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PHP - Having A Little Trouble With A Mysql Result And While $row Loop
I am having a little bit of trouble with this piece of code. I'm sure it's something simple, but I have been working on this thing all day and want to get it finally finished.
Here's the troublesome code: function rrmdir($dir) { if (is_dir($dir)) { $objects = scandir($dir); foreach ($objects as $object) { if ($object != "." && $object != "..") { if (filetype($dir."/".$object) == "dir") rrmdir($dir."/".$object); else unlink($dir."/".$object); } } reset($objects); rmdir($dir); } } $sql_clean = "SELECT * complete WHERE createdate < date_sub(current_date, interval 1 minute)"; $sql_list = mysql_query($sql_clean); while($row = mysql_fetch_assoc($sql_list)) { $directory = "complete/" . $row['fileurl']; rrmdir($directory); } The purpose of this particular bit is to run on a cron every few days. It gets "createdate" and other info from the "complete" table in order to know how old the record is. If the record is older than (in the example, 1 minute; it will be set to several days on public) the defined max age, it removes that directory and everything within it to keep the directory clean and the disk usage down. The error returned is Quote Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home1/latenit2/public_html/kindleprocessor/process/garbagecleaner.php on line 46 Line 46 is Quote while($row = mysql_fetch_assoc($sql_list)) { I may be doing the look-up on the MySQL database incorrectly, too. I haven't discounted that, and I'd be thankful if someone could help me out with this issue. Similar TutorialsHi, I would like to ask, how to display the while loop result 3 per row with the following code below: $startDate = strtotime($datefrom); I have only one result being displayed in a select query result whereas I'm expecting 4 results in an array. This is my code below. $sq2="SELECT course FROM course_reg WHERE userid=?";
$stm =$conn->prepare($sq2);
$stm->bind_param("s",$logged);
$stm->execute ();
$return2= $stm->get_result();
$r2 = $return2->fetch_all();
//print_r($r2);
foreach($r2 as $course){
foreach($course as $courses){
echo $courses;
}
}
If I do print_r($r2); it comes out with array containing all the possible results. When i loop through the array to get individual result, it only comes out with a single result. I.e CME211 I would be glad if you can help me figure where the issue is. Thanks!!!
I want to save the results of a loop as a variable ie $output. I have tried encasing the php within quotes but it does not work. Is there a way to save the complete results as a variable? $result = mysql_query("SELECT * FROM $table2 WHERE $db_item_1 OR $db_item_2", $connection); if (!mysql_num_rows($result)) { echo "Error 13424 - not working"; exit(); } while ($row = mysql_fetch_array($result)) { echo "<tr><td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\" style=\"font-size: 20px;\">" . $row['date'] . "</div></td> <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\"></div></td> <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\" style=\"font-size: 20px;\">" . $row['title'] . "</div></td> <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\"></div></td> <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\" style=\"font-size: 20px;\">" . $row['cost'] . "</div></td> </tr>"; }echo $complete;() The part I want as one result (ie. $complete) is the result of the while loop. There is always at least one result but sometimes 10 which means that it creates 10 table rows. I need to do it this way as later on in the page I use a pdf converter which does not allow loop checks within it otherwise I would just place it within the converter. Hey everyone I've been working on converting my old mysql code to to the mysqli version and am running into some trouble. I am pulling posts form phpbb and displaying them in a script. My first script below works just fine. Code: [Select] <?php require("connect.php"); $query="SELECT * FROM phpbb_topics WHERE (forum_id='13' OR forum_id='14') AND topic_status = '0'"; $query2="SELECT * FROM phpbb_topics WHERE (forum_id='15' OR forum_id='16') AND topic_status = '0'"; $result=mysql_query($query); $result2=mysql_query($query2); echo "<table border=0 cellpadding=2>"; echo "<tr><td>Lost</td><td>Found</td></tr>"; while($row=(mysql_fetch_array($result)) || $row2=(mysql_fetch_array($result2))) { $forum_id=mysql_result($result,$k,"forum_id"); $topic_id=mysql_result($result,$k,"topic_id"); $topic_title=mysql_result($result,$k,"topic_title"); $forum_id2=mysql_result($result2,$k,"forum_id"); $topic_id2=mysql_result($result2,$k,"topic_id"); $topic_title2=mysql_result($result2,$k,"topic_title"); echo "<tr><td width='15%'><a href=/sahbb/viewtopic.php?f=$forum_id&t=$topic_id>$topic_title</a></td> <td width='15%'><a href=/sahbb/viewtopic.php?f=$forum_id2&t=$topic_id2>$topic_title2</a></td> </tr>"; $k++; } echo "</table>"; mysql_close($connect); ?> however its not very clean or memory efficient so i have converted it to the following. Code: [Select] <?php require("includes_database/mysqli_connect.php"); $query="SELECT * FROM phpbb_topics WHERE (forum_id='13' OR forum_id='14')"; $query2="SELECT * FROM phpbb_topics WHERE (forum_id='15' OR forum_id='16')"; $result=mysqli_query($dbc, $query); $result2=mysqli_query($dbc, $query2); echo "<table border=1 cellpadding=2>"; echo "<tr><td>Lost</td><td>Found</td></tr>"; while(($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) || ($row2 = mysqli_fetch_array($result2, MYSQLI_ASSOC))) { echo "<tr><td width='15%'><a href=/sahbb/viewtopic.php?f=".$row['forum_id']."&t=".$row['topic_id'].">".$row['topic_title']."</a></td> <td width='15%'><a href=/sahbb/viewtopic.php?f=".$row2['forum_id']."&t=".$row2['topic_id'].">".$row2['topic_title']."</a></td> </tr>"; } echo "</table>"; mysqli_free_result($result); mysqli_free_result($result2); mysqli_close($dbc); ?> I know the error is in my while loop but i cant seem to figure out exactly where. The new code displays all of the first query in the table and then below that it displays the next query insted of displaying them side by side below would be an example of what is happening [table Lost Found Lost, Australian Sheppard Lost German Shepherded / Collie Mix Lost Golden Retriever/Cocker Mix Lost Female Grey Tiger Lost Pitbull Mix Lost 2 Chihuahua Lost Black Lab Lost - German Shepard Mix and Lab Mix Lost - Female German Sheperd Aprox 8 Week Old Kitten Found! 2 Pugs FOUND Found 1 Corgi female/ 1 Shih Tzu male Found Tan and Black Australian Shepard Found Australian Shepard Found Intact Male Pitbull/Lab Mix Any help that any of you could provide would be greatly appreciated! Hello, New to PHP and trying to loop through rows to and get the name of the "property" in this case. I have tried foreach and while and neither seem to want to return all rows by calling the function. Does it have something to do with echo'ing the table structure or can I not return an array out of a function like this? I commented out the while loop as that wasn't working and put in a foreach loop though that didn't return all the data either. Any help is greatly appreciated Code: [Select] <?php function getProperties(){ $sql = "SELECT propertyName FROM properties;"; $result = mysql_query($sql); //$count = mysql_num_rows($result); $property = mysql_fetch_array($result); foreach($property as $propertyName){ echo "<td width=\"157\" height=\"24\" valign=\"middle\"><div align=\"left\"><input type=\"checkbox\" name=\"$propertyName\" \"id=\"$propertyName\" />$propertyName</div></td>"; } /* for($i=0;$i>=$count;$i++){ while($property = mysql_fetch_array($result)){ $propertyName = $property['propertyName']; echo "<td width=\"157\" height=\"24\" valign=\"middle\"><div align=\"left\"><input type=\"checkbox\" name=\"$propertyName\" \"id=\"$propertyName\" />$propertyName</div></td>"; } } */ } ?> this is only returning one record, when I know there are more, and it's an endless loop crashing my browser, can someone tell me how my brackets are wrong, or what is wrong? Code: [Select] if (isset($_POST['search'])){ $keyword=$_POST['keyword']; } $sql="SELECT id, fname, lname FROM obituaries WHERE lname LIKE '%$keyword%' OR fname LIKE '%$keyword%' ORDER BY id DESC"; $results = mysql_query($sql); $num = mysql_num_rows ($results); if ($num > 0 ) { $i=0; while ($i < $num) { $id = mysql_result($results,$i,"id"); $fname = mysql_result($results,$i,"fname"); $lname = mysql_result($results,$i,"lname"); ?> <a href="view.php?id=<?php echo ($id); ?>"><?php echo($fname); ?> <?php echo($lname); ?></a> <?php } } ?>Thanks in advance Hi guys, Need some help with arrays! The following code puts ALL dates into an array, or should I say should. Because this: do { $start = $row_getCalendarDates['page_date']; $end = $row_getCalendarDates['page_date02']; $init_date = strtotime($start); $dst_date = strtotime($end); $offset = $dst_date-$init_date; $dates = floor($offset/60/60/24) + 1; $booked=0; while ($booked < $dates) { $newDateFinal[] = date("Y-m-d", mktime(12,0,0,date("m", strtotime($start)), (date("d", strtotime($start)) + $booked), date("Y", strtotime($start)))); $booked++; } print_r($newDateFinal); } while ($row_getCalendarDates = mysql_fetch_assoc($getCalendarDates)); Infact prints something like so when i have multiple results: Array ( [0] => 2010-12-01 [1] => 2010-12-02 [2] => 2010-12-03 ) Array ( [0] => 2010-12-01 [1] => 2010-12-02 [2] => 2010-12-03 [3] => 2010-11-01 [4] => 2010-11-02 [5] => 2010-11-03 [6] => 2010-11-04 [7] => 2010-11-05 [8] => 2010-11-06 [9] => 2010-11-07 [10] => 2010-11-08 ) These are 2 results coming from the database, so what it is doing is putting both into arrays, but I 'NEED' them iin a single array, how on earth do I go about doing this? Note however, how the first array has the first set of results but the second has them all ... this is affecting my code and so, need to somehow reference and get rid of the first array. Hope someone can help! Regards, Joe. How can I check if a returned mysql value is equal to '' i.e. nothing? I keep getting an error where the page won't load because the returned value is '' so i need to check for it Hello, i created this script for a client and have ran into an annoying error with the results displaying on a new line for each result instead of side by side, any help is welcome Cheers Code: [Select] <?php $subcat = mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['subcat'])))); $cat = mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['cat'])))); $sql = @mysql_query("SELECT * FROM cakes WHERE category =\"$cat\" AND sub_cat=\"$subcat\" ORDER BY id DESC"); while ($row = mysql_fetch_array($sql)) { $reference = $row['reference']; $image = $row['image']; echo ("<p><img src='./images/cakes/$image' height='289px' width='177px' alt='IMAGE OF CAKE'></img><br />"); echo ("<b>Reference:</b>$reference"."</p>"); } if (!$reference) { echo 'There are no cakes in this category yet.'; } ?> I know the errors only going to be something small i'm missing, but i've been coding all day hello I'm using this code: Code: [Select] $query="SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1";to query first row data when I want see it and echo it I recived : Code: [Select] Resource id #2can anyone help me ? thank you Im not sure where to post this but since it includes php il post it here instead of in the mysql forum. ok so, i have a table and i get the values using while($row = mysql_fetch_array($result)){ and then echo them in rows. that works fine but i need to add a class to the last row of my table. I would need somehow to fetch the last row of the array and make it echo something different. Any help is appreciated Thank you Hi, I don't know anything about php and Mysql but I found a tutorial for a shopping cart and everything is working. The only thing is they don't have the email part in the totorial so I'm kind of stuck with my file. Anyone know how to email the resul to me via email with this code? <? include("includes/db.php"); include("includes/functions.php"); if($_REQUEST['command']=='update'){ $name=$_REQUEST['name']; $email=$_REQUEST['email']; $address=$_REQUEST['address']; $phone=$_REQUEST['phone']; $result=mysql_query("insert into customers values('','$name','$email','$address','$phone')"); $customerid=mysql_insert_id(); $date=date('Y-m-d'); $result=mysql_query("insert into orders values('','$date','$customerid')"); $orderid=mysql_insert_id(); $max=count($_SESSION['cart']); for($i=0;$i<$max;$i++){ $pid=$_SESSION['cart'][$i]['productid']; $q=$_SESSION['cart'][$i]['qty']; $price=get_price($pid); mysql_query("insert into order_detail values ($orderid,$pid,$q,$price)"); } die('Thank You! your order has been placed!'); } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Billing Info</title> <script language="javascript"> function validate(){ var f=document.form1; if(f.name.value==''){ alert('Your name is required'); f.name.focus(); return false; } f.command.value='update'; f.submit(); } </script> </head> <body> <form name="form1" onsubmit="return validate()"> <input type="hidden" name="command" /> <div align="center"> <h1 align="center">Billing information</h1> <table border="0" cellpadding="2px"> <tr><td>Total:</td><td><?=get_order_total()?></td></tr> <tr><td>Name :</td><td><input type="text" name="name" /></td></tr> <tr><td>Address :</td><td><input type="text" name="address" /></td></tr> <tr><td>Email :</td><td><input type="text" name="email" /></td></tr> <tr><td>Phone :</td><td><input type="text" name="phone" /></td></tr> <tr><td> </td><td><input type="submit" value="Place Order" /></td></tr> </table> </div> </form> </body> </html> Hi all, I have a situation where I need to remember what check boxes where checked over pagination, I have managed to do this via the use of this: http://jamesfunk.com/wordpress/?p=65 The problem is that as the code stood: Code: [Select] <input type="checkbox" name="compare" value="<?php echo $list['jobseeker_id'];?>" class="remember_cb"/> It was treating one ticked checkbox as them all because they all have the same name and are in the a while loop! I countered this by changing the code to: Code: [Select] <input type="checkbox" name="compare<?php echo $list['jobseeker_id']?>" value="<?php echo $list['jobseeker_id'];?>" class="remember_cb"/> Which effectively now makes the checkbox name unique... i.e. compare{id}. The problem with this is that I can now no longer process it. This is my processing code: $jobseekers = $_POST['compare']; $i = 0; for($i; $i<count($jobseekers); $i++){ $query = "SELECT * FROM jobseekers WHERE jobseeker_id = '$jobseekers[$i]'"; $result = mysql_query($query) or die (mysql_error()); while ($row = mysql_fetch_array($result)) { // Spit out the required data } } As you can see I am trying to get the data from $_POST['compare'], which will obviously now be blank as I have had to make the name unique.... the trouble is I'm not sure how to actually process this. Can anyone help me out here? any help or advice would be greatly appreciated! many thanks, Greens85 Hi, I am new to php but not to programming. I created a script on a windows platform which connects to the mysql database and returns the results of a table. A very basic script which I wrote to simply test my connection worked. The script works fine on my windows machine but not on my new mac. On the mac it simply does not display any records at all. I know that the database connection has been established because there is no error but I can not see why the result set is not being displayed on screen, as I said it worked fine on my windows machine. The Mac has mysql (with data) and apache running for php. Please could someone help as I have no idea what to do now? Script below: Code: [Select] $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'root'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'test'; mysql_select_db($dbname); mysql_select_db("test", $conn); $result = mysql_query("SELECT * FROM new_table"); while($row = mysql_fetch_array($result)) { echo $row['test1'] . " " . $row['test2'] . " " . $row['test3']; echo "<br />"; } mysql_close($con); I am seeking to learn more about the noted subject, how to use PHP to allow a user to enter search terms and search a database. I have experimented with this with little results save for errors. Please see code listed below: search.php <? //// filename = search.php <form method="post" action="result.php3"> <select name="metode" size="1"> <option value="row_name1">metode1</option> <option value="row_name2">metode2</option> </select> <input type="text" name="search" size="25"> <input type="submit" value="Begin Searching!!"> </form> ?> results.php //// filename = result.php3 <? $hostname = "mysql7.000webhost.com"; // Usually localhost. $username = "a4542527_root"; // If you have no username, leave this space empty. $password = "*******"; // The same applies here. $usertable = "people"; // This is the table you made. $dbName = "a4542527_test1"; // This is the main database you connect to. MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database"); @mysql_select_db( "$dbName") or die( "Unable to select database"); ?> <? //error message (not found message) $XX = "No Record Found"; $query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 "); while ($row = mysql_fetch_array($query)) { $variable1=$row["row_name1"]; $variable2=$row["row_name2"]; $variable3=$row["row_name3"]; print ("this is for $variable1, and this print the variable2 end so on..."); } //below this is the function for no record!! if (!$variable1) { print ("$XX"); } //end ?> Upon viewing search.php I receive the error message: Parse error: syntax error, unexpected '<' in /home/a4542527/public_html/search.php on line 3 I believe I may be missing something and am a bit lost. Thank-you in in advance for any help or suggestions. ~Matty im getting an error: Warning: mysql_fetch_array() [function.mysql-fetch-array]: The result type should be either MYSQL_NUM, MYSQL_ASSOC or MYSQL_BOTH. in C:\www\library\mysql.class.php on line 353 $result_type = "MYSQL_ASSOC"; $row = mysql_fetch_array($result, $result_type) but if i use (below) it works fine...is the constant() function the right way to what im doing? $result_type = "MYSQL_ASSOC"; $row = mysql_fetch_array($result, constant($result_type)) thank you 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 Hello Everyone, I have a quick question for you all, I think its fairly simple... I have created a database and I am using PHP to grab the data: $usera = $_SESSION['username']; $query2 = "SELECT * FROM tracker WHERE id = '$usera', hidden = yes"; mysql_query($query2) or die('Error, query failed : ' . mysql_error()); This hopefully will return multiple rows which look like this in the database. id username date reps hidden 1 supremebeing 2011-01-02 30 yes 4 supremebeing 2011-04-02 46 yes How would i turn each result into a variable eg: $date1 = 2011-01-02; $date2 = 2011-04-02; $reps1 = 30; $reps2 = 46; I think i have explained that well enough for you to understand, please reply if not though and i will provide more information. Thanks in Advance I am getting the row results from mysql as called, but they appear as a straight line instead of new table row when I echo like: echo "TABLETABLETABLETABLE"; I am looking for: echo "TABLE TABLE TABLE TABLE"; Here is my code: Code: [Select] $brand = $_POST['brand']; $city = stripslashes($_POST['city']); $state = $_POST['state']; $zip_code = stripslashes($_POST['zip_code']); if($city !=""){ $query = "SELECT name, bus_type, street, city, state, zip_code, brand, quantity, price1, price2, date FROM `prices` WHERE city ='".$city."' AND state ='".$state."' AND brand = '".$brand."'"; }else{ $query = "SELECT name, bus_type, street, city, state, zip_code, brand, quantity, price1, price2, date FROM `prices` WHERE zip_code ='".$zip_code."'"; } $result = mysql_query($query); $count=mysql_num_rows($result); if($count==0 && $zip_code !=""){?> <td width="100%" class="style9">Sorry! There are no results for that city, state and brand.</td> <? }else{ while($row = mysql_fetch_array($result, MYSQL_ASSOC)){ ?> <td width="100%" class="style9"><?php echo "<table width=\"100%\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\"> <tr> <td{$typ_image}</td> <td><b><a href=\"\" title=\"{$row['street']} {$row['city']}, {$row['state']}. {$row['zip_code']}\" target=\"_blank\">{$row['name']}</a></b><br>Last Updated:{$row['date']}</td> <td><b>{$row['brand']} - {$row['quantity']}</b><br>Before Tax:\${$row['price1']} After Tax:\${$row['price2']}</td> <td{$pago}</td> </tr> </table><BR>"; ?></td> <?php } } ?></table> <p> </p> </td> Hi, I'm trying to make a mysql output to a link so the name will be a link so when you hit this link you will get the full information of this mysql input. Can someone point me in the correct direction? Here is how i get the output from my mysql database into my table. Code: [Select] <td>"; echo $row['name']; echo "</td> |
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