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PHP - If Statment Or Mysql Query I'm Not Sure..
Hi all..
My head just isn't working correctly today I can write this in english but can't think how to put this or even search for the code I'm looking for..
so I have a table such as this..
id - lineno - hitcount
1 - 2 - 0 2 - 2 - 0 3 - 3 - 0 4 - 3 - 1 5 - 4 - 0 6 - 4 - 0 7 - 4 - 0 And want to echo only if the hitcount is 0 for everything with a lineno of x So in this case I should return lines 1,2,5,6 & 7 but not 3. Any help appreciated Andy. Similar TutorialsSo im trying to figure out if a user is already in the table so im checking for the number of rows where the user id shows up on a record and if its greater then one redirect them. My problem is the if statment does not work and I know there is more then one row. Also die ($row2); wont output data but it will in a sentence (see code below for example) Code: [Select] $sql2 = "SELECT * FROM purchases WHERE nid = '$id' AND uid = '$user_id'"; $res2 =mysql_query($sql2) or die (mysql_error()); $row2 = mysql_num_rows($res2); die($row2);//no output die("There are a total of ".$row2." rows");//outputs There are a total of 20 rows if ($row2 > 0){//if statment doesnt work. //already bought the note header("Location: view.php?error=2"); } Here 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, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas? im having a few problems with if and endif see line of code below what have i done wrong <? if($forum_user['g_id'] <= USER_MOD):?><th class="tcr"><?php echo $lang_online['IP'] ?></th><?php echo "\n";endif ?> keeps kicking up error: Parse error: syntax error, unexpected T_ENDIF thanks Hey there fellow PHP coders I am sorta a newbie with php but i am having a heck of a time with this If Else statement and i know someone with experience will be able to tell me why the else if is not working What this is here is on our company intranet our agent love to send out flyers without picture and makes us look bad... so this is an IF Else if there is a picture display the eFlyer options else dont display the flyer options If doesnt seem to do the else can anyone see something that i am not seeing i have been staring at it for 2 hours now trying different little things and havent been able to get it to work PLEASE PLEASE help me thanks for everyone who reads this <?php if( $mlsInfo['PhotoURL'] != "") echo "<a href='emailListing.php?type=rs&mlsn=".$propInfo['MLSNumber']."' target='_blank' onMouseOver='doTooltip(event, 0)' onMouseOut='hideTip()' ><img src='images/buttons/fb_eListing.gif' width='106' height='20' border='0'></a> <a href='emailPostcard.php?type=rs&mlsn=".$propInfo['MLSNumber']."' target='_blank' onMouseOver='doTooltip(event, 2)' onMouseOut='hideTip()' ><img src='images/buttons/fb_ePostcards.gif' width='106' height='20' border='0'></a> <a href='emailFlyer.php?type=rs&mlsn=".$propInfo['MLSNumber']."' target='_blank' onMouseOver='doTooltip(event, 3)' onMouseOut='hideTip()' ><img src='images/buttons/fb_Flyer.gif' width='110' height='20' border='0'></a> <a href='../print_rs.php?type=rs&mlsn=".$propInfo['MLSNumber']."' target='_blank' onMouseOver='doTooltip(event, 5)' onMouseOut='hideTip()' ><img src='images/buttons/fb_printFlyer.gif' width='106' height='20' border='0'></a> <a href='fb_propertyRS_print.php?type=rs&mlsn=".$propInfo['MLSNumber']."' target='_blank' onMouseOver='doTooltip(event, 4)' onMouseOut='hideTip()' ><img src='images/buttons/fb_printListing.gif' width='106' height='20' border='0'></a> "; else echo "You Dont have any photos please add some"; ?> Hi guys, hopfully you can help me write out an if Statment; basically i want to make an "IF" statment to check if the user has purchased a "ticket" within the last 24 hours, if so echo YES if not echo NO the 'ticket' database is setup as follows: auctionID promoID username charID ticketNumber ticketPrice purchaseDate - Example:" 20101228 11:59:25" it will include the WHERE clause of "username =$username" can someone help me please, im not good with time based stuff.. hey guys i have a car dealer script and i want to make the interface look better.. so i want to put icons(door locks-ac-auto-manu-4 door..) my issue is the if.. so im guessing <? if ( $row3['Door_locks'] == 1 ) { echo "<img src="icons/Doors.png" width="32" height="32" border="0">";} ?> Hello... First I should explain what is wrong. I have a database with a table called subs... Within this table I have a unique field called ID, then a fields called member, date(unix timestamp) amount, month, year... HOWEVER for each month and year there is several entries all with different date stamps. How can I extract the entry with the most recent date??? However there is a catch. I want to view payments made since a certain date but only one per month... Below is my code... I thnk I need to add or change something slightly but i am fairly new to PHP and am totally stuck... MANY THANKS IN ADVANCE!!! Code: [Select] [php]$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 a query which when I run in phpmyadmin it returns the results I want. When I put it into PHP I get no results can someone tell me what I'm doing wrong? Code: [Select] <?php include("config.php"); ?> <?php // sending query $sql = mysql_query("SELECT dayname((date(FROM_UNIXTIME(dateline)))) as 'Day Of Week', date((date(FROM_UNIXTIME(dateline)))) as 'Date', count(*) as 'Number of Opened Tickets', ( select count(ticketmaskid) from swtickets where date(FROM_UNIXTIME(swtickets.lastactivity)) = Date and isresolved=1 ) as 'Number of Closed Tickets' from swtickets where ((date(FROM_UNIXTIME(dateline)) between (DATE_SUB(CURDATE(), INTERVAL (IF(DAYOFWEEK(CURDATE())=1, 9, DAYOFWEEK(CURDATE()))) DAY)) and (DATE_ADD(CURDATE(), INTERVAL (6 - IF(DAYOFWEEK(CURDATE())=1, 8, DAYOFWEEK(CURDATE()))) DAY)) )) group by date(FROM_UNIXTIME(dateline))"); ?> <?php echo $sql; ?> All it returns is: Resource id #4 When in phpmyadmin I get: 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 Hi Guys, I have been having problems with a piece of PHP and mysql. I was wondering if you could help me out? The error is = Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /customers/klueless.net/klueless.net/httpd.www/daisysite/home.php on line 120 Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in /customers/klueless.net/klueless.net/httpd.www/daisysite/home.php on line 122 And this is the code which goes with these errors = Code: [Select] $msgquery = "SELECT * FROM spotty_messages WHERE (id_receiver = '" . $userid . "') AND read = '0'"; $messageres = mysql_query($msgquery); $messrow = mysql_fetch_array($messageres); $messagecount = mysql_num_rows($messageres); Hi there, I am executing this query in the code below, it executes as I want it except when it gets the title, it doesnt get the title for that row it just gets it from the first row in the table... if that makes sense... what is going on? Code: [Select] <?php require'styles/top.php'; ?> <br> <center><table border='0' width='100%' style='text-align:center; font-weight:bold;'> <tr> <td width='33%'>Subject</td> <td width='33%'>From</td> <td width='33%'>Date</td> </tr> </table></center> <br> <?php $query = mysql_query("SELECT * FROM messages WHERE to_user='$username' ORDER BY message_id DESC") or trigger_error('Error: ' . mysql_error()); $numrows = mysql_num_rows($query); if ($numrows > 0){ while ($row = mysql_fetch_assoc($query)){ $id = $row['message_id']; $from = $row['from_user']; $to = $row['to_user']; $title - $row['message_title']; $content = nl2br($row['message_content']); $date = $row['date']; echo"<center><table border='0' width='100%' style='text-align:center; font-weight:bold;'> <tr> <td width='33%'>$title</td> <td width='33%'>$from</td> <td width='33%'>$date</td> </tr> </table></center> <br>"; } } else echo ''; ?> </div> <div id='left'> </div> <div id='right'> </div> Hi
I need a sql query help from you guys.
It is a sql query for get all upline referrer details from database for particular person
when new person register his details are storing in wp_members_tbl with uername, password, firstname, lastname, email, phone, address, referrer etc.
Below the query is for user
$wp_aff_members_db = $wpdb->get_row("SELECT * FROM $members_table_name WHERE refid = '".$_SESSION['user_id']."'", OBJECT);
And below the query is for this user's upline referrer
$wp_aff_members_db = $wpdb->get_row("SELECT * FROM $members_table_name WHERE refid = '$referrer'", OBJECT);
Now i need the query for find and get this referrer's upline referrer.
Please help me with a solution
Regards
Edited by rajasekaran1965, 23 October 2014 - 10:09 AM. Query whats wrong with this query? $queryreg = mysql_query("UPDATE application SET employer = '$employer' AND eaddy = '$employer_address' AND ecity = '$employer_city' AND estate = '$employer_state' and ezip = '$employer_zip' AND supervisor = '$employer_supervisor' WHERE appID = '$appID'") or die(mysql_error()); I can't figure it out. 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: $userbiography<br><br> "; } } ?> I have the following MYSQL Query: Basically, the query will not work when i add quotes to the string. I want it to select from e-mails database where subject LIKE '%"$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 I have this query that searches the database based on what the user inputed. However i'm having the following issues with it: The query is supposed to look for the name of a city in a database table that stores the city names along with their state and country CITY STATE COUNTRY Now i have an input field where the user can search a location and it searches the location from the database. When they search the database the input is in the following format: city, state, country I then use the following script to separate them into three fields Code: [Select] $split = explode(',', $location); // $location being the city, state, country $city = $split[0]; $state = $split[1]; $country = $split[2]; Then i use this MySql Query to search the fields to make sure the city and country match Code: [Select] $get_location = mysql_query("SELECT * FROM locations WHERE name LIKE '$city' AND country LIKE '$country'") or die(mysql_error()); $tmp_loc = mysql_fetch_assoc($get_location); This is where the problem beings...if i use "AND" to search both fields i get no results even though there are results in there however if i change it to an "OR" statement it finds teh locations however it doesn't do an accurate search... for example someone searches for Toledo, Ohio, United States, the user will get Toledo, Spain instead of the right Toledo, since Toled, Spain is at the top of the table I've been trying to get this but i can't get this to work at all every time i use AND it gives me no results I hope someone here can check help me out 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 |
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