HoTDaWg Posted March 13, 2006 Share Posted March 13, 2006 Hello,I am creating a photo gallery for the awsome people on my site (if any lol). What I am doing is once the image is uploaded, I want all the info added to a database. Everything is working great. The only problem is that I can not seem to get the location of the file once it is uploaded right. Take a look at the dumping data of my mySql database once I export it:[code]-- phpMyAdmin SQL Dump-- version 2.6.4-pl2-- http://www.phpmyadmin.net-- -- Host: localhost-- Generation Time: Mar 13, 2006 at 05:22 PM-- Server version: 4.1.14-- PHP Version: 4.3.11-- -- Database: `cortez_photoalbums`-- -- ---------------------------------------------------------- -- Table structure for table `images`-- CREATE TABLE `images` ( `id` int(15) NOT NULL auto_increment, `imageurl` blob NOT NULL, `titled` varchar(255) NOT NULL default '', `contributedby` varchar(22) NOT NULL default '', `date` varchar(50) NOT NULL default '0000-00-00', PRIMARY KEY (`id`)) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=6;-- -- Dumping data for table `images`-- INSERT INTO `images` VALUES (1, 0x2275706c6f6164732f22202e20245f46494c45535b2266696c65225d5b226e616d65225d, '4', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (2, 0x2275706c6f6164732f225b2266696c65225d5b226e616d65225d, '1', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (3, 0x2275706c6f6164732f225b2266696c65225d5b226e616d65225d, '7', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (4, 0x2275706c6f6164732f225b226e616d65225d, '6', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (5, 0x282275706c6f6164732f22202e20245f46494c45535b2266696c65225d5b226e616d65225d29, '5', 'asd', '03.13.06');[/code]weird eh? Here is the code for upload.html:[code]<html><head><title>Upload now!</title></head><form action="upload.php" method="post" enctype="multipart/form-data"> <p>File Upload</p> <label for="file">File</label> <input type="file" name="file" id="file" /><br><br> <label for="titled">File Name</label><input type="text" name="titled"><br><br> <label for="contributedby">Contributed by:</label><input type="text" name="contributedby"<br><br> <br /> <input type="submit" name="submit" value="Submit" /></form><body bgcolor="#FFFFFF"></body></html>[/code]here is the code for upload.php:[code]<?phpif (($_FILES["file"]["type"] == "image/gif" OR "images/jpg") &&($_FILES["file"]["size"] < 1500000)) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; echo "Uploading " . $_FILES["file"]["name"]; echo " (" . $_FILES["file"]["type"] . ", "; echo ceil($_FILES["file"]["size"] / 1024) . " Kb).<br />"; if (file_exists("uploads/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; echo "Please delete the destination file and try again."; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "uploads/" . $_FILES["file"]["name"]); echo "The file has been successfully uploaded!:D";$imageurl = '("uploads/" . $_FILES["file"]["name"])';$titled = $_REQUEST['titled'];$contributedby = $_REQUEST['contributedby'];$date = date("m.d.y");$titled=strip_tags($titled);$contributedby=strip_tags($contributedby);$titled=addslashes($titled);$contributedby=addslashes($contributedby);$conn = mysql_connect("localhost","cortez","165745");//now choose the database to usemysql_select_db("cortez_photoalbums");//Save the request in SQL syntax to a string$request = "INSERT INTO images values(0,'".$imageurl."','".$titled."','".$contributedby."','".$date."')";$results = mysql_query($request,$conn); }} else echo "Sorry, we only accept .GIF, and .jpg images under 1500 Kb for upload.";?>[/code]I would love any assistance in solving this problem. HoTDaWg Quote Link to comment Share on other sites More sharing options...
ToonMariner Posted March 13, 2006 Share Posted March 13, 2006 $imageurl = '("uploads/" . $_FILES["file"]["name"])';change to...$imageurl = "uploads/" . $_FILES["file"]["name"];php will not interpret the variable correctly in a single quote string (I see you concatenate but you still have a closing quote). Quote Link to comment Share on other sites More sharing options...
HoTDaWg Posted March 13, 2006 Author Share Posted March 13, 2006 appreciate the help a lot, but according to my export code, my code still doesnt work:[code]-- phpMyAdmin SQL Dump-- version 2.6.4-pl2-- http://www.phpmyadmin.net-- -- Host: localhost-- Generation Time: Mar 13, 2006 at 05:58 PM-- Server version: 4.1.14-- PHP Version: 4.3.11-- -- Database: `cortez_photoalbums`-- -- ---------------------------------------------------------- -- Table structure for table `images`-- CREATE TABLE `images` ( `id` int(15) NOT NULL auto_increment, `imageurl` blob NOT NULL, `titled` varchar(255) NOT NULL default '', `contributedby` varchar(22) NOT NULL default '', `date` varchar(50) NOT NULL default '0000-00-00', PRIMARY KEY (`id`)) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=7;-- -- Dumping data for table `images`-- INSERT INTO `images` VALUES (1, 0x2275706c6f6164732f22202e20245f46494c45535b2266696c65225d5b226e616d65225d, '4', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (2, 0x2275706c6f6164732f225b2266696c65225d5b226e616d65225d, '1', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (3, 0x2275706c6f6164732f225b2266696c65225d5b226e616d65225d, '7', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (4, 0x2275706c6f6164732f225b226e616d65225d, '6', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (5, 0x282275706c6f6164732f22202e20245f46494c45535b2266696c65225d5b226e616d65225d29, '5', 'asd', '03.13.06');INSERT INTO `images` VALUES (6, 0x75706c6f6164732f342e676966, '4', 'some_guy', '03.13.06');[/code]thanks a lot though Quote Link to comment Share on other sites More sharing options...
Barand Posted March 13, 2006 Share Posted March 13, 2006 If imageurl is a string value, why have you defined it as BLOB. Use varchar(255) Quote Link to comment Share on other sites More sharing options...
HoTDaWg Posted March 14, 2006 Author Share Posted March 14, 2006 thank you all so much for your help. The script is working Great! The only problem is I get the following error when using my latest page [b]gallery.php[/b]:[code]Parse error: syntax error, unexpected T_STRING in /home/cortez/public_html/testscripts/photoalbums/gallery.php on line 24[/code]Here is the script:[code]<html><head><title>Homework Administration- Delete Homework</title><META NAME="ROBOTS" CONTENT="NOINDEX, NOFOLLOW"></head><body><?php$conn = mysql_connect("localhost","cortez","165745");//now choose the database to usemysql_select_db("cortez_photoalbums");$id = $_GET['id']; // retrieve id passed via URLif (isset($id)) { $query = "DELETE FROM images WHERE id = '$id' "; $result = mysql_query($query);}$getimages = mysql_query("SELECT * FROM images ORDER BY id DESC");while($r=mysql_fetch_array($getimages)) { extract($r); //remove the $r so its just $variable echo("<hr> Id: $id <br> Image: <image src="images/$imageurl" height="50" width="50"><br> Date: $date<br> <a href='". $_SERVER['PHP_SELF']. "?id=". $id. "'>delete</a>"); }?></body></html>[/code]I would be nowhere today if i had not found out about this site,HotdawgPS: the reason I am adding the uploads/ in my image tag is because take a look at my database when I exported it:[code]-- phpMyAdmin SQL Dump-- version 2.6.4-pl2-- http://www.phpmyadmin.net-- -- Host: localhost-- Generation Time: Mar 14, 2006 at 03:50 PM-- Server version: 4.1.14-- PHP Version: 4.3.11-- -- Database: `cortez_photoalbums`-- -- ---------------------------------------------------------- -- Table structure for table `images`-- CREATE TABLE `images` ( `id` int(15) NOT NULL auto_increment, `imageurl` varchar(255) NOT NULL default '', `titled` varchar(255) NOT NULL default '', `contributedby` varchar(22) NOT NULL default '', `date` varchar(50) NOT NULL default '0000-00-00', PRIMARY KEY (`id`)) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=9;-- -- Dumping data for table `images`-- INSERT INTO `images` VALUES (1, '"uploads/" . $_FILES["file"]["name"]', '4', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (2, '"uploads/"["file"]["name"]', '1', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (3, '"uploads/"["file"]["name"]', '7', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (4, '"uploads/"["name"]', '6', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (5, '("uploads/" . $_FILES["file"]["name"])', '5', 'asd', '03.13.06');INSERT INTO `images` VALUES (6, 'uploads/4.gif', '4', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (7, '2.gif', '2', 'some_guy', '03.13.06');INSERT INTO `images` VALUES (8, '29.gif', 'some image', 'some guy', '03.14.06');[/code] Quote Link to comment Share on other sites More sharing options...
HoTDaWg Posted March 14, 2006 Author Share Posted March 14, 2006 so no one can help me? Quote Link to comment Share on other sites More sharing options...
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