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PHP - How Would Less Than And Greater Than Be Used In A Mysql Query?
i tried > and < but would work very well..
Code: [Select] include("connect.php"); $getbookings = mysql_query("SELECT * FROM phpbb_posts WHERE forum_id='17' ORDER BY post_subject ASC AND post_subject > $date"); just so you know.. that post subjects are dates formatted YYYY/MM/DD.. so they can be sorted. and i would like this code to select a the ones that are greater than or less than todays date.. can someone help me please? thanks Similar TutorialsHere is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 If you also have any feedback on my code, please do tell me. I wish to improve my coding base. Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference. Code: Code: [Select] mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); Full code: Code: [Select] <?php include_once("includes/config.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $name = mysql_real_escape_string($_POST['name']); $server_type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1"); if(mysql_num_rows($check1) < 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas? hello, how would i write if $dollars is greater than 5 or less than -5? thanks! how do I write a statement using a greater than and less than Code: [Select] if($soandso >=4 && <=6) {}how would I do something like this? Hi, My example: $setting = 4 //can challenge if opponent has 4 points lesser than you $team1 = 13; //challenger points $team2 = 7; //points In my above case, it should return false because 13-7 = 6, $team2 has 6 points lesser than team 1. If team2 had 4 points or lesser, it should return true. Can someone write the if() statement for me please? Appreciate that alot. Thanks Hi im pritty new to PHP used to using ASP but found PHP to be more resourcful! Anyway, Im trying to work out how to get my script to define if a prduct fee is higher than an available account balance, but hitting brick walls! CODE BELOW: $Credits = 10; $Fee = 5; if ($Fee > $Credits){ //NEED MORE CREDITS }else{ //YOU CAN PURCHASE THIS ITEM } But returns that 10 is not greater than 5, i guess this is because it is only taking the first digit into consideration, even if it is 12 or 15 is still says its not higher than 5 ??? Ive been at it for hours searched google and come up with nothing... please help I may be acting thick and it may be really simple but im lost! Thanks in advance! $file = 'includes/views.txt'; $f = fopen($file, 'w+'); $views = fread($f, filesize($file)); fwrite($f, $views+1); fclose($f); The contents of views.txt is: Code: [Select] 1 Why is it returning it needs to be greater than 0?...The file isn't empty. Hi, i am trying to write a part of script for google shopping export, and all i want is that in this line, everything that is 0 or smaller, the line "pre-order" krijgt and everything that is 1 or greater the line "in stock" This is what i've got now, but it doesn't work I removed the rest of the long code. Code: [Select] $patterns = array(); $patterns[0] = '=<0'; $patterns[1] = '=>1'; $replacements = array(); $replacements[0] = 'vooraf bestellen'; $replacements[1] = 'op voorraad'; if(OPTIONS_TONEN_FEED_AVAILABILITY == 1) $output .= "~" . preg_replace($patterns, $replacements, $row->quantity); I wrote a code to echo "limit reached" if A is greater than B. But if A is 400030 and B is 400000 it shows no output. If A is further greater than that, let say 400060 or any number higher than that, it shows the output.. Please how do I explain that? The code snippet to demonstrate what I mean is....
<?php include_once('db.php'); error_reporting(E_ALL | E_WARNING | E_NOTICE); ini_set('display_errors', TRUE);
if (session_status() == PHP_SESSION_NONE)
{
session_start();
}
if(!isset($_SESSION['login']))
{
echo ("<script>location.href='../clogin/'</script>");
die();
}
if(isset($_POST['transfer']))
{
$username = $_SESSION['login']; $transAmount = $_POST['transAmount']; $totalTrans = $transAmount + 30; $sql = "SELECT * FROM customer WHERE username = ?";
$stmt = $connection->prepare($sql);
$stmt->bind_param('s', $username);
$stmt->execute();
$result = $stmt->get_result(); if(!$result)
{
die('ERROR:' . mysqli_error($connection));
}
$count = $result->num_rows; if($count == 1)
{
while ($row = $result->fetch_assoc())
{
$accTrans = $totalTrans + $row['dailyTrans'];
$sql2 = "UPDATE customer set dailyTrans=? WHERE username=?"; $stmt = $connection->prepare($sql2); $stmt->bind_param('is', $accTrans,$username);
$stmt->execute(); if(!$stmt)
{
die('network problem');
}
if($row['dailyTrans'] >= $row['dailyLimit'])
{
echo '<script>swal.fire("FAILED!!", "<strong>You have reached the total amount you can send per day.</strong><hr><br/><i>Visit your bank to increase transfer limit.</i>", "error"); window.setTimeout(function(){ window.location.href = "transfer1.php";} , 1700); </script>'; //exit();
}
else {
echo"";
}
}//while loop
}//count
}//submit
?>
My question Summary Again The value for $row['dailyTrans'] is 400030 and the value of $row['dailyLimit'] is 400000 This is suppose to echo out the error but fails... if $row['dailyTrans'] is greater than 400030, it echoes out. What is the logic behind that?. Please be nice with your comments as usual. Thanks . Both Value are integers!! . The code works well just that at 400030 it doesn't output that its greater than 400000
hi dudes how do i write a mysql query with 3 columns, where the first column is 'year', the second is 'month' (integer) and the third is 'day' (integer), ordered by desc, but with an extra quirk, where if any of the three columns is zero (which means there is no data for that date column - assume i have a year and a month, but no day)? my code looks like the following Code: [Select] ORDER BY exhib_date_year DESC, exhib_date_month DESC, exhib_date_day DESC I have 2 tables (see code below) (1) "tbl_users" has all users info, including "user_alias" (VARCHAR) and "user_id" (INT and PRIMARY KEY) (2) "tbl_projects" has all project info, including "project_client" (INT) which i linked to "user_id" from "tbl_users" and displays "user_alias" Code: [Select] CREATE TABLE `tbl_users` ( `user_id` int(11) NOT NULL auto_increment, `user_first` varchar(25) NOT NULL, `user_last` varchar(25) NOT NULL, `user_email` varchar(35) NOT NULL, `user_pw` varchar(12) NOT NULL, `user_role` int(1) NOT NULL, `user_alias` varchar(35) NOT NULL, PRIMARY KEY (`user_id`) ) CREATE TABLE `tbl_projects` ( `project_id` int(11) NOT NULL auto_increment, `project_client` int(1) NOT NULL, PRIMARY KEY (`project_id`) ) So... I managed to display all users and all projects but I want to show all users with ONLY their projects assigned. I mean, when I click on a user, I want to see his or hers contact info and all the projects that specific user is linked to. For instance, i might have project 1, project 2, project 3 and project 4. Project 1, 2 and 4 are linked to user 1 so when i click on user 1 i see projects 1, 2 and 3 (not 4). Code: [Select] <?php require_once('config.php'); mysql_select_db($database, $makeconnection); //this displays all projects $sql_get_projects="SELECT * FROM tbl_projects ORDER BY project_id ASC"; $get_projects = mysql_query($sql_get_projects, $makeconnection) or die(mysql_error()); $row_get_roles = mysql_fetch_assoc($get_projects); $totalRows_get_projects = mysql_num_rows($get_projects); //this displays all users $sql_find_users = "SELECT * FROM tbl_users WHERE user_id = $user_id"; $find_users = mysql_query($sql_find_users, $makeconnection) or die(mysql_error()); $row_get_users = mysql_fetch_assoc($find_users); $totalRows = mysql_num_rows($find_users); ?> Any help would be great!! I am creating a site that has to display 36 images on the screen. The image name is stored in the database. My problem is if I have less than 36 images stored I need to display a default image. here is my current query $sql="SELECT col_image, col_url from tbl_images WHERE col_active='1' and col_bigimage='0' ORDER BY RAND() limit 36"; so If I only have 20 active images. I need to display 16 default images. I hope this makes sense. Bill Hello... First I should explain what is wrong. I have a database with a table called subs... Within this table I have a unique field called ID, then a fields called member, date(unix timestamp) amount, month, year... HOWEVER for each month and year there is several entries all with different date stamps. How can I extract the entry with the most recent date??? However there is a catch. I want to view payments made since a certain date but only one per month... Below is my code... I thnk I need to add or change something slightly but i am fairly new to PHP and am totally stuck... MANY THANKS IN ADVANCE!!! Code: [Select] [php]$query="SELECT * FROM records WHERE section='B' OR section='C' OR section='S' order by section, surname"; $result=mysql_query($query); for ($row=0;$row<mysql_num_rows($result);$row++){ $forename=mysql_result($result,$row,'forename'); $surname=mysql_result($result,$row,'surname'); $id=mysql_result($result,$row,'id'); $ref="19nx".$id.substr($forename,0,2).substr($surname,0,2); $section=mysql_result($result,$row,'section'); $giftAid=mysql_result($result,$row,'giftAid'); if ($giftAid>1){$day=date('d',$giftAid);$month=date('m',$giftAid);$year=date('y',$giftAid);}else{$day="";$month="";$year="";} $giftAidName=mysql_result($result,$row,'giftAidName'); $giftAidComment=mysql_result($result,$row,'giftAidComment'); $subdate=mktime(0,0,0,$submonth,$subday,$subyear); $query="SELECT * FROM subs WHERE member='$id' AND date>$subdate Order BY id DESC"; $subResult=mysql_query($query); $subs=""; for($ss=0;$ss<mysql_num_rows($subResult);$ss++){ $amount=mysql_result($subResult,$ss,'amount'); if ($amount==""){$amount='25';} $date=date("M/Y",mysql_result($subResult,$ss,'date')); $subs=$subs."<a title='$date' alt='$date'>$amount</a>,"; }[/php] This outputs a line of results which is right except it shows 2 or 3 for april, 3 or 4 for may anthoer 2 or 3 for june etc... I hope someone gets my drift! I have the following MYSQL Query: Basically, the query will not work when i add quotes to the string. I want it to select from e-mails database where subject LIKE '%"$stringservername"%' If I do just '%$SERVERNAME%' it works :S - And the mysql data includes the quotes. I use the $strongservername to add the quotes. $stringservername = ""$servername""; $stringjobname = ""$vjobname""; $queryemails = mysql_query("SELECT * FROM `emails` WHERE `subject` like '$searchstatus' AND `subject` like '%$stringservername%' AND `subject` like '%$stringjobname%' AND `fromemail` = '$matchfrom'"); Any suggestions? Thankyou Hi All, Wondered if someone could help me out with a sql query that I am having difficulty with? My database consists of 3 tables, clients, video, category. The video table stores the primary key value of the clients table and the category table as a foreign key. What I am trying to achieve is return all the videos that are associated to a particular client and group them under the relevant category. If there are now videos that match the category then I do not want to display the category. Here is my code so far: Code: [Select] <?php $sql = "SELECT category.cat_id, category.name AS catname FROM category"; $result = mysql_query($sql) or die (mysql_error()); while($categoryrow = mysql_fetch_assoc($result)) { ?> </p> <div class="themeheader"><h5><?php echo $categoryrow['catname']; ?></h5></div> <Br /> <?php $vsql = "SELECT video.video_id, video.title, video.description, video.thumbnail FROM video WHERE video.cat_id = '" . $categoryrow['cat_id'] . "' AND video.client_id = $customerid ORDER BY video.video_id DESC"; $vresult = mysql_query($vsql) or die (mysql_error()); ?> <div class="videos"> <ul> <?php while($videorow = mysql_fetch_assoc($vresult)) { ?> <li id="categoryList"><a href="film-details.php?video_id=<?php echo $videorow['video_id']; ?>"><img src="+_1m4g35/<?php echo $videorow['thumbnail']; ?>" alt="<?php echo $videorow['title']; ?>" title="<?php echo $videorow['title']; ?>" width="291" height="142" border="0" /></a> <h2><?php echo $videorow['title']; ?></h2> <p><?php $limit = 100; if (strlen($videorow['description']) > $limit) $description = substr($videorow['description'], 0, strrpos(substr($videorow['description'], 0, $limit), ' ')) . '... <a href="film-details.php?video_id='.$videorow['video_id'].'">read more</a>'; echo $description; ?> </p> <?php } //end video loop?> </ul> <br class="clearfloat" /> </div> <?php } //end category loop ?> </div> The above code is the closest I have got but it still outputs the categories even when there are no videos that match the category id and the client id. Any help in the right direction gratefully received as I am gradually going insane! Hi all, I have a database with 2 tables, 'users' and 'battles'. The site pulls 2 random peoples pictures and lets the user choose who they think would win the battle. So if user1 is using the site it might show pictures for user5 and user3. If the user1 chooses that user5 wins then an entry is made into the 'battles' table like this : voter win lose user1 user5 user3 Any ideas what query I can use so it only shows 2 people that the user hasnt compared before ? As if its doing this : choose 2 id's from 'users' that user1 hasnt compared before Hope that makes sense. Many thanks, Scott Hello, I have a query where i try to search, but i want to put a limitation, but it doesn't seems to be working :/ Here's my code: $result = mysql_query("SELECT * FROM clients WHERE staff = 0 AND username LIKE '%" . $keyword . "%' OR company LIKE '%" . $keyword . "%' ORcontact LIKE '%" . $keyword . "%' OR address LIKE '%" . $keyword . "%' OR email LIKE '%" . $keyword . "%' OR phone LIKE '%" . $keyword . "%' ORDER BY phone"); Alright, so i basically want the user to search in all of those fields, however i want it to filter all "staff" members, so it should only view the "0" ones, meaning the clients, only problem that when i run this, all clients gets displayed, also the staff accounts. Any suggestion? Hello! Please help... I am trying to use the script below to get results from a mysql database based on a query of the form fields (the names of which are displayed near the top of the script as POST items) When a location or age etc. is entered into the form, I want the script to search for records which meet those criteria. At the moment the script works but only does so if all the values are entered to match what is in the database. e.g. if the location england and the age 22 was entered into the form, and that matched the value in the database, then at the moment, the script will display the result, but if only the location is entered in the form without any value for age/genre etc. then no results are displayed. Any help would be very welcome as I have search high and low for a solution on google... which doesn't seem to exist... I'm not that experienced with php/mysql but am learning on the job so any helpful prompts as to terms etc. would help! Thanks! Lewis <?php if($_POST) { $searchage = $_POST['searchage']; $searchlocation = $_POST['searchlocation']; $searchgenre = $_POST['searchgenre']; $searchinstrument = $_POST['searchinstrument']; $searchexperience = $_POST['searchexperience']; // Connects to your Database mysql_connect("localhost", "user", "pass") or die(mysql_error()); mysql_select_db("DB") or die(mysql_error()); $query = mysql_query("SELECT * FROM table_user WHERE userage = '".$searchage."' AND userlocation = '".$searchlocation."' AND usergenre = '".$searchgenre."' AND userinstrument = '".$searchinstrument."' AND userexperience = '".$searchexperience."'") or die(mysql_error()); $num = mysql_num_rows($query); echo "$num results found!<br>"; while($result = mysql_fetch_assoc($query)) { $username = $result['username']; $useremail = $result['useremail']; $userage = $result['userage']; $userlocation = $result['userlocation']; $usergenre = $result['usergenre']; $userinstrument = $result['userinstrument']; $userexperience = $result['userexperience']; $userbiography = $result['userbiography']; echo " Name: $username<br> Email: $useremail<br> Age: $userage<br> Location: $userlocation<br> Gen $usergenre<br> Instrument: $userinstrument<br> Experience: $userexperience<br> Biography: $userbiography<br><br> "; } } ?> |
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