kikilahooch Posted April 10, 2006 Share Posted April 10, 2006 Hi, I'm new to php and having trouble with it. I am using a MySql database and am trying to display the values that I have stored on a customer into text fields so that they can be edited. I have passed the userId in from my login.php page and want to be able to update the details relevent to the userId that was passed in.Here is the code I have so far, it could be completly wrong:<?php include("db.php");$userId= $_GET['id']; $sql = "select password, name, surname, address, address2, county, country, telNo, paypalEmail from customer where userName = '$userId';";//db$result = mysql_query($sql,$conn) or die(mysql_error());//get the number of rows in the result set; should be 1 if a matchif (mysql_num_rows($result) == 1) { //if authorized, get the values of name, surname... $password = mysql_result($result, 0, 'password'); $name = mysql_result($result, 0, 'name'); $surname = mysql_result($result, 0, 'surname'); $address = mysql_result($result, 0, 'address'); $address2 = mysql_result($result, 0, 'address2'); $county = mysql_result($result, 0, 'county'); $country = mysql_result($result, 0, 'country'); $telNo = mysql_result($result, 0, 'telNo'); $paypalEmail = mysql_result($result, 0, 'paypalEmail');}Any help would be appreciated Link to comment https://forums.phpfreaks.com/topic/7029-passing-values-from-a-database-into-a-text-field/ Share on other sites More sharing options...
craygo Posted April 10, 2006 Share Posted April 10, 2006 you could do it in an array.here is a sample[code]<? // all previous stuff hereif (mysql_num_rows($result) == 1) {//if authorized, get the values of name, surname...// Fetch results into an array$r = mysql_fetch_array($result);// print out formprint '<form name=somename method=POST action=somepage.php><table><tr><td width=40%>Name</td><td width=60%><input type=text name=uname value="'.$r['name'].'"></td>// rest of rows here...<tr><td colspan=2 align=center><input type=submit value=Edit></td></tr></table></form>';}?>[/code]Ray Link to comment https://forums.phpfreaks.com/topic/7029-passing-values-from-a-database-into-a-text-field/#findComment-25528 Share on other sites More sharing options...
wildteen88 Posted April 10, 2006 Share Posted April 10, 2006 Now nothings wrong so far. But I dont why but when I see stuff like this:[code] $password = mysql_result($result, 0, 'password');$name = mysql_result($result, 0, 'name');$surname = mysql_result($result, 0, 'surname');$address = mysql_result($result, 0, 'address');$address2 = mysql_result($result, 0, 'address2');$county = mysql_result($result, 0, 'county');$country = mysql_result($result, 0, 'country');$telNo = mysql_result($result, 0, 'telNo');$paypalEmail = mysql_result($result, 0, 'paypalEmail');[/code]It confuses the crap out of me, however It might not to you. But there is a simple to the above which is:[code]//put our results into an array ($row);$row = mysql_fect_array($result);$name = $row['name'];$surname = $row['surname'];$address = $row['address'];$address2 = $row['address2'];$county = $row['county'];$country = $row['country'];$telNo = $row['telNo'];[/code]Much clearner and easier to follow! However this is my personal opinion. But you dont have to change your code if you dont want to. Link to comment https://forums.phpfreaks.com/topic/7029-passing-values-from-a-database-into-a-text-field/#findComment-25530 Share on other sites More sharing options...
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