ExosSho
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i dont know if u solved the problem, but maybe the mysql part isnt right, i think it should be while($row = mysql_fetch_assoc($result)) insted of mysql_fetch_array. but maybe im wrong
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php picture up-loading, with mysql-support
ExosSho replied to ExosSho's topic in Editor Help (PhpStorm, VS Code, etc)
[!--quoteo(post=386554:date=Jun 21 2006, 02:40 PM:name=cmgmyr)--][div class=\'quotetop\']QUOTE(cmgmyr @ Jun 21 2006, 02:40 PM) [snapback]386554[/snapback][/div][div class=\'quotemain\'][!--quotec--] which line is 60 [/quote] [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--] $num = mysql_affected_rows($result); [/quote] ok i got it this is what it was [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]$num = mysql_affected_rows($result);[/quote] but if u delete the "$result" in the brackets it works fine thank u again for your help ^^ -
php picture up-loading, with mysql-support
ExosSho replied to ExosSho's topic in Editor Help (PhpStorm, VS Code, etc)
ok, first thank u for quick(and helpful) replie, now the data is beeing send to the data base,... but now there appears a error [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--] Warning: mysql_affected_rows(): supplied argument is not a valid MySQL-Link resource in C:\apachefriends\xampp\htdocs\upload\showit2.php on line 60 [/quote] -
i started writing a little program, which uploads files (till now pics prefered) to the folder, which contains the PHP-file.(i know; noobish) on the first page, u select the file u want to upload, and may enter a Title( in order to enable a gallery-database with mysql); second page shows u wether the data has been sent to the database(THIS IS THE MAIN PROBLEM because it wont save anything to the database), info about the file(like type etc.), where its gonna be saved, or if a file(pic) alrady exists with the same name, in this case the script should view the "new-version"(which cant be uploaded) and the "old-version"(which alrady exists) [b]this is the formular where u enter the stuff[/b] [code] <html> <head> <link rel="stylesheet" type="text/css" href="style_1.css"> </head> <body> <div style="background-color:#efefef;width:480px;border:5px solid #00c000;padding:1px;margin:0px auto;text-align:center;"> <form enctype="multipart/form-data" action="showit.php" method="post"> data: <input name="upfile" type="file" size="25"> <p>titel: <input name="titel" size="25"> <p><input type="submit" value="send" name="send"> </form> </div> </body> </html> [/code] [b]this file shows the pics, and should send data to mysql[/b] [code]<html> <head> <link rel="stylesheet" type="text/css" href="style_1.css"> </head> <body> <div style="background-color:#efefef;width:480px;margin:0px auto;border:5px solid #00c000;padding:1px;text-align:center;"> <?php echo "<h3><ul><k>control:</k></ul></h3><p>"; echo "original-name: <zitat>$upfile_name</zitat><br>"; echo "size: <zitat>$upfile_size Byte</zitat><br>"; echo "type: <zitat>$upfile_type</zitat><br>"; $dname = explode(".",$upfile_name); $dnamesize = count($dname); $upfile_ext = $dname[$dnamesize-1]; echo "data-extension: <zitat>" . $upfile_ext . "</zitat><br>"; list($headw, $headh) = getimagesize($upfile); echo "original-properties: <zitat>" . $headw . "x" . $headh . "</zitat><br>"; $verz = getcwd(); $exi = $verz ."/". $upfile_name; if($upfile_size>0 && !file_exists($exi)) { echo "new home of " . $upfile_name . " : <zitat>" . $verz. "</zitat><br>"; copy($upfile,$upfile_name); echo "data has been copied in: <zitat>$upfile_name</zitat><br>"; if($headw>470) {; echo '<img src="'.$upfile_name.'" width="470" alt="'.$title.'" />'; } else { echo '<img src="'. $upfile_name .'"><br>'; } } else if(file_exists($exi)) { copy ($upfile, "redrasterblue.png"); echo "<r>a file with the name: <zitat>" . $upfile_name . "</zitat> does already exist!</r><br>"; echo "<table>"; echo "<tr><td><k>new-" . $upfile_name . "</k></td><td><k>old-" . $upfile_name . "</k></td></tr>"; echo '<tr><td><img src="redrasterblue.png" width="200"></td><td><img src="'.$upfile_name.'" width="200"></td></tr>'; echo "</table><br>"; } else { echo "copy-error: data does not exist"; echo " or false data-type"; } if(isset($_POST['send'])) { mysql_connect(localhost,root) or die("no connection to mysql"); mysql_select_db("daten") or die("not able to select database"); $sqlab = "insert into upload (name, typ, titel) values ($upfile_name, $upfile_type, $titel)"; mysql_query($sqlab); $num = mysql_affected_rows(); } if ($num>0) { echo "<g>one data has been added<p></g>"; } else if($num<=0) { echo "<r>an Error appeared, "; echo "no data has been added to mysql<p></r>"; } ?> </div> </body> </html> [/code] thanks for any help^^