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PHP - Help: Checking Before Doing Query In Php With Mysql
Hello,
I have a problem. I run this query: $query = "SELECT column1 FROM table WHERE column2='$kk'"; $which = $link1; mysql_query($query,$which); some codes.... It is displaying all data correct but only when $kk exists in column2 and when $kk does not exist, it display blank page. So, i want to do a check before running above query to make sure that $kk exists in column2. How to check it before running query? Similar TutorialsI don't know precisely how to put it so i will give an example: suppose you have a search form with input fields for: -an age range, -a data range, -for the sex, -postal code as soon as the user would fill in one of these input fields, a query should be generated including a condition doing something with the inputted value. For example, the user fills in maxAge, the the query should say: SELECT * FROM table WHERE age >= maxAge; Filling more than one criterion then would do .= " AND criteria (condition) inputvalue " etc... My question comes to the part about: how to check if there is already a WHERE in the clause and how to put WHERE or AND for every criterion. Since doing like above, i won't be able to search for another criterion if i don't select maxAge first because then there would be no WHERE in the sql clause. Hopefully i've put it clear thanks in advance Hi there, I am working on a PHP website. And I am running a query which checks for the expired products. We check to see if the expiry date is greater than the current systems date then display the record. It looks something like this: Code: [Select] select * from products WHERE products.productExpiry) >= '2010-10-13 00:00:00' 'productExpiry' field is of 'Datetime' Datatype And Expiry date is stored in database as: 2010-10-19 04:39:35 Now it works fine. But the problem only arises (no record shows) when there is no date in the 'productExpiry' field. How can I check or set a condition in my SQL query that whenever the 'productExpiry' is Null or Blank don't compare it with the current date. Please reply All comments and feedbacks are welcomed Thank you! In an .html, before : <!-- BEGIN myquery --> I need to stop output and check the content of "myquery" Is there a command for that ? Thank You. Okay I need some help. I want to check a MySQL DB and then if it says 0 then don't do anything if it says anything else echo out what's there. I have tried for about a hour now. hey guys, i need some help with my php/mysql iplogger. My code: <?php //finds out ip $ip = $_SERVER['REMOTE_ADDR']; //conects to the mysql server $connection = mysql_connect('localhost', 'root', ''); //sellects the database mysql_select_db('iplog', $connection); //looks for duplacute ips $dup = mysql_query("SELECT COUNT(number) FROM logged_ips WHERE ip_address = '$ip'",$connection); $count = mysql_result($dup, 0); //checks to see if there is a duplecate name if ($count == 0){ //inserts the ip in to the database $string = 'INSERT INTO `logged_ips` (`aid`, `ip_address`, `ip_visits`) VALUES (\'' . null . '\', \'' . $ip . '\', \'0\')'; mysql_query($string, $connection); }else{ //adds a visit to the database $string2 = "UPDATE `logged_ips` SET `ip_visits` = '++1' WHERE `ip_address` = $ip LIMIT 0,1"; mysql_query($string2, $connection); } //outputs the ip echo $ip; ?> error: Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\randoms\logger\test.php on line 11 127.0.0.1 It also doesnt put anything in to the mysql database. Please help. Thanks jragon Is there an alert method for the MySQL database or something similar?
What is the proper way to check if the MySQL database connection is working?
I mean other than "creating" and "retrieving" with "insert" and "select".
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? 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 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 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, Relative newbie here pulling his hair out! I've just started to attempt to build a stock control system for my parent's business and am having problems. I'm building a (what I though) simple page that will just return all the info from a table. Each stock item has it's own stock code (5 characters) and also a barcode. There may be more than one barcode assigned to each stock item so I created a seperate table with just two fields - barcode and stockcode - the barcode obviously has to be unique but the stockcode doesn't. The way I have done it is with a form where you can enter the stock code - this then becomes $STOCKCODE and the rest of the page displays the result for that code - so far so good. Now I want it to see if the data entered into the form is greater than 5 characters. If it is it will then assume that you have entered a barcode and look for the stockcode from the other table and assign that to $STOCKCODE instead. The code below is the part that isn't working. Entering a stock code brings up the correct result, but enter a barcode and it just returns an incorrect product (always the same incorrect product irrespective of what you type in). Executing the barcode lookup query directly in mysql on it's own works and generates the correct stock code but there's something awry in the way I've done it that I can't seem to work out. The "echo "not recognised" at the end doesn't work either. Googling the problem had just confused me even more! Any help would be appreciated! Cheers Code: [Select] //retrieve form result $getcode=$_POST['item_entry']; //Count characters in form result $num_char=strlen($getcode); //barcode lookup query $result = mysql_query("SELECT barcode.SKU FROM barcode WHERE $getcode = 'barcode.BARCODE'"); //se if the entered data was a stockcode or a barcode if ($num_char == 5) {$STOCKCODE=$getcode;} elseif ($num_char > 5) {$STOCKCODE=mysql_fetch_assoc($result);} else echo "not recognised"; 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 '%" . $keyword . "%' ORDER BY phone"); Alright, so i basically want the user to search in all of those fields, however i want it to filter all "staff" members, so it should only view the "0" ones, meaning the clients, only problem that when i run this, all clients gets displayed, also the staff accounts. Any suggestion? Hello... First I should explain what is wrong. I have a database with a table called subs... Within this table I have a unique field called ID, then a fields called member, date(unix timestamp) amount, month, year... HOWEVER for each month and year there is several entries all with different date stamps. How can I extract the entry with the most recent date??? However there is a catch. I want to view payments made since a certain date but only one per month... Below is my code... I thnk I need to add or change something slightly but i am fairly new to PHP and am totally stuck... MANY THANKS IN ADVANCE!!! Code: [Select] [php]$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! 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 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. $loc="SELECT * FROM table WHERE username='a', id='1' AND uin='123'"; $get=myspl_query($loc) or die(mysql_error()); Please somone tell me the right command as how to use multiple AND in mysql query. Thanks 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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