truck7758

yes i am using php5 and i have also inserted extension=php_mysqli.dll into my php.ini. any ideas? Cheers

Hi, when trying to view my web page im getting the following error: Fatal error: Class 'mysqli' not found in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\neworder.php on line 42 Line 42 is the first line of sql within the page which is $mysqli = new mysqli('localhost','username','password'); Any ideas, Thanks

$rCrime = mysql_query("SELECT id,auto,geld,wapen,chauffeur,wapenexpert FROM ".$oc." WHERE gamename='".$prefix."' AND id = '.intval($_GET['id']).' AND leader = '".$data->login."';"); try that. this is almost the same as a query i have running and that works fine

Parse error: syntax error, unexpected '"' in is that the whole error message?

$rCrime = mysql_query('SELECT id,auto,geld,wapen,chauffeur,wapenexpert FROM ".$oc." WHERE gamename='".$prefix."' AND id = '.intval($_GET['id']).' AND leader = "'.$data->login.'"'); try that

sorted now. dont know how lol heres the code i used for anyone else who may find it useful <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'orders'; $table = 'orders'; $file = 'order'; $link = new mysqli(localhost, username, password) or die("Can not connect1." . mysql_error()); mysqli_select_db($link, "orders") or die("Can not connect2."); $result = mysqli_query($link, "SHOW COLUMNS FROM ".$table."", MYSQLI_STORE_RESULT); $i = 0; if (mysqli_num_rows($result) > 0) { while ($row = mysqli_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysqli_query($link, "SELECT * FROM ".$table.""); while ($rowr = mysqli_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= $rowr[$j].", "; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d_H-i",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> Thanks All

iv now managed to get to this <?php $mysqli = new mysqli('localhost','root','newr00t'); $mysqli->select_db('orders'); $query_rsReport = "select * from orders"; $rsReport = $mysqli->query("SELECT * FROM orders"); $totalRows_rsReport = mysqli_num_rows($rsReport); if (mysqli_field_count($rsReport)) { /* this was a select/show or describe query */ $fields = mysqli_store_result($rsReport); //$fields = mysqli_field_count($rsReport); for ($i = 0; $i < $fields; $i++) { $header .= $mysqli->field_name($rsReport, $i) . ","; } while ($row = $mysqli->fetch_row($rsReport)) { $line = ''; foreach ($row as $value) { $value = '"' . $value . '"' . ","; $line .= $value; } $data .= trim($line) . "\n"; } $title = "Will appear in the top left cell"; $data = str_replace("\r", " ", $data); header("Content-type: application/x-msdownload"); header("Content-Disposition: attachment; filename=cdbg_r1.csv"); header("Pragma: no-cache"); header("Expires: 0"); print "$title\n\n$header\n$data"; mysqli_free_result($fields); $rsReport->close; ?> and now im getting the error: Parse error: syntax error, unexpected $end in c:\webs\test\export.php on line 36 which is my last line. Any ideas, Mike

iv now sorted that by making it: $rsReport = $mysqli->query($query_rsReport, $mysqli); now im getting: Fatal error: Call to undefined method mysqli::num_rows() in c:\webs\test\export.php on line 7 line 7 = $totalRows_rsReport = $mysqli->num_rows($rsReport); any ideas, cheers

ok, heres what iv managed to get <?php $mysqli = new mysqli('localhost','username','password'); $mysqli->select_db('orders'); $query_rsReport = "select * from orders"; $rsReport = mysqli->query($query_rsReport, $mysqli); $totalRows_rsReport = mysqli->num_rows($rsReport); $fields = mysqli->field_count($rsReport); for ($i = 0; $i < $fields; $i++) { $header .= mysqli->field_name($rsReport, $i) . ","; } while ($row = mysqli->fetch_row($rsReport)) { $line = ''; foreach ($row as $value) { $value = '"' . $value . '"' . ","; $line .= $value; } $data .= trim($line) . "\n"; } $title = "Will appear in the top left cell"; $data = str_replace("\r", " ", $data); header("Content-type: application/x-msdownload"); header("Content-Disposition: attachment; filename=cdbg_r1.csv"); header("Pragma: no-cache"); header("Expires: 0"); print "$title\n\n$header\n$data"; $rsReport->close; ?> this gives the error message: Parse error: syntax error, unexpected T_OBJECT_OPERATOR in c:\webs\test\export.php on line 6 line 6 = $rsReport = mysqli->query($query_rsReport, $mysqli); any ideas? Thanks

yeh i re-read your post after i replied and realised that how do you go about changing the mysql to mysqli as i may not know much but i know its not just a case of adding the 'i' onto the end of all the mysql's thanks

Hi, Whats $CDBG? also i dont think this will work as only mysqli queries seem to work on mine (have no idea why) cheers

i have created a php site where you can add/view the database but now i want to add a link where you can download the contents of a particular table into an csv file Thanks

its saved locally. to connect i use $mysqli = new mysqli('localhost','username','password');

nope

any ideas, tips, prod in the right direction. I seem to be able to find instructions on how to do it using mysql but for some reason i only seem to be able to use mysqli (excuse my noobiness is this doesn't make sense) Thanks

Hi, would it be possible to add a export function in php so it exports all my sql results to a csv or something similar (just to be able to be viewed in excel). Thanks, Mike

$createQuery1 = "INSERT INTO tblOwnerBK (`title`, `ownerName`,`address`,`phoneNumber`) VALUES ('".$title."','".$ownerName."','".$address."','".$phoneNumber."')"; try that

can anyone see whats wrong with this query: <? $supp = $_POST["supp"]; $supp = addslashes($supp); // Connecting, selecting database $mysqli = new mysqli('host','username','password'); $mysqli->select_db('orders'); // Performing SQL query $result = $mysqli->query("SELECT * FROM supplier where `name` = '".$supp."'"); while($line = $result->fetch_assoc()) { echo "\t<tr>\n"; foreach ($line as $col_value) { if (empty($col_value)) { echo "\t\t<td> </td>\n"; } else { echo "\t\t<td>$col_value</td>\n"; } } echo "\t</tr>\n"; } // Free resultset mysqli_free_result($result); // Closing connection $mysqli->close(); ?> Thanks, Mike

Hi, i have created a page that displays a list of suppliers but what i want to be able to do is filter the results to specify individual suppliers. I currently have the following code which displays the full table: <title>View Suppliers</title> <BODY TEXT="red" LINK="yellow" BGCOLOR="black"> <a href="http://www.google.co.uk"> <img src="new_logo2.jpg" alt="logo" align="center" width="210" height="95" /> </a> <? print strftime("%A %dth %B %Y %H:%M"); ?> <h1 align="center">Suppliers<h1> <table border=\"7\"> <tr> <th>Supplier ID</th> <th>Name</th> <th>Address</th> <th>Telephone Number</th> <th>Fax Number</th> <th>Website</th> <th>Account Number</th> <th>Account Manager</th> <th>Email Address</th> <th>Account Login</th> <th>Account Password</th> <th>Other Info</th> </tr> <?php // Connecting, selecting database $mysqli = new mysqli('host','username','password'); $mysqli->select_db('orders'); // Performing SQL query $result = $mysqli->query("SELECT * FROM supplier"); while($line = $result->fetch_assoc()) { echo "\t<tr>\n"; foreach ($line as $col_value) { if (empty($col_value)) { echo "\t\t<td> </td>\n"; } else { echo "\t\t<td>$col_value</td>\n"; } } echo "\t</tr>\n"; } // Free resultset mysqli_free_result($result); // Closing connection $mysqli->close(); ?> </table> <p> <a href="index.php"> <INPUT TYPE="SUBMIT" VALUE="Home" STYLE="font-family:sans-serif; font-size:large; font-style:italic; background:red; color:black; width:6em; height:1.5em"> </a> and i also have the following which creates the drop down list of suppliers: <? $mysqli = new mysqli('localhost','root','newr00t'); $mysqli->select_db('orders'); $result = $mysqli->query("SELECT * FROM supplier"); echo "<SELECT name='supp'>\n"; while($row = $result->fetch_assoc()) { echo "<option value='{$row['suppid']}'>{$row['name']}</option>\n"; } echo "</select>\n"; $result->close(); ?> what i need now is to join the two with sum sort of 'submit' button but i dont really know how to do this. Any ideas, Thanks

seems fine to me using firefox

resolved using previous suggestion by ansarka. the reason it didnt work straight away was i had an error on my insert script Thanks All

Any ideas??? Thanks, Mike

i have now done a few more tests and it is now just submitting the value '1' and i cannot see where it is getting this value from. Any ideas, Mike

i have just done another test and this timme it submitted '2' ??? Thanks, Mike

This no displays 'Select' as the default value although i have changed this to '0' which is great but now when i submit my form it submits the value '3' ??? Any ideas? Thanks, Mike