j05hr Posted November 1, 2009 Share Posted November 1, 2009 I'm using some code to load and view images, when I try to upload images, it says Table 'image-store.images' doesn't exist which really confuses me because my table is called image-store and I've got a field called images so what could be the reasons for this? My code if it helps... index.php <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <?php if(isset($_POST['imageInsert'])) { function checkUrl($url) { if(@file_get_contents($url)) { return TRUE; } else { return FALSE; } } $con = mysql_connect("localhost", "root", ""); $db = mysql_select_db("image-store", $con); $url = $_POST['url']; if(preg_match("#((\.jpeg)|(\.jpg)|(\.gif)|(\.png))$#is", $url, $match)) { if($match[1]==".jpg") { $ext = ".jpeg"; } else { $ext = trim($match[1]); } $ext = substr($ext, 1); if(checkUrl($_POST['url'])) { $image = mysql_real_escape_string(file_get_contents($_POST['url'])); if(mysql_query("INSERT INTO `images` (image, ext) VALUES ('$image', '$ext') ") or die (mysql_error())) { echo "insert complete"; } else { echo "could not insert error occured"; } } else { echo "url incorrect"; } } else { echo "invalid file type"; } } elseif(isset($_POST['imageShow'])) { $src = "imageLoader.php?id=".$_POST['id']; echo "<h2>Show Image: ID#".$_POST['id']."</h2>\n"; echo "<img src=\"".$src."\" />\n"; } else { echo "<h2>No image Selected To Show</h2>\n"; } ?> <h2>Upload Image</h2> <form method="post" action="index.php"> url: <input type="text" name="url" /> <input type="submit" name="imageInsert" /> </form> <h2>Show image</h2> <form method="post" action="index.php"> id: <input type="text" name="id" /> <input type="submit" name="imageShow" /> </form> </body> </html> imageLoader.php <?php if(isset($_GET['id'])) { $con = mysql_connect("localhost", "root", ""); $db = mysql_select_db("image-store", $con); $id = $_GET['id']; $get_query = mysql_query("SELECT * FROM `images` WHERE id = '$id'") or die(mysql_error()); $get_array = mysql_fetch_array($get_query); $image_contents = $get_array['images']; header('Content-Type: image/gif'); echo $image_contents; } ?> Quote Link to comment https://forums.phpfreaks.com/topic/179817-solved-database-error/ Share on other sites More sharing options...
j05hr Posted November 1, 2009 Author Share Posted November 1, 2009 Database screenshot Quote Link to comment https://forums.phpfreaks.com/topic/179817-solved-database-error/#findComment-948643 Share on other sites More sharing options...
Mchl Posted November 1, 2009 Share Posted November 1, 2009 Your DATABASE is called image-store. Inside this database there is one TABLE called image-store as well. A FIELD (a.k.a. COLUMN) in this table is called 'images'. So you need to "INSERT INTO `image-store` (images, ext) VALUES ('$image', '$ext') " "SELECT * FROM `image-store` WHERE id = '$id'" Quote Link to comment https://forums.phpfreaks.com/topic/179817-solved-database-error/#findComment-948644 Share on other sites More sharing options...
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