PHP - How To Get This Variable Into A Query
I have this variable that I need to get into my database and I just can't think through it. Here is the code.
$sql = "INSERT INTO names (id, name, color, uploaddate) VALUE (NULL,'$master[2]['dbid']','$color','$time')"; I know I need to escape those quotes around dbid somehow, but I can't seam to get it figured out. If I have to I will set it = to a different variable, but I don't want to do it that way if I don't have to Similar Tutorials$id = '9'; $MyQuery = "SELECT * FROM bookings WHERE id = '$id'"; $retrieve = mysql_query($MyQuery) or die(mysql_error()); if(mysql_num_rows($retrieve) != 0): $row = mysql_fetch_assoc($retrieve); else: echo 'No Information'; endif; echo 'Full Name: ' . ($row['fullname']) . '<br><br>'; The above code is working for me however when I change it to what I do... $id = ($row['id']); It tells me I have an undefined row variable... What will fix this? Thanks in advance... okay i have table named ZAMJENE with 2 rows only row that i need to setup is named id_zamjena and have values 15 and 16 when i do echo on page it work great but when i press button i get only 16 for both entries full cde is too long so i paste only important parts $query_zamjene = mysql_query ("SELECT * FROM zamjene WHERE id_user_1='$moj_id' AND zamjenjeno='0' ORDER BY id_event_1"); echo "<form method='post' action='event_zamjenjen.php'>"; while ($rows = mysql_fetch_array($query_zamjene)) { $id_zamjena = $rows['id_zamjena']; $id_event_1 = $rows['id_event_1']; $id_event_2 = $rows['id_event_2']; $id_user_2 = $rows['id_user_2']; echo "$id_zamjena"; // THIS ROW WORKS CORRECTLY echo "<input name='id_zamjena' type='hidden' value=$id_zamjena>"; // output of this is always 16 echo "<input name='response' type='submit' value='Prihvati zamjenu' /> <input name='response' type='submit' value='Odbaci zamjenu' /> <br>"; } echo"</form>"; i have a small search profile search and need to set $getuname variable to users username if((trim($country) !== ''&&($state) !== ''&&($city) !== '')){ $data = mysql_query("SELECT * FROM users WHERE users.state='$state' AND users.country='$Country' AND users.city='$city' ORDER BY RAND() DESC LIMIT 1 "); Echo "Country city state search"; $getuname = mysql_real_escape_string($_GET['username']); am i meant to use the get or is that only for forms? Hello, So I have a content management site set up, but I'm stuck on one part. I have a query. Code: [Select] Select `email` FROM author WHERE author.id IN(SELECT authorid FROM job WHERE id = '$_POST[id]') I'm a beginner in PHP and SQL. Basically I need the query to run, it will come out with one result and I need that to be where the email is sent to. This is the existing code I have for sending out the email $subject = " Job Application Confirmation"; $message = "Hello $_POST[name] , \r\rYou, or someone using your email address, has applied for the job: $text Using the Resource Locator at Prahan.com. Keep this as future reference. \r\r Full Name: $_POST[name] \r Location: $_POST[location] \r Email: $_POST[email] \r Additional Information: $_POST[info] \r Job ID: $id \r Job Name: $text \r \r Expect a response shortly. \r\r Regards, Prahan.com Team"; $headers = 'From: Prahan.com Team <noreply@prahan.com>'; 'Reply-To: noreply@prahan.com' . "\r\n" . 'X-Mailer: PHP/' . phpversion(); $to1 = 'pratik@prahan.com'; mail($to1, $result, $message, $headers); Thanks in Advance! Hi there, im trying to have a form show up when user clicks "add joke". I need the variable to be retrieved from the url query string. I cant get the form to show up. I think its either an issue with the GET function at the top or the link down at the bottom. Please help! <?php // If the user wants to add a joke $_GET['addjoke'] = $addjoke; if (isset($addjoke)): ?> <FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST> <P>Type your joke he <BR> <TEXTAREA NAME="joketext" ROWS=10 COLS=40 WRAP> </TEXTAREA><BR> <INPUT TYPE=SUBMIT NAME="submitjoke" VALUE="SUBMIT"> </FORM> <?php else: // Connect to the database server $dbcnx = @mysql_connect("servername", "username", "password"); if (!$dbcnx) { echo( "<P>Unable to connect to the " . "database server at this time.</P>" ); exit(); } // Select the jokes database if (! @mysql_select_db("jhodara2") ) { echo( "<P>Unable to locate the joke " . "database at this time.</P>" ); exit(); } // If a joke has been submitted, // add it to the database. $joketext = $_POST['joketext']; $submitjoke = $_POST['submitjoke']; if ("SUBMIT" == $submitjoke) { $sql = "INSERT INTO jokes SET " . "JokeText='$joketext', " . "JokeDate=CURDATE()"; if (mysql_query($sql)) { echo("<P>Your joke has been added.</P>"); } else { echo("<P>Error adding submitted joke: " . mysql_error() . "</P>"); } } echo("<P> Here are all the jokes " . "in our database: </P>"); // Request the text of all the jokes $result = mysql_query( "SELECT JokeText FROM jokes"); if (!$result) { echo("<P>Error performing query: " . mysql_error() . "</P>"); exit(); } // Display the text of each joke in a paragraph while ( $row = mysql_fetch_array($result) ) { echo("<P>" . $row["JokeText"] . "</P>"); } // When clicked, this link will load this page // with the joke submission form displayed. echo("<P><A HREF='$PHP_SELF?addjoke=1'>Add a Joke!</A></P>"); endif; ?> see the problem live at http://www.freewaycreative.com/insert2.php I have a template page that opens with a GO click from a drop-down in a different page. The variable is being passed through to the URL and in the GET statement. However, my PHP code is producing the following error: Quote Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /data/23/2/100/53/2263053/user/2487001/htdocs/myalaskacenter/events/venues/test_venues-results.php on line 52 . Here is my code: Code: [Select] <?php //Assigns the venueCode tothe variable with a more convenient name $venueURL = $_GET['VenueCode']; @$DB = mysqli_connect('server','username','password','database'); if (mysqli_connect_errno()){ echo '</blockquote><br/><br /> Sorry, this webpage is temporarily unavailable.<br /> <a href="http://alaskapac.centertix.net/">Click here to search for events.</a>'; } else { ?> <?php $Query = "SELECT Events.EventTitle, DATE_FORMAT(Performance.startDateTime, '%W, %M %e, %Y') AS startDate, DATE_FORMAT(Performance.startDateTime, '%h:%i %p') AS startTime, Events.EventID, Events.thumb, Events.ShoWareEventLink, Events.tagline, Performance.category_id, Promoters.Presenter, Promoters.website, Events.startDATE, Events.endDATE, Events.EventOnSaleDate, venues.VenueName FROM ((Events LEFT JOIN Performance ON Events.EventID = Performance.EventID) LEFT JOIN Promoters ON Events.PromoterCode = Promoters.PromoterCode) LEFT JOIN venues ON Performance.VenueCode = venues.VenueCode WHERE venues.VenueCode=".$venueURL." AND Events.group_id=1 AND Performance.category_id!=2 AND Performance.category_id!=5 AND Performance.category_id!=7 AND Performance.category_id!=8 AND Events.EventOnSaleDate IS NOT NULL AND (Performance.PerfType='Public Event' OR Events.EventID='79') AND Performance.endDateTime >= now()-INTERVAL 1 DAY AND Events.PublishDate <= now() AND Events.startDATE IS NOT NULL ORDER BY Performance.startDateTime"; $Result = mysqli_query($DB,$Query); $NumResults = mysqli_num_rows($Result); ?> <?php if($NumResults=0){ echo "<p class='submenu'>$NumResults Performances</p>"; } while ($Row = mysqli_fetch_assoc($Result)){ $eventtitle = $Row['EventTitle']; $eventDate = $Row['startDate']; echo '<p><a href="' . $Row['ShoWareEventLink'] . '"><img src="https://alaskapac.centertix.net/UPLImage/' . $Row['thumb'] . '" alt="' . $Row['EventTitle'] . '" title="' . $Row['EventTitle'] . '" align="left" border="0"><span class="Heading3_blue">' . $Row['EventTitle'] . '</span>'; if($Row['FreeEvents']==TRUE){ echo '<img src="/images/free.gif" alt="Free Event" title="Free Event" width="80" height="80" align="right" border="0"></a><br />'; } elseif ($Row['EventOnSaleDate'] <= date("Y-m-d g:i a")){ /** IF ONSALEDATE<=NOW**/ echo '<img src="/images/logos/ctx/BUY_Tickets_gold.gif" alt="Buy Tickets" title="Buy Tickets" width="85" height="32" align="right" border="0"></a><br />'; } else {/** IF ONSALEDATE!<NOW**/ echo '<img src="/images/logos/ctx/AvailableSoon.png" alt="Available Soon" title="Available Soon" align="right" border="0"></a><br /><i>Tickets available ' . date("l, F j, Y", strtotime($Row['EventOnSaleDate'])) . ' at ' . date("g:i a", strtotime($Row['EventOnSaleDate'])) . '.</i>'; /** +ONSALEDATE **/ } echo '<br />Presented by <a href="' . $Row['website'] .'" target="_blank">' . $Row['Presenter'] . '</a>'; echo '<br />'.$Row['startDate']. ' at ' . $Row['startTime'].' - '.$Row['VenueName']; if ($Row['FreeEvents']==TRUE){ echo '<br /><br />'.$Row['BriefDescription']; } echo '<br /><br /></p><hr>'; } ?> <?php mysqli_free_result($Result); mysqli_close($DB); } ?> Here is my sample Results page: http://www.myalaskacenter.com/events/venues/test_venues-results.php?VenueCode=ACH Hi This does not make sense to me, I hope someone can tell me why this query don't work: Code: [Select] safe_query("UPDATE ".PREFIX."cup_challenges SET reply_date='$postdate', $select_map $select_date challenged_info='".$_POST['info']."', status='2' WHERE chalID='".$_GET['challID']."'");; and this query does work: Code: [Select] safe_query("UPDATE ".PREFIX."cup_challenges SET reply_date='$postdate', map1='".$_POST['map1']."', map2='".$_POST['map2']."', map3='".$_POST['map3']."', date1='".$_POST['date1']."', date2='".$_POST['date2']."', date3='".$_POST['date3']."', date4='".$_POST['date4']."', challenged_info='".$_POST['info']."', status='2' WHERE chalID='".$_GET['challID']."'"); Variables in $select_map and $select_date makes the query fail but when I copy/paste this (bold) in query it works: $select_map = map1='".$_POST['map1']."', map2='".$_POST['map2']."', map3='".$_POST['map3']."', $select date = date1='".$_POST['date1']."', date2='".$_POST['date2']."', date3='".$_POST['date3']."', date4='".$_POST['date4']."', am I using the variable in query incorrectly? I got my search page error and the server pop up below message:
Notice: Undefined variable: query in /home/tz005/public_html/COMP1687/search.php on line 64 Minimum length is 3 Where should I make a correction in the script? Here is my php script: <?php Can anyone point out how to write a MySQL query with a PHP variable in the WHERE clause. I've tried {} {'xx'} and () and it still doesn't work. Here is the code <?php ini_set('display_errors',1); error_reporting(E_ALL|E_STRICT); include ("include/connect.php"); include ("include/session.php"); $username = $session->userinfo['username']; $result = mysql_query("SELECT email FROM customer WHERE user = {'$username'} "); while($row = mysql_fetch_array($result)) { $custemail = $row['email']; } echo "Session username: " . $username . ""; echo "Session customer email: " . $custemail . ""; ?> So I'm trying to show the email address for a record that matches the username of the user logged in. I really appreciate the help. Hi Guys. I need your help. Is it possible to store results from sql query as a variable. I am basically doing a few joins to get this result and I wanted to know if this was possible. I am using the following code in PDO query $this->query('SELECT * FROM users WHERE id = :id'); // line 11 $this->bind(':id', $_SESSION['users']['id']); // line 12 it is working on local server, but on live server its giving some error or warning (i am not sure) PHP Notice:  Undefined index: users in ... on line 12 but the query is still working correctly. Please guide me what should i do to stop getting this error or warning or notice. Thanks🙂 I am working on a room availability calendar that has links for each day. When you click on a link it passes the day month and year in the URL to another page like this: echo "<td class=\"today\"> <a href=\"status.php?month=$month&day=$day_num&year=$year\">$day_num</a> </td>\n"; I can see that the correct information is being passed through the URL like this: Code: [Select] .../status.php?month=12&day=9&year=2010 Then the information is supposed to be passed to the MySQL query, but here is my question: How do I do this? I have a DB table set up, but the query is currently returning a blank page. Here is my current query: // connects to server and selects database. include ("../includes/dbconnect.inc.php"); // table name $table_name = "availability"; // query database for events $result = mysql_query ("SELECT id FROM $table_name WHERE month=$month AND year=$year AND day=$day_num LIMIT 1") or die(); if (mysql_num_rows($result) > 0 ) { while($row = mysql_fetch_array($result)) { extract($row); echo "<h1>Current availability for ".$row['month'] . "/" . $row['day'] . "/" . $row['year'] . "</h1>"; echo " <ul>"; echo " <li>Earth Room: " . $row['earth_room'] . "</li>"; echo " <li>Air Room: " . $row['air_room'] . "</li>"; echo " <li>Fire Room: " . $row['fire_room'] . "</li>"; echo " <li>Water Room: " . $row['water_room'] . "</li>"; echo " </ul>"; } } else { echo " <ul>"; echo " <li>Currently no reservations.</li>"; echo " </ul>"; } Any help is appreciated. Thanks, kaiman hello all! my query, when executed, generates an error i don't understand. i've seen similar entries, but i couldn't find one that related to this particular problem. either that, or i'm a retard and didn't see it. apologies if that's the case. here's the error: --> Database Query Failed: Unknown column 'client_id' in 'where clause' <-- i triple checked the database table and, YUP, the column is correctly named "client_id" and it is an INT column, so it doesn't need to be encapsulated with single quotes. here's the code that makes it happen: <CODE> $clientID = $_SESSION['clientID']; $query = "SELECT * "; $query .= "FROM client_data "; $query .= "WHERE client_id = {$clientID} "; $query .= "LIMIT 1"; $results = mysql_query($query); </CODE> what. the. heck. thanks in advance gang. WR! Hi i have created a query where i want some columns to be retrieved. i get these 2 errors and i dont understand why.
Notice: Undefined variable: db
Fatal error: Call to a member function query() on a non-object in
this is my query
function get_products_all() { $connect = get_connected_db(); try { $results = $db->query(" ***** error on this line**** SELECT id, product, description, price, picc FROM products ORDER BY id DESC "); } catch (Exception $e) { echo "Data could not be retrieved from the database."; exit; } $result = $results->fetchALL(PDO::FETCH_ASSOC); return $result; }what am i doing wrong? Hi, My issue here is that I cant get my query string variables to ONLY feed into if/else statement on my secondary page. I include my secondary page (fine.php) from my index page. My query string variables keep being fed back into my original if/ese statement on index.php. here is the if/else on index.php (these links work fine): <?php if($_SERVER['QUERY_STRING']=='/index.php' || $_SERVER['QUERY_STRING']=='') { include 'port.php'; } elseif (isset($_GET['pos'])){ include 'pos.php'; } elseif (isset($_GET['web'])){ include 'web.php'; } elseif (isset($_GET['fine'])){ include 'fine.php'; } else {include '404.php';} here is the if/else on my secondary page (fine.php). These links are supposed to alert the if/else in the next table cell. However, they instead alert the if/else in index.php. <td><br/> <a href="?backset"><img src="fine/thumbs/x-backset.jpg" border="0"></a><br/><br/> <a href="?backside"><img src="fine/thumbs/x-backside.jpg" border="0"></a><br/><br/> <a href="?bannerprint"><img src="fine/thumbs/x-bannerprint.jpg" border="0"></a><br/><br/> <a href="?chopu"><img src="fine/thumbs/x-chopu.jpg" border="0"></a><br/><br/> </td> <td><br/> <div id="DivPiece" align="left"> <?PHP if (isset($_GET['backset'])){ include 'fine/backset.php'; } elseif (isset($_GET['backside'])){ include 'fine/backside.php'; } elseif (isset($_GET['bannerprint'])){ include 'fine/bannerprint.php'; } elseif (isset($_GET['chopu'])){ include 'fine/chopu.php'; } ?> </div> </td> How can I get the links on the secondary page to only alert the if/else statement on that page, and BLOCK the if/else statement on index.php from seeing them? I still want to use the query string though. Thanks! [/quote] Hi, In some cases I need to execute a default query , so I am storing the SELECT statement in a variable like this and executing it: Code: [Select] $default_query="SELECT * from user where userid='$userid'"; $user_query=mysql_query($default_query); The above code returns an empty results. But if I execute it without the variable it works fine. Code: [Select] $user_query=mysql_query("SELECT * from user where userid='$userid'"); Am I syntatically wrong somewhere? Hi I'm having a problem getting a query to work. I have a simple form with user input for start and end date with format: 2009-03-19 (todays date): $Startdate = $_POST['date']; This works well when something is entered into the form, and afterwards using my query: SELECT COUNT(*) as total FROM mydb WHERE Date BETWEEN '$Startdate' AND '$EndDate' ........ Problem is if user submits the form without entering anything in the date input fields, which makes sense. I want to check if inputs has been made, and if not set af default date, but can't make it work: if (isset($_POST['date']) && $_POST['date'] !='') { $Startdate = $_POST['date'];} else { $Startdate = '1980-01-01';} How can I set $Startdate to something that can be used in the query as below doesn't work? I am trying to pass in a $string variable into my query like so but it is returning a warning: Code: [Select] $string = "clientName == '$input'"; $input = "Sam"; $table_id = 'booking'; $query ="SELECT * FROM booking WHERE. '$string' "; $test = mysql_query($query); echo $test; I know the following simple demonstration fails and what to do to rectify it (though with a lot of additional code is necessary), but I don't know why this is the behaviour. It would be great if someone could explain to me why, and what the neatest way of fixing it would be. Code: [Select] +----------+--------------+------+-----+---------+ | Field | Type | Null | Key | Default | +----------+--------------+------+-----+---------+ | Id | int(11) | NO | PRI | NULL | | Name | varchar(45) | YES | | NULL | | Number | int(9) | YES | | NULL | Code: [Select] $name = "Bradley Cooper"; $number = ""; $query = "INSERT INTO table (Id, Name, Number) VALUES (1, '$name', $number)"; mysql_query($query); Because the $number variable is empty, this simple query fails. Logically, one would think that nothing or NULL gets inserted to that field since the MySQL structure rule states that default is NULL, and since nothing is fed to it in the query, I think that it would automatically be NULL since if the column is omitted, that is exactly the value that will be inserted. Enclosing the variable with quotes is not preferable since this field accepts integers. Also, converting the variable to an integer by (int)$number results in a 0, which is not what I want either. Checking if the variable is empty of not, and if it is, assign $number = "''" so that the query would succeed by enclosing the empty string with single quotes. But this creates a lot of unnecessary code (some of my tables are 80 columns wide and the query therefore have equal amount of variables). So to summarise, why is this happening and how do I neatly avoid assigning quotes to the empty strings? In other words, how do I insert nothing when variables that usually contain an integer is on occation empty? Hi Members,
I am search for the reason for the problem why my mysql query cannot fetch data and store in file based on id in $variable form. For example, $sql="SELECT * FROM mytable WHERE mine_id='1234'"; works for me. But when i use $sql="SELECT * FROM mytable WHERE mine_id='$id'";, files are created as empty. I chanaged the quotes and could not store the data in file. So anyone please help me.
For more clear, i attach the part of my code
for ($i=0;$i<=10;$i++) { $id=$seqs[$i]; $dbo = new PDO($dbc, $user, $pass); echo $sql = "SELECT * FROM mine_id WHERE locus_id='$id'"; $qry = $dbo->prepare($sql); $qry->execute(); $data = fopen('file.csv', 'w'); while ($row = $qry->fetch(PDO::FETCH_ASSOC)) { fputcsv($data, $row); } } Edited by phpnewbie007, 20 November 2014 - 02:17 AM. |