genista Posted November 4, 2006 Share Posted November 4, 2006 Hopefully this should be a quick fix...The first query in my script pulls information from two tables, one of which could be empty:[code=php:0]$query = "SELECT s.username, si.image1FROM suppliers AS sINNER JOIN supplierimages AS si ON s.username = si.usernameWHERE s.username = '$id'"; [/code]$id = username (part of the session). Now, the problem I am having is when I need to update the supplierimages table with a username, supplierid and the details of the image:[code=php:0] $query = "UPDATE `supplierimages` as si SET `si.image1 = '$image1', `si.username` = '$id' INNER JOIN `suppliers` as s ON `s.supplierid` = si.supplierid WHERE s.username = '$id'"; [/code]I get an error on this query:"check the manual that corresponds to your MySQL server version for the right syntax to use near 'si.username` = '' INNER JOIN `suppliers` as s ON `s.supplierid` = si.supplierid ' at line 2"Any ideas?Thanks,G Quote Link to comment Share on other sites More sharing options...
Psycho Posted November 4, 2006 Share Posted November 4, 2006 You don't have a quote mark (`) at the end of [b]si.image1[/b] right after the SET Quote Link to comment Share on other sites More sharing options...
genista Posted November 4, 2006 Author Share Posted November 4, 2006 Ok I have changed that but still no joy, here is the complete code that I have right now which may help answer:[code=php:0]<?php session_start();include_once ("../../private/supplierconfig.php"); checkLoggedIn("yes");error_reporting (E_ALL); //Retrieve details from database to start (image1 also part of suppliers table - need to have this only )$id = $_SESSION['username'];echo $id;/*$query1 = "SELECT username FROM suppliers WHERE username = '$id'";echo $query1; $result=mysql_query($query1, $link) or die("MySQL query $query1 failed. Error if any: ".mysql_error());*//*$query = "select username, supplierid, image1 from suppliers where username='$id'";*/$query = "SELECT s.username, si.image1FROM suppliers AS sINNER JOIN supplierimages AS si ON s.username = si.usernameWHERE s.username = '$id'"; $result=mysql_query($query, $link) or die("MySQL query $query failed. Error if any: ".mysql_error());echo $query; //get the first (and only) row from the result $row = mysql_fetch_array($result, MYSQL_ASSOC); $username=$row['username']; $image1 = $row['image1']; if(isset( $submit )) { //If the Submitbutton was pressed do: if ($_FILES['imagefile']['type'] == "image/jpeg"){ $_FILES['imagefile']['name'] = basename($_FILES['imagefile']['name']);copy ($_FILES['imagefile']['tmp_name'], "files/".$username.$_FILES['imagefile']['name']) or die ("Could not copy"); echo ""; echo "Name: ".$_FILES['imagefile']['name'].""; echo "Size: ".$_FILES['imagefile']['size'].""; echo "Type: ".$_FILES['imagefile']['type'].""; echo "Upload Complete...."; $image1 = $username.$_FILES['imagefile'] ['name']; echo "$image1"; /*$query = "UPDATE `supplierimages` SET `image1` = '$image1', `username` = '$username' WHERE `username` = '". mysql_real_escape_string($_SESSION['username']). "' LIMIT 1";*/ $query = "UPDATE `supplierimages` as si,suppliers as s,SET `si.image1 = '$image1', `si.username` = '$id' WHERE`s.supplierid` = si.supplierid AND s.username = '$id'"; /* $query = "UPDATE `supplierimages` as si SET `si.image1` = '$image1', `si.username` = '$id' INNER JOIN `suppliers` as s ON `s.supplierid` = si.supplierid WHERE s.username = '$id'";*/ /* s.username, si.image1FROM suppliers AS sINNER JOIN supplierimages AS si ON s.supplierid = si.supplieridWHERE s.supplierid = '$id'";*//*$query = "UPDATE `suppliers` SET `image1` = '$image1' WHERE `username` = '". mysql_real_escape_string($_SESSION['username']). "' LIMIT 1"; $result = mysql_query($query, $link) or die('Update failed: ' . mysql_error()); */ $result = mysql_query($query, $link) or die('Update failed: ' . mysql_error()); echo $query; //print_r($query); mysql_info($link) ; if(mysql_affected_rows($link) == 0); } } else { echo "<br><br>"; } ?><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN "http://www.w3.org/TR/html4/strict.dtd"> <html lang="en"> <form name="form1" method="post" action="#" enctype="multipart/form-data"> <input type=text name="username" value="<?php echo $username; ?>"> <input type="file" name="imagefile"> <p>*</td><td><input name="submit" type="submit" value="Submit"></p> </table> </form>[/code]Excuse all the comments, have been ttrying numerous ways to get this to work. The interesting thing is that when this query is run the $id field is blank, when it should contain the session's username. Thanks. Quote Link to comment Share on other sites More sharing options...
fenway Posted November 5, 2006 Share Posted November 5, 2006 And what error message are you getting now? Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.