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  • Session is working in Localhost Properly but not Online (Cpanel)

    - by nando pandi
    Hello guys Sorry for my stupid question regarding to my yesterday question its not solved yet even the advice you have given but still not working. i have removed all of spaces but still showing the problem for me. it's working perfect in localhost but not in CPANEL. Here is the errors which give: Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at /home/scalepro/public_html/Admin Panel/Remote Employee/main.php:1) in /home/scalepro/public_html/Admin Panel/Remote Employee/main.php on line 1 Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/scalepro/public_html/Admin Panel/Remote Employee/main.php:1) in /home/scalepro/public_html/Admin Panel/Remote Employee/main.php on line 1 Warning: Cannot modify header information - headers already sent by (output started at /home/scalepro/public_html/Admin Panel/Remote Employee/main.php:1) in /home/scalepro/public_html/Admin Panel/Remote Employee/main.php on line 13 Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively in Unknown on line 0 ANY ONE PLEASE ??? Here is my code: <?php session_start(); require_once('../../Admin Panel/db.php'); if(isset($_POST['email']) && !empty($_POST['email']) && isset($_POST['password']) && !empty($_POST['password'])) { $email = $_POST['email']; $password = $_POST['password']; $query="SELECT RemoteEmployeeFullName, RemoteEmployeeEmail, RemoteEmployeePassword FROM remoteemployees WHERE RemoteEmployeeEmail='".$email."' AND RemoteEmployeePassword='".$password."'"; $queryrun=$connection->query($query); if($queryrun->num_rows > 0) { $_SESSION['email']=$RemoteEmployeeFullName; header("Location: /home/scalepro/public_html/Admin Panel/Remote Employee/REPLists.php"); } else { echo 'Email: <b>'.$email. '</b> or Password <b>'. $password.'</b> Is Not Typed Correctly Try Again Please!.'; header( "refresh:5;url= /home/scalepro/public_html/spd/myaccount.php" ); } } else { header( "refresh:5;url= /home/scalepro/public_html/spd/myaccount.php" ); } ?> if the condition gets true this will be redirected to a page by the name of REPLists.php here is the page. <?php session_start(); require_once('../../Admin Panel/db.php'); ?> <html> <head> <style> .wrapper { width:1250px; height:auto; border:solid 1px #000; margin:0 auto; padding:5px; border-radius:5px; -webkit-border-radius:5px; -moz-border-radius:5px; -ms-border-radius:5px; } .wrapper .header { width:1250px; height:20px; border-bottom:solid 1px #f0eeee; margin:auto 0; margin-bottom:12px; } .wrapper .header div { text-decoration:none; color:#F60; } .wrapper .header div a { text-decoration:none; color:#F60; } .wrapper .Labelcon { width:1250px; height:29px; border-bottom:solid 1px #ccc; } .wrapper .Labelcon .Label { width:125px; height:20px; float:left; text-align:center; border-left:1px solid #f0eeee; font:Verdana, Geneva, sans-serif; font-size:14.3px; font-weight:bold; } .wrapper .Valuecon { width:1250px; height:29px; border-bottom:solid 1px #ccc; color:#F60; text-decoration:none; } .wrapper .Valuecon .Value { width:125px; height:20px; float:left; text-align:center; border-left:1px solid #f0eeee; font-size:14px; } </style> </head> <body> <div class="wrapper"> <div class="header"> <div style="float:left;"><font color="#000000">Email: </font> <?php if(isset($_SESSION['email'])) { echo $_SESSION['email']; } ?> </div> <div style="float:right;"> <a href="#">My Profile</a> | <a href="logout.php">Logout</a></div> </div> <div class="Labelcon"> <div class="Label">Property ID</div> <div class="Label">Property Type</div> <div class="Label">Property Deal Type</div> <div class="Label">Property Owner</div> <div class="Label">Proposted Price</div> </div> <?php if(!isset($_SESSION['email'])) { header('Location:../../spd/myaccount.php'); } else { $query = "SELECT properties.PropertyID, properties.PropertyType, properties.PropertyDealType, properties.Status, properties.PropostedPrice, remoteemployees.RemoteEmployeeFullName, propertyowners.PropertyOwnerName, propertydealers.PropertyDealerName FROM remoteemployees, propertyowners, propertydealers, properties WHERE properties.PropertyOwnerID=propertyowners.PropertyOwnerID AND properties.PropertyDealerID=propertydealers.PropertyDealerID AND remoteemployees.RemoteEmployeeID=properties.RemoteEmployeeID ORDER BY properties.PropertyID "; $query_run = $connection->query($query); if( $connection->error ) exit( $connection->error ); while($row=$query_run->fetch_assoc()) { ?> <div class="Valuecon"> <div class="Value"><?php echo $row['PropertyID'] ?></div> <div class="Value"><?php echo $row['PropertyType'] ?></div> <div class="Value"><?php echo $row['PropertyDealType']?></div> <div class="Value"><?php echo $row['PropertyOwnerName'] ?></div> <div class="Value"><?php echo $row['PropostedPrice'];?></div> </div> <?php } }?> </div> </body> </html>

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  • Insert names into table if not already existing

    - by John
    I am trying to build an application that allows users to submit names to be added to a db table (runningList). Before the name is added tho, I would like to check firstname and lastname against another table (controlList). It must exist in (controlList) or reply name not valid. If the name is valid I would like to then check firstname and lastname against (runningList) it make sure it isnt already there. If it isnt there, then insert firstname & lastname. If it is there, reply name already used. Below is code that worked before I tried to add the test against the controlList, I somehow broke it altogether while trying to add the extra step. if (mysql_num_rows(mysql_query('SELECT * FROM runninglist WHERE firstname=\'' .$firstname . '\' AND lastname=\'' . $lastname . '\' LIMIT 1')) == 0) { $sql="INSERT INTO runninglist (firstname, lastname) VALUES ('$_POST[firstname]','$_POST[lastname]')"; } else { echo "This name has been used."; } Any direction would be appreciated. Thanx John

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  • Force reload/refresh when pressing the back button

    - by FlyingCat
    I will try my best to explain this. I have an application that show the 50+ projects in my view page. The user can click the individual project and go to the update page to update the project information. Everything works fine except that after user finish updating the individual project information and hit 'back' button on the browser to the previous view page. The old project information (before update) is still there. The user has to hit refresh to see the updated information. It not that bad, but I wish to provide better user experience. Any idea to fix this? Thanks a lot.

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  • Loop results executing twice

    - by ozzysmith
    I creating a simple site with PHP where the users can submit blogs and other users (who are logged in) can post comments on them. I have made a link called "comments" below each blog that when clicked will show / hide all the comments relevant to the specific blog (also if the user is logged in, it will show a form field in which they can submit new comments). So basically each blog will have multiple comments. I have done two different codes for this but they both have the same problem that each comment appears twice (everything else works fine). Could anyone point out why? mysql_select_db ("ooze"); $result = mysql_query ("select * from blog") or die(mysql_error()); $i = 1; while($row = mysql_fetch_array($result)) { echo "<h1>$row[title]</h1>"; echo "<p class ='second'>$row[blog_content]</p> "; echo "<p class='meta'>Posted by .... &nbsp;&bull;&nbsp; $row[date] &nbsp;&bull;&nbsp; <a href='#' onclick=\"toggle_visibility('something$i'); return false\">Comments</a><div id='something$i' style='display: none;'>"; $i++; $a = $row["ID"]; $result2 = mysql_query ("select * from blog, blogcomment where $a=blogID") or die(mysql_error()); while($sub = mysql_fetch_array($result2)) { echo "<p class='third' >$sub[commentdate] &nbsp;&bull;&nbsp; $sub[username]</p><p>said:</p> <p>$sub[comment]</p>"; } if ( isset ($_SESSION["gatekeeper"])) { echo '<form method="post" name="result_'.$row["ID"].'" action="postcomment.php"><input name="ID" type = "hidden" value = "'.$row["ID"].'" /><input name="comment" id="comment" type="text" style="margin-left:20px;"/><input type="submit" value="Add comment" /></form>'; } else { echo '<p class="third"><a href="register.html">Signup </a>to post a comment</p>'; } echo "</div>"; } mysql_close($conn); //second version of inner loop:// if ( isset ($_SESSION["gatekeeper"])) { while($sub = mysql_fetch_array($result2)) { echo "<p class='third' >$sub[commentdate] &nbsp;&bull;&nbsp; $sub[username] said:</p> <p>$sub[comment]</p>"; } echo '<form method="post" name="result_'.$row["ID"].'" action="postcomment.php"><input name="ID" type = "hidden" value = "'.$row["ID"].'" /><input name="comment" id="comment" type="text" style="margin-left:20px;"/><input type="submit" value="Add comment" /></form>'; } else { while($sub = mysql_fetch_array($result2)) { echo "<p class='third' >$sub[commentdate] &nbsp;&bull;&nbsp; $sub[username] said:</p> <p>$sub[comment]</p>"; } echo '<p class="third"><a href="register.html">Signup </a>to post a comment</p>'; } echo "</div>"; } mysql_close($conn);

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  • Saving data that is in a table efficiently and also is easy to retrieve and echo back out

    - by Harry
    Information I currently have a table which image is below, Problem I have made this table using ul & li Here is the code http://jsfiddle.net/8j2qe/1/ Question What would be the best way of storing the data in the image and easily displaying it? Keeping in mind that each column can only have 1 entry. Thank you! And any questions will be answered ASAP! EDIT Sorry, I dont think I was clear enough in my initial question. What I am asking is, what is the best way to store and then display this type of data. I want to DISPLAY data from my database to show like it would in the image. Should I have a column in my database for each column on the table, then say either A,B,C or D depending on what column it is in but then how would I display it using PHP in my code provided? Im struggling to find a good way of explaining this, I am sorry.

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  • 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

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  • Is there a fast way to change all the collation to utf8_unicode?

    - by Mark
    I have realised after making about 20 tables that I need to user utf8_unicode as opposed to utf8_general. What is the fastest way to change it using PHPMyAdmin? I had one idea: Export the database as SQL then using notepad run a find and replace on it and then reimport it... but it sounds like a bit of a headache. Is there a better way?

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  • Is a clear and replace more efficient than a loop checking all records?

    - by Matt
    I have a C# List, that is filled from a database.. So far its only 1400 records, but I expect it to grow a LOT.. Routinely I do a check for new data on the entire list.. What I'm trying to figure out is this, is it faster to simply clear the List and reload all the data from the table, or would checking each record be faster.. Intuition tells me that the dump and load method would be faster, but I thought I should check first...

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  • Cannot add or update a child row: a foreign key constraint fails

    - by myaccount
    // Getting the id of the restaurant to which we are uploading the pictures $restaurant_id = intval($_GET['restaurant-id']); if(isset($_POST['submit'])) { $tmp_files = $_FILES['rest_pics']['tmp_name']; $target_files = $_FILES['rest_pics']['name']; $tmp_target = array_combine($tmp_files, $target_files); $upload_dir = $rest_pics_path; foreach($tmp_target as $tmp_file => $target_file) { if(move_uploaded_file($tmp_file, $upload_dir."/".$target_file)) { $sql = sprintf(" INSERT INTO rest_pics (branch_id, pic_name) VALUES ('%s', '%s')" , mysql_real_escape_string($restaurant_id) , mysql_real_escape_string(basename($target_file))); $result = mysql_query($sql) or die(mysql_error()); } I get the next error: Cannot add or update a child row: a foreign key constraint fails (rest_v2.rest_pics, CONSTRAINT rest_pics_ibfk_1 FOREIGN KEY (branch_id) REFERENCES rest_branches (branch_id) ON DELETE CASCADE ON UPDATE CASCADE However, this error totally disappears and everything goes well when I put directly the restaurant id (14 for example) instead of $restaurant_id variable in the sql query. The URL am getting the id from is: http://localhost/rest_v2/public_html/admin/add-delete-pics.php?restaurant-id=2 Any help please?

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  • Get a unique data in a SQL query

    - by Jensen
    Hi, I've a database who contain some datas in that form: icon(name, size, tag) (myicon.png, 16, 'twitter') (myicon.png, 32, 'twitter') (myicon.png, 128, 'twitter') (myicon.png, 256, 'twitter') (anothericon.png, 32, 'facebook') (anothericon.png, 128, 'facebook') (anothericon.png, 256, 'facebook') So as you see it, the name field is not uniq I can have multiple icons with the same name and they are separated with the size field. Now in PHP I have a query that get ONE icon set, for example : $dbQueryIcons = mysql_query("SELECT * FROM pl_icon WHERE tag LIKE '%".$SEARCH_QUERY."%' GROUP BY name ORDER BY id DESC LIMIT ".$firstEntry.", ".$CONFIG['icon_per_page']."") or die(mysql_error()); With this example if $tag contain 'twitter' it will show ONLY the first SQL data entry with the tag 'twitter', so it will be : (myicon.png, 16, 'twitter') This is what I want, but I would prefer the '128' size by default. Is this possible to tell SQL to send me only the 128 size when existing and if not another size ? In an another question someone give me a solution with the GROUP BY but in this case that don't run because we have a GROUP BY name. And if I delete the GROUP BY, it show me all size of the same icons. Thanks !

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  • 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

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  • 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..

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  • sql query without subquery

    - by user1285737
    I need to rewrite this query and I'm not allowed to use a subquery. I need to select the name and color of the parts that are heavier than the wheel. SELECT name, color FROM parts WHERE weight > (SELECT weight FROM parts WHERE name="wheel"); This is the table: PARTS ID NAME COLOR WEIGHT 1 wheel black 100 2 tire black 50 3 gear red 20 Thanks in advance

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  • 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

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  • Finding comma separated values with a colon delimiter

    - by iconMatrix
    I am setting values in my database for tourneyID,Selected,Paid,Entered,date then separating each selection with a colon So I have a string that may look like this 187,S,,,09-21-2013:141,S,,,06-21-2013:144,S,,,05-24-2013 but it also could look like this 145,S,,,07-12-2013:142,S,,,05-24-2013:187,S,,,09-21-2013 and some times is looks like this 87,S,,,07-11-2013:125,S,,,06-14-2013 I am trying to find this sequence: 187,S,,,09-21-2013 I have data stored like that because I paid a programmer to code it for me. Now, as I learn, I see it was not the best solution, but it is what I have till I learn more and it is working. My problem is when using LIKE it returns both the 187 and 87 values $getTeams = mysql_query("SELECT * FROM teams WHERE (team_tourney_vector LIKE '%$tid,S,P,,$tourney_start_date%' OR team_tourney_vector LIKE '%$tid,S,,,$tourney_start_date%') AND division='$division'"); I tried this using FIND_IN_SET() but it would only return the the team id for this string 187,S,,,09-21-2013:141,S,,,06-21-2013:144,S,,,05-24-2013 and does not find the team id for this string 145,S,,,07-12-2013:142,S,,,05-24-2013:187,S,,,09-21-2013 SELECT * FROM teams WHERE FIND_IN_SET('187',team_tourney_vector) AND (team_tourney_vector LIKE '%S,,,09-21-2013%') Any thoughts on how to achieve this?

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  • 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

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  • 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(); } ?>

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  • need an sql query

    - by CKeven
    I currently have two tables: 1. car(plate_number, brand, cid) 2. borrow(StartDate, endDate, brand, id) I want to write a query to get all available brand and count of available cars for each brand

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  • 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.

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  • 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

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  • Return a result in the original page.

    - by Josh
    When the user submit his data, I take him to a different page where plenty of calculations are made, then I redirect him to the original page with a simple : <?php header("Location:http://mysite.com/index.php"); The problem is I need all the variables and results to be show in this page, but they are obviously stored in the other one, so I need your help please :) ! Thank you for reading this !

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