PHP - Variable In Query Works, But Php Does Not
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 Similar Tutorialsso my situation is something like this , i'm trying to fetch user details based on `id` that isset is getting, but some how the `variable that contains the $_GET value doesn't work` in query but when i put an static value to pdo query then it works and show the result. i have checked by doing `var_dump` of variable `$user` before query and it shows the correct value but not working in query. Below is the code i'm working with: I'm trying to update every record where one field in a row is less than the other. The code gets each row i'm looking for and sets up the query right, I hope I combined the entire query into one string each query seperated by a ; so it's like UPDATE `table` SET field2= '1' WHERE field1= '1';UPDATE `table` SET field2= '1' WHERE field1= '2';UPDATE `table` SET field2= '1' WHERE field1= '3';UPDATE `table` SET field2= '1' WHERE field1= '4';UPDATE `table` SET field2= '1' WHERE field1= '5'; this executes properly if i run the query in phpMyAdmin, however when I run the query in PHP, it does nothing... Any advice? I have a need to do a special sorting for a bunch of leads. The person who sees the leads (l.user) is looking at leads that are for a few different people (l.team_id). When I run my query in sqlyog, it gives me the results I am looking for, where the rank is used to sort. When I run my query through phpMyAdmin, php, or the MySQL command line, all of the ranks are NULL, so nothing gets sorted. All of these attempts are on the same database, on the same computer (Windows 7). Any ideas or help is appreciated.
SELECT x.lead_id, x.rank, x.team_id, x.status FROM (SELECT l.lead_id, l.team_id, l.status, CASE WHEN @ps != l.team_id THEN @rownum := 0 WHEN @ps = l.team_id THEN @rownum := @rownum + 1 ELSE @rownum END AS rank, @ps := l.team_id FROM `system-leads` l WHERE (l.user = 189905706) AND l.status != "closed" LIMIT 100000000 OFFSET 0) `x` ORDER BY x.status ASC, x.rank ASC, x.lead_id ASCI wonder why it works in sqlyog and not the others, but I need it to work regardless of what client I am using. I don't understand why "if(isSet($_GET['sid'])) " works before it is declared in the variable name(" $name = $_GET['sid'];"). 1. How does php store the date in this get statement? 2. Is GET an array? Code: [Select] if(isSet($_GET['sid'])) { echo "<h2>Exercise</h2>"; $name = $_GET['sid']; $timein = time(); $quest=1; $nextquest=2; } else { echo "<h2>Enter Student Number:</h2>"; $quest=0; $nextquest=0; } Good Afternoon Team, Am sitting with something simple using the language below. If I copy the echo output of my query as included below it works perfectly in phpmyadmin but doesn't work on a website. Variables all seem to echo consistently/correctly and POST checks seem to verify this is working correctly as well. I worry the error comes with the syntax I used in combining the sql queries. That, or perhaps LAST_INSERT_ID does not work in the php script as well as it does in phpmyadmin. All help appreciated.
if(isset($_POST[`region_id`])) {
I'm having trouble with a simple SELECT query. I just cannot figure out what the problem is... <?php //Include database connection details include 'login/config.php'; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } $qry="SELECT * FROM members"; $result = mysqli_query($link, $qry); echo "<table>"; while($row = mysqli_fetch_array($result, MYSQL_ASSOC)) { $getid = ($row['member_ID']); $firstname = ($row['firstname']); $lastname = ($row['lastname']); $email = ($row['email']); echo "<tr><td>$firstname</td><td>$email</td></tr>"; } echo "</table>"; ?> I know I have a connection to the DB, and I know that the query will return values as I have tested in in phpmyadmin. Can anyone see anything obvious I am missing? Thanks I have a really weird problem. I'm trying to run a mysql query that works fine in phpmyadmin but in php script is giving me an error. The query is: Code: [Select] (SELECT DISTINCT art.`TEMPLATE`,gal.`ARTICLE_ID`,art.`TITLE`,art.`DESCRIPTION`,MATCH(art.`TITLE`,art.`DESCRIPTION`,gal.`CONTENT`) AGAINST ('WORD*' IN BOOLEAN MODE) AS score FROM articles art,galeries gal WHERE gal.`ARTICLE_ID`=art.`ARTICLE_ID` AND MATCH(art.`TITLE`,art.`DESCRIPTION`,gal.`CONTENT`) AGAINST ('WORD* ' IN BOOLEAN MODE)) UNION (SELECT DISTINCT `TEMPLATE`,`ARTICLE_ID`,`TITLE`,`DESCRIPTION`,MATCH(`TITLE`,`DESCRIPTION`,`CONTENT`) AGAINST ('WORD* ' IN BOOLEAN MODE) AS score FROM articles WHERE (MATCH(`TITLE`,`DESCRIPTION`,`CONTENT`) AGAINST ('WORD* ' IN BOOLEAN MODE))) ORDER BY score DESC LIMIT 0,30 Snipset from php script code: function search($start_row,$ammount,$search_key,$pages){ $start_row = intval($start_row) * $ammount; $return_val = ""; $symbols = array('/','\\','\'','"',',','.','<','>','?',';',':','[',']','{','}','|','=','+','-','_',')','(','*','&','^','%','$','#','@','!','~','`' );//this will remove punctuation $pattern = "#[^(\w|α|β|γ|δ|ε|ζ|η|θ|ι|κ|λ|μ|ν|ξ|ο|π|ρ|σ|τ|υ|φ|χ|ψ|ω|Α|Β|Γ|Δ|Ε|Ζ|Η|Θ|Ι|Κ|Λ|Μ|Ν|Ξ|Ο|Π|Ρ|Σ|Τ|Υ|Φ|Χ|Ψ|Ω|ς|ά|έ|ό|ί|ύ|ώ|ή|ϊ|ϋ|ΐ|ΰ|Ά|Έ|Ό|Ί|Ύ|Ώ|Ή|Ϊ|Ϋ|\d|\'|\"|\.|\!|\?|;|,|\\|\/|\-|:|\&|@)]+#"; $search_key = greek_text::to_upper($search_key); $wc = strip_tags($search_key); $wc = preg_replace($pattern, " ", $wc); for ($i = 0; $i < sizeof($symbols); $i++) { $wc = str_replace($symbols[$i],' ',$wc); } $wc = str_replace("΄", " ", $wc); $wc = str_replace(chr(162), " ", $wc); if( !$keep_numbers ) { $wc = preg_replace('#(^|\s+)[\d\s]+(\s+|$)#',' ',$wc); $pattern = '#(^|\s+)([0-9]+[a-zA-ZαβγδεζηθικλμνξοπρστυφχψωΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩςάέόίύώήϊϋΐΰΆΈΌΊΎΏΉΪΫ]+\s*)+(\s+|$)#'; preg_match($pattern,$wc,$aa); $wc = preg_replace($pattern, " ", $wc); } $wc = trim(preg_replace("/\s\s+/", " ", $wc)); $wc = explode(" ", $wc); $cleaned_keyword = array_filter($wc); $cleaned_keyword = greek_text::removeStopWordsFromArray($cleaned_keyword); $stemmed_keywords = greek_text::stemWordsArray($cleaned_keyword); $query = "(SELECT DISTINCT art.`TEMPLATE`,gal.`ARTICLE_ID`,art.`TITLE`,art.`DESCRIPTION`,MATCH(art.`TITLE`,art.`DESCRIPTION`,gal.`CONTENT`) AGAINST ('"; while(list($key,$val)=each($stemmed_keywords)){ if($val<>" " and strlen($val) > 1){ $query .= $val."*"; $search_keys .= $val."* "; } } $query .= "' IN BOOLEAN MODE) AS score FROM articles art,galeries gal WHERE gal.`ARTICLE_ID`=art.`ARTICLE_ID` AND MATCH(art.`TITLE`,art.`DESCRIPTION`,gal.`CONTENT`) AGAINST ('".$search_keys."' IN BOOLEAN MODE))"; $query .= " UNION (SELECT DISTINCT `TEMPLATE`,`ARTICLE_ID`,`TITLE`,`DESCRIPTION`,MATCH(`TITLE`,`DESCRIPTION`,`CONTENT`) AGAINST ('".$search_keys."' IN BOOLEAN MODE) AS score FROM articles WHERE (MATCH(`TITLE`,`DESCRIPTION`,`CONTENT`) AGAINST ('".$search_keys."' IN BOOLEAN MODE))) ORDER BY score DESC LIMIT ".intval($start_row).",".$ammount; $rs = $this->dbActions->execQuery($query); $this->dbActions->execQuery("INSERT INTO searches (`KEY`,`DATE`,`RESULTS`) VALUES ('".$search_key."',NOW(),".$pages.")"); $search_results = "<div id='results'>"; while($row = mysql_fetch_array($rs)){ $search_results.= "<div id='result'>"; $search_results.= "<div class='result_title'><h4><a href='article.php?articleId=".$row["ARTICLE_ID"]."'>".$row['TITLE']."</a></h4></div>"; $search_results.= "<div class='result_description'>".$row['DESCRIPTION']."</div>"; $search_results.= "</div>"; $search_results.= "<div class='result_seperator'></div>"; } $search_results .= "</div>"; $return_val = $search_results; return $return_val; } dbactions class: require 'includes/errors.php'; error_reporting(0); class DBActions{ var $dbCon; var $errorHandler; function DBActions(){ $this->dbCon = $this->dbCon(); $errorHandler = new errors(1); } private function dbCon(){ require 'conf/configuration.php'; $dbcon = mysql_connect($dbUrl,$dbUser,$dbPass); if(!$dbcon)trigger_error("Unable to connect to database $dbUrl for user $dbUser",E_USER_ERROR); mysql_select_db($dbName); mysql_query("SET NAMES 'UTF8'"); mysql_query('set character set utf8'); return $dbcon; } function dbClose(){ if($this->dbCon) mysql_close($this->dbCon); } function execQuery($query){ $result = mysql_query($query,$this->dbCon); $msg = "Unable to execute query ".$query; if(mysql_num_rows($result) > 0) return $result; else if($result == false) trigger_error($msg,E_USER_ERROR); else return $result; } function send_error_mail(){ $this->errorHandler->sendErrorEmail(); } } I've try everything but can't get this working.. Any healp whould be really appreciated. Thanks in advance.
Hi, this query runs fine when I run it from PHPMyAdmin: UPDATE `tran_term_taxonomy` SET `description` = (SELECT keyword from `good_keywords` ORDER BY RAND() LIMIT 1,1) WHERE `tran_term_taxonomy`.`taxonomy` = 'post_tag' AND `tran_term_taxonomy`.`description` = "" LIMIT 1 However, when I run the same query in a PHP file on my server, the page doesn't load at all. The message I get is: www.somesite.com is currently unable to handle this request. HTTP ERROR 500. This is my PHP code: <?php include("/database/connection/path/db_connect.php"); $result4 = mysqli_query($GLOBALS["___mysqli_ston"], "UPDATE `tran_term_taxonomy` SET `description` = (SELECT keyword from `good_keywords` ORDER BY RAND() LIMIT 1,1) WHERE `tran_term_taxonomy`.`taxonomy` = 'post_tag' AND `tran_term_taxonomy`.`description` = "" LIMIT 1"); echo $result4; ?> So how do I make this query work please? Thanks for your guidance. Hi, I'm new to PHP/MySQL and need some help getting my query to work for my selection list: The selection list is built with: <form action='processformmissing.php' method='POST'> <fieldset> <legend>Choose Department</legend> <select name='depart'> <option value=''></option> <?php while ($row = mysqli_fetch_array($result)) { extract($row); echo "<option value='$department'>$department</option>\n"; } ?> </select> <p><input type='submit' value='Select Department' /></p> </fieldset> </form> The data is then sent to: $depart = $_POST['depart']; $deptlike = "%".$depart."%"; echo "<p>$depart</p>"; echo "<p>$deptlike</p>"; $query = "SELECT * FROM lifecerts INNER JOIN employees ON lifecerts.cid = employees.cid WHERE department LIKE '$deptlike' ORDER BY employees.name"; Hitting the submit button from my selection list form seems to be working fine because when I echo my data ($depart and $deptlike) it is giving me the correct value, but the query doesn't give me any results. However, if my post data comes from a text box instead of a selection list, my query works fine. Any thoughts on what I'm doing wrong??? Many thanks! hi i have the script below which copies data from one table to another but will only insert new data update current data or delete old data from tempproducts to products then it will delete the tempproducts from the db however i keep getting this error: Warning: mssql_query() [function.mssql-query]: Query failed in E:\UpdateProducts.php on line 33 updateproducts.php Code: [Select] <?php include('../../otherscripts/functions.php'); $log = new Logging(); // create DB connection $host = "localhost"; $user = "user"; $pass = "pass"; $mydb = "db"; $db = mssql_connect($host,$user,$pass); //Select Database mssql_select_db($mydb); // delete all old data $sql0 = "SELECT * FROM tempproduct"; $sql1 = "INSERT INTO products SELECT * FROM tempproduct WHERE manf_part_no NOT IN (SELECT manf_part_no FROM products) AND supp_id NOT IN (SELECT supp_id FROM products)"; $sql2 = "DELETE FROM products WHERE manf_part_no NOT IN (SELECT manf_part_no FROM tempproduct) AND supp_id NOT IN (SELECT supp_id FROM tempproduct)"; $sql3 = "UPDATE p1 SET p1.avail_qty = t1.avail_qty, p1.cost_price = t1.cost_price, p1.rrp = t1.rrp, p1.date_added = t1.date_added, p1.description = t1.description FROM Products p1 INNER JOIN tempproduct t1 ON (p1.manf_part_no = t1.manf_part_no AND p1.supp_id = t1.supp_id)"; $sql4 = "TRUNCATE TABLE tempproduct"; //If tempproduct is empty done Execute Commands if it is full then execute commands $query = mssql_query($sql0) or die($log->lwrite('Failed to select for count from db')); $rowcount = mssql_num_rows($query); if($rowcount == 0){ $log->lwrite('Teh tempproduct am emptyish'); } else{ mssql_query($sql1) or die($log->lwrite('Failed to insert to db'.$sql1)); mssql_query($sql2) or die($log->lwrite('Failed to Delete from db')); mssql_query($sql3) or die($log->lwrite('Failed to Update db')); mssql_query($sql4) or die ($log->lwrite('Failed to TRUNCATE db')); } ?> if i run $sql1 command in the sql manager it runs fine and no errors occur? 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 $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=$id_zamjena>"; // output of this is always 16 echo "<input name='response' type='submit' value='Prihvati zamjenu' /> <input name='response' type='submit' value='Odbaci zamjenu' /> <br>"; } echo"</form>"; I got my search page error and the server pop up below message:
Notice: Undefined variable: query in /home/tz005/public_html/COMP1687/search.php on line 64 Minimum length is 3 Where should I make a correction in the script? Here is my php script: <?php Hello, So I have a content management site set up, but I'm stuck on one part. I have a query. Code: [Select] Select `email` FROM author WHERE author.id IN(SELECT authorid FROM job WHERE id = '$_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, $headers); Thanks in Advance! Hi Guys. I need your help. Is it possible to store results from sql query as a variable. I am basically doing a few joins to get this result and I wanted to know if this was possible. Hi This does not make sense to me, I hope someone can tell me why this query don't work: Code: [Select] safe_query("UPDATE ".PREFIX."cup_challenges SET reply_date='$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 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? 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 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. |