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PHP - Upload Image, And Add Path To Mysql Database
Hello freaks!
Im new to this forum, but im not all that new to PHP and MySQL. Although there's been some years since the last time I used it, so don't go all freaky on me if I dont do this right Let's go on-topic: Im in progress of making an internal web-page for me and my colleagues to make things a bit easier for us. I am making an database of our different projects, and I need some help with the input form - as I need to upload an image to the server, and store the path in the MySQL database. In my input form, I need to store information from text fields, and I need to upload an image to the server and store the path in the database. Before I can even start to code this (although I have coded the input forum without the upload), I need to know what would be the best way to do this. I guess there are several ways.. What would the expert do (That's you right?)? Should I have the information input, and image upload in the same form, or should I make a second form (maybe on a different page) for the upload? Is it necessary with two tables, one for the info and one for the image path, and then tie them together with the imageID, or is it fine to use just one table? Any thoughts would be appreciated! <!-- TechThat --> Similar TutorialsI am in the process of writing a CMS for a friend that is looking to add a classified section to his site. He will be the only one that will ever use it. I got the code to work with one image when he asked me if I could do it so he could post 6 images for each item. I am unable to figure out how to do this. I was told to do this with a while loop. This is the code that I have written so far. <?php //Check to make sure title is filled in. If not redirects to show_add.php if (!$_POST[title]) { header("Location: show_add.php"); exit; } else { //check and see if a session has started session_start(); } //if session has not been properly started redirects back to the administration menu if ($_SESSION[valid] != "yes") { header("Location: admin_menu.php"); exit; } include('includes/connection.php'); //check and see if the type of uploaded file is an image function is_valid_type($file) { $valid_types = array("image/jpg", "image/jpeg", "image/gif", "image/bmp"); if (in_array($file['type'], $valid_types)) return 1; return 0; } //Set Constants $TARGET_PATH = "/home/content/m/i/k/mikedmartiny/html/db_images/"; $title = $_POST['title']; $year = $_POST['year']; $make = $_POST['make']; $model = $_POST['model']; $descript = $_POST['descript']; $image = $_FILES['image']; //Sanitize the inputs $title = mysql_real_escape_string($title); $year = mysql_real_escape_string($year); $make = mysql_real_escape_string($make); $model = mysql_real_escape_string($model); $descript = mysql_real_escape_string($descript); $image['name'] = mysql_real_escape_string($image['name']); //$target_path full string $TARGET_PATH .= $image['name']; //make sure that all fields from form are filled in if ( $title == "" || $year == "" || $make =="" || $model == "" || $descript == "" || $image['name'] == "") { $_SESSION['error'] = "ALL FIELDS ARE REQUIRED!"; header ("Location: show_add.php"); exit; } //check to make sure it has the right file type if (!is_valid_type($image)){ $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header ("Location: show_add.php"); exit; } //check to see if a file with that name exsists if (file_exists($TARGET_PATH)){ $_SESSION['error'] = "A FILE WITH THAT NAME ALL READY EXIST!"; header ("Location: show_add.php"); exit; } //move the image - write path to database while($image <=2) { move_uploaded_file($image['tmp_name'], $TARGET_PATH) } else { // Make sure you chmod the directory to be writeable $_SESSION['error'] = "COULD NOT UPLOAD FILE. CHECK WRITE/REWRITE PERMISSIONS ON THE FILE DIRECTORY!"; header ("Location: show_add.php"); exit; } $sql = "INSERT INTO $table (id, title, year, make, model, descript, image, image_two) VALUES ('', '$_POST[title]', '$_POST[year]', '$_POST[make]', '$_POST[model]', '$_POST[descript]', '" . $image['name'] . "', '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); ?> I only have it set up to work with 2 images right now. I thought it would be easier to get it to work properly. Then I could just add in the rest of the info later. Any help would be greatly appreciated. Hi, I have a form that allows a user to enter details and also to upload a image. The form works great and inserts into the database if i dont include the file input. What i would like happen is: The user enters details and selects a file The file then gets uploaded to ../images and the path is then written to bedroom1 variable The addresses and the bedroom1 variable that now holds the uploaded file path e.g. images/test.jpg is written to the database Form Code: [Select] <form action="admin.php" method="POST"> <table> <tr><td> Address 1: </td><td> <input type="text" name="address1" id="address1" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Address 2: </td><td> <input type="text" name="address2" id="address2" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> County: </td><td> <input type="text" name="county" id="county" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Bedroom 1: </td><td> <input type="file" name="bedroom1" id="bedroom1" input name="uploadedfile"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> <input type="submit" name="submitaddapartment" value="Add" id="registerbutton"> </td></tr> </table </form> And the php code to add it to the database Code: [Select] <?php //get variables $submitaddapartment = $_POST['submitaddapartment']; $address1 = strip_tags($_POST['address1']); $address2 = strip_tags($_POST['address2']); $county = strip_tags($_POST['county']); $bedroom1 = strip_tags($_POST['bedroom1']); if ($submitaddapartment) //if submit button was pressed { if ($address1&&$address2&&$county&&$bedroom1&&) //if fields arn't blank { include 'dbase.php'; //connect to database if ((($_FILES["bedroom1"]["type"] == "image/gif") || ($_FILES["bedroom1"]["type"] == "image/jpg") || ($_FILES["bedroom1"]["type"] == "image/pjpeg")) && ($_FILES["bedroom1"]["size"] < 20000000)) { if ($_FILES["bedroom1"]["error"] > 0) { echo "Return Code: " . $_FILES["bedroom1"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["bedroom1"]["name"] . "<br />"; echo "Type: " . $_FILES["bedroom1"]["type"] . "<br />"; echo "Size: " . ($_FILES["bedroom1"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["bedroom1"]["tmp_name"] . "<br />"; if (file_exists("../images/" . $_FILES["bedroom1"]["name"])) { echo $_FILES["bedroom1"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["bedroom1"]["tmp_name"], "/" . $_FILES["bedroom1"]["name"]); echo "Stored in: " . "../images/" . $_FILES["bedroom1"]["name"]; $bedroom1=".'images/.' $_FILES["bedroom1"]["name"]"; } } } else { echo "Invalid file"; } mysql_query ("INSERT INTO user VALUES(NULL,'$address1','$address2','$county','$bedroom1')"); } else { } } ?> Any help would be greatly appreciated. Thanks Fred At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. I have a PHP script that uploads images to a folder on my server (attachments folder). Currently the folder sits within my webroot and is publicly accessible (I have to use chmod 777 due to permissions issue). So, I created the "attachments" folder outside of my webroot (so that it is not publicly accessible), but I do not know how to set the path in the PHP code to upload it to that "attachments" folder outside of the webroot. As you see in the snippet of PHP code below, the code currently uploads the the "attachments" folder within the www (webroot) directory. How do I make it upload to the "attachments" folder OUTSIDE of the www (webroot) directory? foreach($files[$form] as $file){ $str = $file[1]; if (eval("if($str){return true;}")) { $_values[$file[0]] = $_FILES[$file[0]]["name"]; $dirs = explode("/","attachments//"); $cur_dir ="."; foreach($dirs as $dir){ $cur_dir = $cur_dir."/".$dir; if (!@opendir($cur_dir)) { mkdir($cur_dir, 0777);}} $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; copy($_FILES[$file[0]]["tmp_name"],$_values[$file[0]."_real-name"]); @unlink($_FILES[$file[0]]["tmp_name"]); }else{ $flag=true; if ($_isdisplay) { //$ExtFltr = $file[2]; //$FileSize = $file[4]; if (!eval("if($file[2]){return true;}")){echo $file[3];} if (!eval("if($file[4]){return true;}")){echo $file[5];} $_ErrorList[] = $file[0]; } } } hi i want to store url to images in database for logged in users (where id = $id) and recall the image hopefully using --------------------- <img src="<?php echo row['link']; ?>" /> or similar and need help with the sql update string any ideas please help i been stuck with this for some time and now decided to ask around in this forum for help, please help if you can. hi- i'm trying to browse for a CSV file and then upload it to my mysql database. i found the code below and not working. i don't get any errors. the connection to the database is ok bcz i get the existing results but not the ones from the CSV file added to the db.. any ideas? Code: [Select] <?php ob_start(); require_once('../../connections/congif.php'); mysql_select_db($dbname, $db); $sql_get_project="SELECT * FROM gifts_tbl ORDER BY autoID DESC LIMIT 25"; $get_project = mysql_query($sql_get_project, $db) or die(mysql_error()); $row_get_project = mysql_fetch_assoc($get_project); //database connect info here //check for file upload if(isset($_FILES['csv_file']) && is_uploaded_file($_FILES['csv_file']['tmp_name'])){ //upload directory $upload_dir = "csv_dir/"; //create file name $file_path = $upload_dir . $_FILES['csv_file']['name']; //move uploaded file to upload dir if (!move_uploaded_file($_FILES['csv_file']['tmp_name'], $file_path)) { //error moving upload file echo "Error moving file upload"; } //open the csv file for reading $handle = fopen($file_path, 'r'); //turn off autocommit and delete the product table mysql_query("BEGIN"); while (($data = fgetcsv($handle, 1000, ',')) !== FALSE) { //Access field data in $data array ex. $name = $data[0]; //Use data to insert into db $sql = sprintf("INSERT INTO gifts_tbl (player_id) VALUES ('%s)", mysql_real_escape_string($name) ); mysql_query($sql) or (mysql_query("ROLLBACK") and die(mysql_error() . " - $sql")); } unlink($file_path); } ob_flush(); ?> <!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>Gifts</title> </head> <body> <h1>Testing for CSV upload</h1> <form action="" method="post"> <input type="file" name="csv_file"> <input type="submit" name="csv_submit" value="Upload CSV File"> </form> <h2>Results</h2> <?php do { ?> <ul> <li><?php echo $row_get_project['player_id']; ?></li> </ul> <?php } while ($row_get_project = mysql_fetch_assoc($get_project)); ?> </body> </html> hello friends, while clicking the form all the information goes to database, I have one image upload field, when cliking the submit button, i would like 'image name' to go in database and file to go in /upload folder, i have tried this for hours and gave up, if anyone help me in this, i would be very greatful hi all i need a script that will allow me upload images to 2 dif folders on my server and then add the info to my database along with some other form data. iv been looking all over for code or scripts for days now and have been playing with cut and copyed code but no look again any help i will be greatful for as im a noob to php but al learning quick here is my html form Code: [Select] <html> <body> <form action="add.php" method="post"> Project Name: <input type="text" name="pro_name" /><br> Thumbnail: <input type="file" name="thumbnail" /><br> ////// this image to ../thum Short Details: <input type="text" name="short_details" /><br> Full Details: <input type="text" name="full_details" /><br> Category: <input type="text" name="cat" /><br> Image1: <input type="file" name="image1" /><br>//// and image1,2,3,4 to ../images Image2: <input type="file" name="image2" /><br> Image3: <input type="file" name="image3" /><br> Image4: <input type="file" name="image4" /><br> <input type="submit" /> </form></body></html> here is my code for add.php witch only adds the info to the DB Code: [Select] <?php error_reporting(E_ALL); include ("../includes/db_config.php"); $con = mysql_connect($db_hostname,$db_username,$db_password); @mysql_select_db($db_database) or die( "Unable to select database"); $sql="INSERT INTO $db_table (pro_name, thumbnail, short_details, full_details, cat, image1, image2, image3, image4) VALUES ('$_POST[pro_name]','$_POST[thumbnail]','$_POST[short_details]','$_POST[full_details]','$_POST[cat]','$_POST[image1]','$_POST[image2]','$_POST[image3]','$_POST[image4]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> hi, i have a form that requires image to be uploaded. i know how to store the form values in a database but i don't know how to store the image into the database. i want the user to upload picture and preview the data provided review (including the image) before submitting to the database. thanks Hello I am having problems uploading an image through a HTML form. I want the image to be uploaded to the server and the image name to be written to the mysql database. Below is the code I am using: Code: [Select] <?php if (isset($_POST['add'])){ echo "<br /> add value is true"; $name = $_POST['name']; $description = $_POST['description']; $price = $_POST['price']; $category_id = $_POST['category_name']; $image = $_FILES['image']['name']; //file path of the image upload $filepath = "../images/"; //mew name for the image upload $newimagename = $name; //new width for the image $newwidth = 100; //new height for the image $newheight = 100; include('../includes/image-upload.php'); mysql_query("INSERT INTO item (item_name, item_description, item_price, item_image) VALUES ('$name','$description','$price','$image')"); ?> Here is the image-upload.php file code: Code: [Select] <?php //assigns the file to the image $image =$_FILES["image"]["name"]; $uploadedfile =$_FILES["image"]["tmp_name"]; if ($image) { //retrieves the extension type from image upload $extension = getextension($image); //converts extension to lowercase $extension = strtolower($extension); //create image from uploaded file type if($extension=="jpg" || $extension=="jpeg") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefromjpeg($uploadedfile); }else if($extension=="png") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefrompng($uploadedfile); }else{ $src = imagecreatefromgif($uploadedfile); } //creates a list of the width and height of the image list($width,$height)=getimagesize($uploadedfile); //adds color to the image $tmp = imagecreatetruecolor($newwidth,$newheight); //create image imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height); //set file name $filename = $filepath.$newimagename.".".$extension; $imagename = $newimagename.".".$extension; //uploads new file with name to the chosen directory imagejpeg($tmp,$filename,100); //empty variables imagedestroy($src); imagedestroy($tmp); } ?> Any help would be appreciated, fairly new to all this! Thanks!!! Hi Guys, I am trying to use a script that is available on the net to upload an image into a mysql using BLOB. However it wont upload it keep getting Warning: fread() [function.fread]: Length parameter must be greater than 0 in /home/theacidf/public_html/attendance/upload.php on line 17 Here is the code... <?php // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","theacidf_admin","password")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("theacidf_attendanc"); // Insert any new image into database if ($_REQUEST[completed] == 1) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($image) < 149000) { mysql_query ("insert into pix (title, imgdata) values (\"". $_REQUEST[whatsit]. "\", \"". $image. "\")"); } else { $errmsg = "Too large!"; } } // Find out about latest image $gotten = @mysql_query("select * from pix order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../images/logo_png.png","rb"); $bytes = fread($instr,filesize("../images/logo_png.png")); } // If this is the image request, send out the image if ($_REQUEST[gim] == 1) { header("Content-type: image/jpeg"); print $bytes; exit (); } ?> <html><head> <title>Upload an image to a database</title> <body bgcolor=white><h2>Here's the latest picture</h2> <font color=red><?= $errmsg ?></font> <center><img src=?gim=1><br> <?= $title ?></center> <hr> <h2>Please upload a new picture and title</h2> <form enctype=multipart/form-data method=post> <input type=hidden name=MAX_FILE_SIZE value=150000> <input type=hidden name=completed value=1> Please choose an image to upload: <input type=file name=imagefile><br> Please enter the title of that pictu <input name=whatsit><br> then: <input type=submit></form><br> <hr> </body> </html> Why wont it run properly? Thanks in advance... S
I have a form on which the filepond plugin send the file manually
Look here
if(empty($_POST['image'])) {
echo 'add file!';
}
$myimage = $_POST['image'];
$myimage = str_replace('data:image/png;base64,', '', $myimage);
$myimage = str_replace(' ', '+', $myimage);
$decode = base64_decode($myimage);
$myfile = $_SERVER['DOCUMENT_ROOT'].'/mages/' . uniqid() . '.png';
//now put the file
file_put_contents($myfile, $decode);
Okay, so I found this code on-line that uploads images via a form, inserts them in to MySQL and displays the images on the page. It works great! Problem is, I spent a long time hacking this script to make it work with my site just to realize it did not re-size the image after it uploaded (dumb me). I have found other code that do this but I really do not want to start over at this point and my level of expertise with PHP are a step below what I need to figure this out. I am attaching the "original" code. If anyone knows how to add the necessary lines of code to make every image re-size (reduce) itself. I would very much appreciate it. Thanks <?php $con=mysql_connect("localhost", "root", "rootwdp")or die("cannot connect"); mysql_select_db("student",$con)or die("cannot select DB"); if(isset($_POST['upload'])) { $img=$_FILES["image"]["name"]; foreach($img as $key => $value) { $name=$_FILES["image"]["name"][$key] ; $tname=$_FILES["image"]["tmp_name"][$key]; $size=$_FILES["image"]["size"][$key]; $oext=getExtention($name); $ext=strtolower($oext); $base_name=getBaseName($name); if($ext=="jpg" || $ext=="jpeg" || $ext=="bmp" || $ext=="gif"){ if($size< 1024*1024){ if(file_exists("upload/".$name)){ move_uploaded_file($tname,"upload/".$name); $result = 1; list($width,$height)=getimagesize("upload/".$name); $qry="select id from img where `img_base_name`='$base_name' and `img_ext`='$ext'"; $res=mysql_fetch_array(mysql_query($qry)); $id=$res['id']; $qry="UPDATE img SET `img_base_name`='$base_name' ,`img_ext`='$ext' ,`img_height`='$height' ,`img_width`='$width' ,`size`='$size' ,`img_status`='Y' where id=$id"; mysql_query($qry); echo "Image '$name' updated<br />"; } else{ move_uploaded_file($tname,"upload/".$name); $result = 1; list($width,$height)=getimagesize("upload/".$name); $qry="INSERT INTO `img`(`id` ,`img_base_name` ,`img_ext` ,`img_height` ,`img_width`, `size` ,`img_status`)VALUES (NULL , '$base_name', '$ext', '$height', '$width', '$size', 'Y');"; mysql_query($qry); echo "Image '$name' uploaded<br />"; } } else{ echo "Image size excedded.<br />File size should be less than 1Mb<br />"; } } else{ echo "Invalid file extention '.$oext'<br />"; } } } function getExtention($image_name){ return substr($image_name,strrpos($image_name,'.')+1); } function getBaseName($image_name){ return substr($image_name,0,strrpos($image_name,'.')); } ?> <!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> <script type="text/javascript"> function addItems() { var table1 = document.getElementById('tab1'); var newrow = document.createElement("tr"); var newcol = document.createElement("td"); var input = document.createElement("input"); input.type="file"; input.name="image[]"; newcol.appendChild(input); newrow.appendChild(newcol); table1.appendChild(newrow); } function remItems() { var table1 = document.getElementById('tab1'); var lastRow = table1.rows.length; if(lastRow>=2) table1.deleteRow(lastRow-1); } </script> <style type="text/css"> <a class="tooltip" and them url href="/http.www.anyurl.com" </a> a.tooltip:hover span{display:inline; position:absolute; border:2px solid #cccccc; background:#efefef; color:#333399;} a.tooltip span {display:none; padding:2px 3px; margin-left:8px; width:150px;} </style> </head> <body> <form method="post" action="" enctype="multipart/form-data"> <table align="center" border="0" id="tab1"> <tr> <td width="218" align="center"><input type="file" name="image[]" /></td> <td width="54" align="center"><img src="Button-Add-icon.png" alt="Add" style="cursor:pointer" onclick="addItems()" /></td> <td><img src="Button-Delete-icon.png" alt="Remove" style="cursor:pointer" onclick="remItems()" /></td> </tr> </table> <table align="center" border="0" id="tab2"> <tr><td align="center"><input type="submit" value="Upload" name="upload" /></td></tr> </table> </form> <table border="0" style="border:solid 1px #333; width:800px" align="center"><tr><td align="center"> <iframe style="display:none" name="if1" id="if1"></iframe> <? $qry="select * from img where img_status='Y' order by id"; $res=mysql_query($qry); $i=0; if(mysql_num_rows($res)){ ?> <div align="center"><ul style="width:650px; border: 0px"> <? while($fetch=mysql_fetch_array($res)){ $hratio=120/$fetch['img_height']; $wratio=120/$fetch['img_width']; $ratio=($hratio < $wratio) ? $hratio : $wratio; $hth=$fetch['img_height']*$ratio; $wth=$fetch['img_width']*$ratio; ?> <li style="width:120px; height:180px; border:0px solid #333;float:left;list-style:none outside none; padding-right:5px;"><img src="upload/<? echo $fetch['img_base_name'].'.'.$fetch['img_ext']; ?>" width="<? echo $wth; ?>" height="<? echo $hth; ?>" title="<? echo "image : ".$fetch['img_base_name'].".".$fetch['img_ext']; ?>" /><br /> <? if($i==0) $fp=fopen("fileInfo.txt",'w'); else $fp=fopen("fileInfo.txt",'a'); fwrite($fp,"Image : ".++$i ."\r\n"); fwrite($fp,"Name : ".$fetch['img_base_name'].".".$fetch['img_ext']."\r\n"); fwrite($fp,"width X height : ".$fetch['img_width']." X ".$fetch['img_height']."\r\n"); fwrite($fp,"Size : ".round($fetch['size']/1024,1)."Kb\r\n"); fwrite($fp,"____________________________________\r\n"); fclose($fp); echo $fetch['img_base_name'].".".$fetch['img_ext'].'<br />'; echo $fetch['img_width'].' X '; echo $fetch['img_height'].'<br />'; echo round($fetch['size']/1024,1) .'Kb'; ?> </li> <? }?> </ul> </div><? }?> </td></tr></table> </body> </html> what's wrong with this ? i can't upload Hi! So I'm working for someone, and they want me to fix this error in a PHP file.. Here is the code: <?php
include_once('config.php');
$online = mysql_query("SELECT * FROM bots WHERE status LIKE 'Online'");
$offline = mysql_query("SELECT * FROM bots WHERE status LIKE 'Offline'");
$dead = mysql_query("SELECT * FROM bots WHERE status LIKE 'Dead'");
$admintrue = mysql_query("SELECT * FROM bots WHERE admin LIKE 'True'");
$adminfalse = mysql_query("SELECT * FROM bots WHERE admin LIKE 'False'");
$windows8 = mysql_query("SELECT * FROM bots WHERE so LIKE '%8%'");
$windows7 = mysql_query("SELECT * FROM bots WHERE so LIKE '%7%'");
$windowsvista = mysql_query("SELECT * FROM bots WHERE so LIKE '%vista%'");
$windowsxp = mysql_query("SELECT * FROM bots WHERE so LIKE '%xp%'");
$unknown = mysql_query("SELECT * FROM bots WHERE so LIKE 'Unknown'");
$totalbots = mysql_num_rows(mysql_query("SELECT * FROM bots"));
$onlinecount = 0;
$offlinecount = 0;
$deadcount = 0;
$admintruecount = 0;
$adminfalsecount = 0;
$windows8count = 0;
$windows7count = 0;
$windowsvistacount = 0;
$windowsxpcount = 0;
$unknowncount = 0;
while($row = mysql_fetch_array($online)){
$onlinecount++;
}
while($row = mysql_fetch_array($offline)){
$offlinecount++;
}
while($row = mysql_fetch_array($dead)){
$deadcount++;
}
while($row = mysql_fetch_array($admintrue)){
$admintruecount++;
}
while($row = mysql_fetch_array($adminfalse)){
$adminfalsecount++;
}
while($row = mysql_fetch_array($windows8)){
$windows8count++;
}
while($row = mysql_fetch_array($windows7)){
$windows7count++;
}
while($row = mysql_fetch_array($windowsvista)){
$windowsvistacount++;
}
while($row = mysql_fetch_array($windowsxp)){
$windowsxpcount++;
}
while($row = mysql_fetch_array($unknown)){
$unknowncount++;
}
$statustotal = $onlinecount + $offlinecount + $deadcount;
$admintotal = $admintruecount + $adminfalsecount;
$sototal = $windows7count + $windowsvistacount + $windowsxpcount + $unknowncount;
?>
Can anyone tell me the error here, can how to fix it? Is there a way to get the source path for an uploaded file? It seems that's a function of the browser and the two I've tried FF and Safari don't return it in $_FILES['file']['name']. This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=328623.0 Hi Guys, I need help for in storing data from PHP from array in mysql. I'm very new to PHP/Mysql and have started learing it just few weeks back. I'm tryting to build a website for myself. I'm having tough time trying to insert the data from Form in mysql. I have a form where user update the company name and insert there product name with there image. I want to rename the image with the corresponding text field value and insert the upload path in the table. my FORM Code: [Select] <!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> <style type="text/css"> body,td,th { color: #000; font-family: Tahoma, Geneva, sans-serif; font-size: 80%; } body { background-color: #9CF; } table { background: #CCF; } th { font: bold normal 16px/normal "Times New Roman", Times, serif; text-transform: capitalize; color: #00F; background: #FFC; } td { font: bold 14px Georgia, "Times New Roman", Times, serif; text-transform: capitalize; color: #3A00FF; background: #FF9; } </style> </head> <body> <form action="upload-file1.php" method="post" enctype="multipart/form-data" name="form"> <strong>Company Name:</strong> <input name="product" type="text" /> <br /> <table> <tr bgcolor="#FF9900"> <th colspan="3" bgcolor="#CCFFFF">filter </th> <th colspan="5" bgcolor="#CCFFFF">heater</th> </tr> <tr><td>filter1</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater1: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter2</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater2: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter3</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater3: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter4</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater4: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter5</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater5: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> </table> <br /> <br /> <input name="submit" type="submit" value="submit" /> </form> </body> </html> my PHP Code Code: [Select] <?php require("connect.php"); ?> <?php $product = $_POST['product']; echo $product; if(isset($_POST["submit"])){ $sql = "INSERT INTO company (product) VALUES ('$product')"; $query = mysql_query($sql) OR DIE(mysql_error()); $comp_id = mysql_insert_id(); echo $sql; } ?> <?php function filter() { if(isset($_POST['submit'])) { foreach($_POST['filter'] as $key => $val) { if(trim($val) != '') $filter[] = $val; } $total_records_filter = count($filter); for($i = 0; $i < $total_records_filter; $i++) { $prod_val = $filter[$i]; $sql_prod = "INSERT INTO filter(filter_id, filter) VALUES('', '$prod_val')"; echo $sql_prod.'<br>'; mysql_query($sql_prod) OR die(mysql_error()); } } } filter(); function heater() { if(isset($_POST['submit'])) { foreach($_POST['heater'] as $key => $val) { if(trim($val) != '') $heater[] = $val; } $total_records_heater = count($heater); for($i = 0; $i < $total_records_heater; $i++) { $dir_val = $heater[$i]; $sql_dir = "INSERT INTO heater(heater_id, heater) VALUES('', '$dir_val')"; echo $sql_dir.'<br>'; mysql_query($sql_dir) OR die(mysql_error()); } } } heater(); ?> <?php mysql_close($link)?> My Table structure company comp_id int(5) PK product varchar(50) No filter filter_id int(5) PK filter varchar(25) No filter_path varchar(100) No heater heater_id int(5) PK heater varchar(25) No heater_path varchar(100) No company_filter id int(5) PK comp_id int(5) PK FK (From company table) filter_id int(5) PK FK (From filter table) company_heater id int(5) PK comp_id int(5) PK FK (From company table) heater_id int(5) PK FK (From heater table) Regards BW so what i have going on is that i need help writing a basic script to upload original and copy with resize for thumbnail. then also need to rename both image files with content in the form. My server supports GD Library and that imagemagik or whatever it is lol. upload dirs orig: ../media/photos/ thumb: ../media/photos/thumb/ mySQL DB: name: m_photos fields: id(INT) m_cat(varchar) m_sub_cat(varchar) pic(varchar) description(varchar) p_group(varchar) I have the form written up for how it should look like: Code: [Select] <form action="upload.php" method="post" enctype="multipart/form-data"> Upload an image for processing<br /> <input type="file" name="Image"><br /> <select name="sub_group"> <option value="Photo Shoot">Photo Shoot</option> <option value="Live Performances">Live Performances</option> <option value="Randoms">Randoms</option> <option value="Fan Photos">Fan Photos</option> </select><br /> Location: <input type="text" name="p_group" /><br /> Date of Pic: <input type="text" name="date" /><br /> Description: <input type="text" name="description" /><br /> <input type="submit" value="Upload"> </form> so what needs to happen is for the file upload name. it needs to grab a count from "p_group" database to give it a starting number in a "00" pattern and then the date of pic needs to go in there next then the p_group name. so when the files gets uploaded it would look something like this after upload. "03 - Oct 21, 2011 - The Mex.jpg" NOTE: all pics must be converted to ".jpg" extension. both the orig and thumb will use that same file name cuz they just go into different dirs re-sizing for the thumbnail should be done by aspect ratio and needs to either be 300px width or 400px height. so if anyone would like to help me out please do. im not the greatest at writing in php yet Hi guys, I have this code that I created. It is suppost to get a file from a form and upload it to a directory that is outside the root folder of my webserver. then the image name is added to the preset location of the directory that contains all of the uploaded images and is placed into a mysql database so I can call it later. the problem I have is that when some one else signs up through this form and the file they upload has the same file name it replaces the one that is already in the images directory that has all the other pictures. I need to add something to this code that adds $username to the end of the original file name and then uploads it to the directory with the new name. Since every username is unique it will make every username unique and will eliminate my problem. I am not sure how to do this. Please, anyone that can help. Thanks. This is my code. // get file attributes!!!!! $photoname = $_FILES['photo']['name']; $tmp_name = $_FILES['photo']['tmp_name']; if ($photoname) { // start upload process $location = "../userpictures/$photoname"; move_uploaded_file($tmp_name,$location); } else { $location = "../userpictures/nophoto.png"; } $queryreg = mysql_query(" INSERT INTO users VALUES ('','$fullname','$username','$password','$date','$location','$email','$phone') "); If there is a better way to upload images to a mysql database and outputting it later then please let me know as well. Thanks. |
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