christo Posted May 3, 2006 Share Posted May 3, 2006 Once again i'm passing values from html to php but the request is not working and gives [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]Problem with query: SELECT * FROM object WHERE city = Montpellier AND transaction = Location AND accomodation = Maison AND nb_rooms = 6 AND price = 1650000 AND age_par = AgenceUnknown column 'Montpellier' in 'where clause'[/quote] i'm passing the values with [code]$city = $HTTP_POST_VARS['city'];$transaction = $HTTP_POST_VARS['transaction'];$accomodation = $HTTP_POST_VARS['accomodation'];$nb_rooms = $HTTP_POST_VARS['nb_rooms'];$age_par = $HTTP_POST_VARS['age_par'];$price = $HTTP_POST_VARS['price'];[/code] at the top of the script and my request is [code]$host = "???";$user = "???";$pass = "???";$base = "???";$db_conx = mysql_connect($host,$user,$pass) or print("error connection MySQL !");mysql_select_db($base, $db_conx) or print("Error connection DB !");$query = "SELECT * FROM object WHERE city = $city AND transaction = $transactionAND accomodation = $accomodation AND nb_rooms = $nb_rooms AND price = $price ANDage_par = $age_par";$rs = mysql_query($query) or die('Problem with query: '.$query.'<br><br>'.mysql_error());[/code]I need to check if the values i'm passing exist in the DB and if yes show results, but i wonder if linking all the vals with AND is the correct way to go... ? Link to comment https://forums.phpfreaks.com/topic/8962-sql-request/ Share on other sites More sharing options...
Orio Posted May 3, 2006 Share Posted May 3, 2006 When using the "WHERE" you got to use single quotes- [!--sql--][div class=\'sqltop\']SQL[/div][div class=\'sqlmain\'][!--sql1--][span style=\'color:blue;font-weight:bold\']SELECT[/span] * [color=green]FROM[/color] [color=orange]sometable[/color] [color=green]WHERE[/color] column1[color=orange]=[/color][color=red]'123'[/color] [!--sql2--][/div][!--sql3--]Notice the single quotes around the "123".This also applys about variables, (IE "WHERE coulmn1='$city' ").Irui. Link to comment https://forums.phpfreaks.com/topic/8962-sql-request/#findComment-32926 Share on other sites More sharing options...
christo Posted May 3, 2006 Author Share Posted May 3, 2006 That's it ;-)When i try to display the data from the DB i get an endles While loop with an error...[!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\foreign\easyphp1-8\www\agence\test\search.php on line 31[/quote][code]$query = "SELECT * FROM object WHERE city = '$city' AND transaction = '$transaction'AND accomodation = '$accomodation' AND nb_rooms = '$nb_rooms' AND price = '$price' ANDage_par = '$age_par'";$rs = mysql_query($query) or die('Probleme avec la requette :'.$query.'<br><br>'.mysql_error());while(($data=mysql_fetch_array($query))!==false) { //I get the Warning here echo "<tr> <td>".$data['city']."</td> <td>".$data['transaction']."</td> <td>".$data['accomodation']."</td> <td>".$data['nb_rooms']."</td> <td>".$data['price']."</td> <td>".$data['age_par']."</td> </tr>";} [/code]Any ideas??? Link to comment https://forums.phpfreaks.com/topic/8962-sql-request/#findComment-32968 Share on other sites More sharing options...
.josh Posted May 3, 2006 Share Posted May 3, 2006 change this:while(($data=mysql_fetch_array($query))!==false) { to this:while ($data=mysql_fetch_array($rs)) { Link to comment https://forums.phpfreaks.com/topic/8962-sql-request/#findComment-32977 Share on other sites More sharing options...
christo Posted May 3, 2006 Author Share Posted May 3, 2006 I didn't think that would wok !!!1000 Thanks :-)christos Link to comment https://forums.phpfreaks.com/topic/8962-sql-request/#findComment-32982 Share on other sites More sharing options...
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