mike12255 Posted June 2, 2009 Share Posted June 2, 2009 So i got a script that should upload a resized image and then throw some data in a database however im getting the following error; Unknown column 'random' in 'field list' Im not sure what its doing this i tried running the query through phpmyadmin and i was able to do it. Also i dont know why its skippin the if statment that checks to see if file is null either because i went through browse and selected a file. Anyway heres the code: <?php include ("connect.php"); if (isset($_POST['submit'])){ $pic = $_POST['file']; $name = $_POST['name']; if ($_POST['file'] != NULL){ $idir = "./images/"; // Path To Images Directory $tdir = "./images/thumbs/"; // Path To Thumbnails Directory $twidth = "125"; // Maximum Width For Thumbnail Images $theight = "84"; // Maximum Height For Thumbnail Images $url = $_FILES['imagefile']['name']; // Set $url To Equal The Filename For Later Use if ($_FILES['imagefile']['type'] == "image/jpg" || $_FILES['imagefile']['type'] == "image/jpeg" || $_FILES['imagefile']['type'] == "image/pjpeg") { $file_ext = strrchr($_FILES['imagefile']['name'], '.'); // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php $copy = copy($_FILES['imagefile']['tmp_name'], "$idir" . $_FILES['imagefile']['name']); // Move Image From Temporary Location To Permanent Location $path = "$idir" . $_FILES['imagefile']['name']; $desc = $_POST['desc']; $thpath = "$tdir" . $_FILES['imagefile']['name']; if ($copy) { // If The Script Was Able To Copy The Image To It's Permanent Location // print 'Image uploaded successfully.<br />'; // Was Able To Successfully Upload Image $simg = imagecreatefromjpeg("$idir" . $url); // Make A New Temporary Image To Create The Thumbanil From $currwidth = imagesx($simg); // Current Image Width $currheight = imagesy($simg); // Current Image Height if ($currheight > $currwidth) { // If Height Is Greater Than Width $zoom = $twidth / $currheight; // Length Ratio For Width $newheight = 100; // Height Is Equal To Max Height $newwidth = 150; // Creates The New Width } else { // Otherwise, Assume Width Is Greater Than Height (Will Produce Same Result If Width Is Equal To Height) $zoom = $twidth / $currwidth; // Length Ratio For Height $newwidth = 150; // Width Is Equal To Max Width $newheight = 100; // Creates The New Height } $dimg = imagecreate($newwidth, $newheight); // Make New Image For Thumbnail imagetruecolortopalette($simg, false, 256); // Create New Color Pallete $palsize = ImageColorsTotal($simg); for ($i = 0; $i < $palsize; $i++) { // Counting Colors In The Image $colors = ImageColorsForIndex($simg, $i); // Number Of Colors Used ImageColorAllocate($dimg, $colors['red'], $colors['green'], $colors['blue']); // Tell The Server What Colors This Image Will Use } imagecopyresized($dimg, $simg, 0, 0, 0, 0, $newwidth, $newheight, $currwidth, $currheight); // Copy Resized Image To The New Image (So We Can Save It) imagejpeg($dimg, "$tdir" . $url); // Saving The Image imagedestroy($simg); // Destroying The Temporary Image imagedestroy($dimg); // Destroying The Other Temporary Image // die($name); $sql = "INSERT INTO catagories (Name,picture,thumb) VALUES ('$name','$path','$thpath')"; mysql_query ($sql) or die (mysql_error());?> <script type="text/javascript"> <!-- alert ("Item Added Sucsesfully.") // --> </script> <?php } else { print '<font color="#FF0000">ERROR: Unable to upload image.</font>'; // Error Message If Upload Failed } } else { print '<tr><td><font color="#FF0000">ERROR: Wrong filetype (has to be a .jpg or .jpeg. Yours is </td></tr>'; // Error Message If Filetype Is Wrong print $file_ext; // Show The Invalid File's Extention print '.</font>'; } }else{ //die($name); $sql = "INSERT INTO catagories (Name) VALUES ($name)"; mysql_query($sql) or die (mysql_error()); } } ?> Quote Link to comment https://forums.phpfreaks.com/topic/160582-solved-php-wont-shove-my-query-through-unkown-column-in-field-list/ Share on other sites More sharing options...
Philip Posted June 2, 2009 Share Posted June 2, 2009 Place single quotes around $name: $sql = "INSERT INTO catagories (Name) VALUES ($name)"; //so it looks like this: $sql = "INSERT INTO catagories (Name) VALUES ('$name')"; Quote Link to comment https://forums.phpfreaks.com/topic/160582-solved-php-wont-shove-my-query-through-unkown-column-in-field-list/#findComment-847473 Share on other sites More sharing options...
mike12255 Posted June 2, 2009 Author Share Posted June 2, 2009 arh! cure my blindnesss, thanks Quote Link to comment https://forums.phpfreaks.com/topic/160582-solved-php-wont-shove-my-query-through-unkown-column-in-field-list/#findComment-847476 Share on other sites More sharing options...
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