]['field']." = '".$audit3[$j]['newvalue']."', "; } } /*echo "<pre>"; print_r($audit3); exit;*/ $primarykey = "SHOW INDEXES FROM `".$t_table[$n]."` WHERE Key_name = 'PRIMARY'"; $prime = mysql_query($primarykey); $pkey = mysql_fetch_array($prime); $updatefield .= "]"; echo $updatefield = str_replace(", ]"," WHERE ".$pkey['Column_name']." = '".$t_ids[$n]."'",$updatefield); } In the above code I am fetching ou the distinct IDs in which update operation has been done, and then accordingly query is fired to get all the changes done on different fields of the selected distinct ids... Here I am creating the Update query by fetching the records from the initially described table which is here mentioned as audittrail table... Therefore I need the last made change in the field so that only latest change can be selected in the select queries i have used... please go through the code.. and see how can i make the required change i need finally..

    Read the article

  • SELECT product from subclass: How many queries do I need?

    - by Stefano
    I am building a database similar to the one described here where I have products of different type, each type with its own attributes. I report a short version for convenience product_type ============ product_type_id INT product_type_name VARCHAR product ======= product_id INT product_name VARCHAR product_type_id INT -> Foreign key to product_type.product_type_id ... (common attributes to all product) magazine ======== magazine_id INT title VARCHAR product_id INT -> Foreign key to product.product_id ... (magazine-specific attributes) web_site ======== web_site_id INT name VARCHAR product_id INT -> Foreign key to product.product_id ... (web-site specific attributes) This way I do not need to make a huge table with a column for each attribute of different product types (most of which will then be NULL) How do I SELECT a product by product.product_id and see all its attributes? Do I have to make a query first to know what type of product I am dealing with and then, through some logic, make another query to JOIN the right tables? Or is there a way to join everything together? (if, when I retrieve the information about a product_id there are a lot of NULL, it would be fine at this point). Thank you

    Read the article

  • INSERT..ON DUPLICATE KEY UPDATE - but NOT using the duplicate key to compare.

    - by calumbrodie
    I am trying to solve a problem I have inherited with poor treatment of different data sources. I have a user table that contains BOTH good and evil users. create table `users`( `user_id` int(13) NOT NULL AUTO_INCREMENT , `email` varchar(255) , `name` varchar(255) , PRIMARY KEY (`user_id`) ); In this table the primary key is currently set to be user_id. I have another table ('users_evil') which contains ONLY the evil users (all the users from this table are included in the first table) - the user_id's on this table do NOT correspond to those in the first table. I want to have all my users in one table, and simply flag which are good and which are evil. What I want to do is alter the user table and add a column ('evil') which defaults to 0. I then want to dump the data from my 'users_evil') table and then run an INSERT..ON DUPLICATE KEY UPDATE with this data into the first table (setting 'evil'=1 where the emails match) The problem is that the 'PK' is set to the user_id and not the 'email'. Any suggestions, or even another strategy to successfully achive this. Can I run this statement but treat another column as PK only for the duration of the statement.

    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

  • Dynamically Insert Variables into DB Table using PreparedStatement

    - by gran_profaci
    I was working with PreparedStatement today and noticed that it used setString() setTimestamp() etc. to insert variables into the DB. I basically have 20 tables each with at least 15 columns and it would not be feasible for me to manuallt write down all the setters. Considering that I have an ArrayList "Vals" which contains all the variables to be inputted in String format (obtained by getString() using PreparedStatement itself), is there any way I can do an insert without using expressly using the Setters? That would save me a lot of time.

    Read the article

  • Cakephp query doesn't render correct data

    - by user2915012
    I'm totally new in cakephp and fetching problem at the time of query to render data I tried this to find out categories/warehouses table info but failed.. $cart_products = $this->Order->OrdersProduct->find('all', array( 'fields' => array('*'), 'contain' => array('Category'), 'joins' => array( array( 'table' => 'products', 'alias' => 'Product', 'type' => 'LEFT', 'conditions' => array('Product.id = OrdersProduct.product_id') ), array( 'table' => 'orders', 'alias' => 'Order', 'type' => 'LEFT', 'conditions' => array('Order.id = OrdersProduct.order_id') ) ), 'conditions' => array( 'Order.store_id' => $store_id, 'Order.order_status' => 'in_cart' ) )); I need the result something like this... Array ( [0] => Array ( [OrdersProduct] => Array ( [id] => 1 [order_id] => 1 [product_id] => 16 [qty] => 10 [created] => 2013-10-24 08:04:33 [modified] => 2013-10-24 08:04:33 ) [Product] => Array ( [id] => 16 [part] => 56-987xyz [title] => iPhone 5 battery [description] => iPhone 5c description [wholesale_price] => 4 [retail_price] => 8 [purchase_cost] => 2 [sort_order] => [Category] => array( [id] => 1, [name] => Iphone 5 ) [Warehouse] => array( [id] => 1, [name] => Warehouse1 ) [weight] => [created] => 2013-10-22 12:14:57 [modified] => 2013-10-22 12:14:57 ) ) ) How can I find this? Can anybody help me? thanks

    Read the article

  • Determine week number based on starting date

    - by kreetiv
    I need help to create a function to determine the week number based on these 2 parameters: Starting date Specified date For example, if I specify April 7, 2010 as the starting date & passed April 20, 2010 as the date to lookup, I would like the function to return WEEK 2. Another example, if I set March 6, 2010 as starting date and looked up April 5, 2010 then it should return WEEK 6. I appreciate your time and help.

    Read the article

  • Difficults on sql query

    - by João Madureira Pires
    I have the following tables: TableA (id, tableB_id, tableC_id) TableB (id, expirationDate) TableC (id, expirationDate) I want to retrieve all the results from TableA ordered by tableB.expirationDate and tableC.expirationDate. thanks

    Read the article

  • User welcome message in php

    - by user225269
    How do I create a user welcome message in php. So that the user who has been logged on will be able to see his username. I have this code, but it doesn't seem to work. <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM users WHERE Username='$username'"); while($row = mysql_fetch_array($result)) { echo $row['Username']; echo "<br />"; } ?> I'm trying to make use of the data that is inputted in this login form: <form name="form1" method="post" action="verifylogin.php"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Login User</strong></td> </tr> <tr> <td width="30" height="35"><font size="2">Username:</td> <td width="30"><input name="myusername" type="text" id="idnum" maxlength="5"></td> </tr> <tr> <td width="30" height="35" ><font size="2">Password:</td> <td width="30"><input name="mypassword" type="password" id="lname" maxlength="15"></td> </tr> <td align="right" width="30"><td align="right" width="30"><input type="submit" name="Submit" value="Submit" /></td> <td align="right" width="30"><input type="reset" name="Reset" value="Reset"></td></td> </tr> </form> But this, verifylogin.php, seems to be in the way. <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="school"; $tbl_name="users"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $myusername=$_POST['myusername']; $mypassword=$_POST['mypassword']; $myusername = stripslashes($myusername); $mypassword = stripslashes($mypassword); $myusername = mysql_real_escape_string($myusername); $mypassword = mysql_real_escape_string($mypassword); $sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'"; $result=mysql_query($sql); $count=mysql_num_rows($result); if($count==1){ session_register("myusername"); session_register("mypassword"); header("location:userpage.php"); } else { echo "Wrong Username or Password"; } ?> How do I do it? I always get this error when I run it: Notice: Undefined variable: username in C:\wamp\www\exp\userpage.php on line 53 Can you recommend of an easier on how I can achieve the same thing?

    Read the article

  • SELECT set of most recent id, amount FROM table, where id occurs many times

    - by Jon Cram
    I have a table recording the amount of data transferred by a given service on a given date. One record is entered daily for a given service. I'd like to be able to retrieve the most recent amount for a set of services. Example data set: serviceId | amount | date ------------------------------- 1 | 8 | 2010-04-12 2 | 11 | 2010-04-12 2 | 14 | 2010-04-11 3 | 9 | 2010-04-11 1 | 6 | 2010-04-10 2 | 5 | 2010-04-10 3 | 22 | 2010-04-10 4 | 17 | 2010-04-19 Desired response (service ids 1,2,3): serviceId | amount | date ------------------------------- 1 | 8 | 2010-04-12 2 | 11 | 2010-04-12 3 | 9 | 2010-04-11 Desired response (service ids 2, 4): serviceId | amount | date ------------------------------- 2 | 11 | 2010-04-12 4 | 17 | 2010-04-19 This retrieves the equivalent as running the following once per serviceId: SELECT serviceId, amount, date FROM table WHERE serviceId = <given serviceId> ORDER BY date DESC LIMIT 0,1 I understand how I can retrieve the data I want in X queries. I'm interested to see how I can retrieve the same data using either a single query or at the very least less than X queries. I'm very interested to see what might be the most efficient approach. The table currently contains 28809 records. I appreciate that there are other questions that cover selecting the most recent set of records. I have examined three such questions but have been unable to apply the solutions to my problem.

    Read the article

  • ASP, My SQL & case sensitity suddenly broken

    - by user131812
    Hi There, We have an old ASPsite that has been working fine for years with a MY SQL database. All of a sudden last week lots fo SQL queries stopped working. The database has a table called 'members' but the code calls 'Members'. It appears the queries used to work regardless of case sensitivity on the table names, but something has changed recently somewhere to enforce case. This has me stumped as the site has not been touched in years, the server config hasn't changed & the database provide has not changed anything. Is there any simple way to ignore case for an ASP site (without editing lots fo files :) Thanks Ben

    Read the article

  • JQUERY - Find all Elements with Class="X" and then POST all those elements to the server to INS into

    - by nobosh
    Given a large text block from a WYSIWYG like: Lorem ipsum dolor sit amet, <span class="X" id="12">consectetur adipiscing elit</span>. Donec interdum, neque at posuere scelerisque, justo tortor tempus diam, eu hendrerit libero velit sed magna. Morbi laoreet <span class="X" id="13">tincidunt quam in facilisis.</span> Cras lacinia turpis viverra lacus <span class="X" id="14">egestas elementum. Curabitur sed diam ipsum.</span> How can I use JQUERY to find the following: <span class="X" id="12">consectetur adipiscing elit</span> <span class="X" id="13">tincidunt quam in facilisis.</span> <span class="X" id="14">egestas elementum. Curabitur sed diam ipsum.</span> And post it to the server as follows 12, consectetur adipiscing 13, tincidunt quam in facilisis. 14, egestas elementum. Curabitur sed diam ipsum. In a way where in Coldfusion it can loop through the results and make 3 inserts into the DB? Thanks

    Read the article

  • what's wrong with this code?

    - by user329820
    Hi this is my code which will not work correctly ! what is wrong with its data type :( thanks CREATE TABLE T1 (A INTEGER NOT NULL); CREATE TABLE T3 (A SMALLINT NOT NULL); INSERT T1 VALUES (32768.5); SELECT * FROM T1; INSERT T3 SELECT * FROM T1; SELECT * FROM T3;

    Read the article

  • Find the closest locations to a given address

    - by xtine
    I have built an application in CakePHP that lists businesses. There are about 2000 entries, and the latitude and longitude coordinates for each business is in the DB. I now am trying to tackle the search function. There will be an input box where the user can put a street address, city, or zipcode, and then I would like it to return the 11 closest businesses as found from the database. How would I go about doing this?

    Read the article

< Previous Page | 487 488 489 490 491 492 493 494 495 496 497 498  | Next Page >