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PHP - Overwrite Sql Entry
Hello there,
I have some code here which sends a number of variables from flash to SQL... I would simply like to add the functionality to overwrite records which have the same 'name' or 'pseudo'... can anyone help me please ? Thanks in advance Martin <?php $pseudo=$_POST['var1']; $score=$_POST['var2']; $table = $_POST['tab']; $dategame = $_POST['tempjoueur']; //$micro = microtime(); //$dategame = time()."".substr($micro, 2, 6); $_COOKIE['User'] = $_SERVER['REMOTE_ADDR']; $envoie = InsertDatas($table, "name, score, dategame", "'".$pseudo."','".$score."','".$dategame."'"); if ($envoie) { print_r("OK, $pseudo, $score, $dategame,$ipclient"); } else { echo "BAD, $pseudo, $score, $dategame,$ipclient"; } ?> Similar TutorialsAlright, wasn't quite sure how to summarize this in the title, but I want to: Check if a user status is "active" or not based on the UserName input. I have a table witch holds: Code: [Select] VarChar Username Var CharPassWord int Active Ted TedsPW 1 something like the above(assuming it formatted correctly. In my php script I will want to input a variable for Username to check for: inputUN in this example would be "Ted". $UserNameToCheck = $_GET['inputUN']; Then I want to check for that UserName in the database, if it exists, I want pull the value for the "Active" field for just that UserName and echo it. Thanks for any help. I am using php to upload a file to my server, and at the same time inserting the files name and url into my mysql database.
$sql = "UPDATE uploads SET name = '$name', url='$target_path'";
$statement = $dbh->prepare($sql);
$statement->execute();
This is working, however, when I upload a new file, rather than making a new entry in my database, it just overwrites the first one.I'm quite new at mysql so was wondering how I would make it add new entrys instead of overwriting the current one? hi all, i have written a seating plan page for my website which allows users to click on a seat and that seat is then allocated to there username, however theres nothing stopping a user changing the URL to a seat thats been taken and "taking" that seat from someone else. heres my code for my reserve_seat.php <?PHP session_start(); /* get user id */ $user_id = $_SESSION['username']; /* get the seat number */ $seat_id = $_GET['seat']; /* connect to data base */ require ('./secure/connect.php'); /* create query */ $query = "UPDATE seats_table set taken = FALSE, user_id = '0' WHERE user_id = '$user_id'"; $query2 = "UPDATE seats_table set taken = TRUE, user_id = '$user_id' WHERE id = '$seat_id'"; /* $query3 = "UPDATE users set signed_up = '3' WHERE username = '$user_id'"; */ /* execute the query */ $result = mysql_query($query); $result2 = mysql_query($query2); /* $result3 = mysql_query($query3); */ /* advise user their seat has been reserved */ include 'seating.php'; ?> so if a user reserves seat 28 the url is http://localhost/reserve_seat.php?seat=26, but theres nothing stopping another user typing this url and "stealing" this seat i need some kind of IF statement before the sql queries that checks to see if the seats "taken" column is 1 or 0 bearing in mind there is 49 seats but im lost at how to write it. any help would be great Lee Hi, I've got a FORM Setup. The form has a current image and an option to add a new image (In place of that image) The form fields in question are : Code: [Select] <input type="hidden" name="image_url" value="<?php echo $row_select_propertyimages['image_url']; ?>" /> <input type="file" name="new_image_url" class="inputfile" /> I've setup an IF Statement, to check that if new_image_url ISNT set, then to keep the current image. Otherwise, overwrite it with new_image_url. When I dont choose to upload a new file, The code works correctly. The code is as follows. Code: [Select] if(empty($_POST['new_image_url'])) { $image_url = $row_select_propertyimages['image_url']; } else { $image_url = $_FILES['new_image_url']; } The data then gets input to a Database. But at the moment, It is not overwriting what is currently there. Can anyone help me on this? My full code is : Code: [Select] <?php if (!function_exists("GetSQLValueString")) { function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") { if (PHP_VERSION < 6) { $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue; } $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue); switch ($theType) { case "text": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "long": case "int": $theValue = ($theValue != "") ? intval($theValue) : "NULL"; break; case "double": $theValue = ($theValue != "") ? doubleval($theValue) : "NULL"; break; case "date": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "defined": $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue; break; } return $theValue; } } $colname_select_propertyimages = "-1"; if (isset($_GET['image_id'])) { $colname_select_propertyimages = $_GET['image_id']; } mysql_select_db($database_db_connect, $db_connect); $query_select_propertyimages = sprintf("SELECT * FROM image WHERE image_id = %s", GetSQLValueString($colname_select_propertyimages, "int")); $select_propertyimages = mysql_query($query_select_propertyimages, $db_connect) or die(mysql_error()); $row_select_propertyimages = mysql_fetch_assoc($select_propertyimages); $totalRows_select_propertyimages = mysql_num_rows($select_propertyimages); mysql_select_db($database_db_connect, $db_connect); $query_select_property = "SELECT property.property_id, property.property_name FROM property "; $select_property = mysql_query($query_select_property, $db_connect) or die(mysql_error()); $row_select_property = mysql_fetch_assoc($select_property); $totalRows_select_property = mysql_num_rows($select_property); $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "addproperty")) { if(empty($_POST['new_image_url'])) { $image_url = $row_select_propertyimages['image_url']; } else { $image_url = $_FILES['new_image_url']; } echo $image_url; $updateSQL = sprintf("UPDATE image SET image_url=%s, image_desc=%s, image_disphome=%s, image_property_id=%s WHERE image_id=%s", GetSQLValueString($image_url, "text"), GetSQLValueString($_POST['image_desc'], "text"), GetSQLValueString(isset($_POST['image_disphome']) ? "true" : "", "defined","'Y'","'N'"), GetSQLValueString($_POST['image_property_id'], "int"), GetSQLValueString($_POST['image_id'], "int")); mysql_select_db($database_db_connect, $db_connect); $Result1 = mysql_query($updateSQL, $db_connect) or die(mysql_error()); // $updateGoTo = "success_editproperty.php"; // if (isset($_SERVER['QUERY_STRING'])) { // $updateGoTo .= (strpos($updateGoTo, '?')) ? "&" : "?"; // $updateGoTo .= $_SERVER['QUERY_STRING']; // } // header(sprintf("Location: %s", $updateGoTo)); } ?> Many Thanks in advance. Hi, I have a page where a user can change his password what i want to acheive is a way of checking the database if the text the user has entered in the textboxes already exist in the db, and if it does exist change a certain part. for example the user goes to the address, types in the email, user name and password twice. if the username and email match in the db i would like the password to write over the old password that was in the db. i keep confusing myself when i think i know what im doing but i keep stumbling. i know this sounds a bit confusing so please ask if you need more understanding. so far ive got: Code: [Select] <?php $n=$_POST['uname']; $e=$_POST['email']; if( $_POST['submitted'] == 'yes' ) { if( $_POST['pass_1'] != $_POST['pass_2'] ) { // fields don't match, so do something to indicate the error . . . echo '<p>Passwords Do Not Match</p>'; } // connect to the db include('config.php'); $query="select * from user where uname='$n' and email='$e'"; $result=mysql_query($query); } ?> <form action="" method="post"> <input type="text" name="uname" id="uname" size="30"> <input type="text" name="email" id="email" size="30"> <input type="password" name="pass_1" /> <input type="password" name="pass_2" /> <input type="hidden" name="submitted" value ="yes" /> <input type="submit" name="submit" value="Change Password" /> </form> just by looking at my code again i think its not right at all, the first php bit is only checking the password textboxes. i think i need to remove my email and username out of that form and put them in another. am i right? sorry im not that good at php or mysql thanks in advance and sorry for such a long read Hello. i am building a facility for members of my website to upload profile pictures. It works, but what I want to do is have the below script check and overwrite the previous file if they want to update it. As you will see it enforces the image to be named after the username. Code: [Select] function findexts ($filename) { $filename = strtolower($filename) ; $exts = split("[/\\.]", $filename) ; $n = count($exts)-1; $exts = $exts[$n]; return $exts; } $ext = findexts ($_FILES['uploaded']['name']); //$ran = rand () ; $ran2 = $eo_user_name."."; $target = "../images/eoprofile/"; //This assigns the subdirectory you want to save into... make sure it exists! $target = $target . $ran2.$ext; //This combines the directory, the random file name, and the extension if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) { echo "The file has been uploaded as ".$ran2.$ext; } else { echo "Sorry, there was a problem uploading your file."; } Help appreciated. Thanks... Hi every body.I'm new to php and this site.this code I've written is for uploading file and I want it to don't overwrite files with the same name.I thought I told it to with file exists.but doesn't work.can't figure out where the problem is.Sorry for my English and Thanks!
<?php function upload($file,$dest){ $a=explode('.', $file['name']); $filename=$a[0]; $ext=$a[1]; $add=microtime(); if (file_exists($file['name'])) { $filename=$add.$filename.$ext; } if(move_uploaded_file($file['tmp_name'],$dest.$file['name'])){ echo 'File Uploaded'; } print_r($file['name']); } /* Array ( [picture] => Array ( [name] => Chrysanthemum.jpg [type] => image/jpeg [tmp_name] => C:\Users\NOVINP~1\AppData\Local\Temp\php\upload\phpFA89.tmp [error] => 0 [size] => 879394 ) */ ?> <html> <head> <title>File Upload</title> </head> <body> <?php if($_FILES['picture']['name']){ upload($_FILES['picture'],'upload/'); } ?> <form action="" method="post" enctype="multipart/form-data"> <table width="500" align="center"> <tr> <td><input type="file" name="picture"></td> </tr> <tr> <td><input type="submit" value="Upload" name="submit"></td> </tr> <tr> <td><input type="hidden" name="form" value="1"></td> </tr> </table> </form> </body> </html> hey guys i have this code FILE NAME "index.php" Code: [Select] <?php // make a connection to the database mysql_connect ("localhost", "root", "vertrigo") or die ('Error: I Failed To Connect To The Database ' . mysql_error()); mysql_select_db ("test"); // Get Data $query = mysql_query("SELECT * FROM TestTable"); // display the data and loop while ($row = mysql_fetch_array($query)) { echo "<br /> ID: ".$row['ID']."<br /> First Name: ".$row['FName']."<br /> Last Name: ".$row['LName']."<br /> Contact Number: ".$row['CNumber']."<br />";} ?> <form method="post" action="update.php"> <table border="1" align="center"> <tr> <td align="right" width="220">ID: </td> <td align="left" width="220"> <input type="text" name="ID" size="30" /></td> </tr> <tr> <td align="right" width="220">First Name: </td> <td align="left" width="220"> <input type="text" name="FName" size="30" /></td> </tr> <tr> <td align="right" width="220">Last Name: </td> <td align="left" width="220"> <input type="text" name="LName" size="30" /></td> </tr> <tr> <td align="right" width="220">Contact Number: </td> <td align="left" width="220"> <input type="text" name="CNumber" size="30" /></td> </tr> <tr> <td align="right" width="220"><input type="reset" value="Reset" /> </td> <td align="left" width="220"> <input type="submit" value="Update Database" /></td> </tr> </table> </form> and i also have this code FILE NAME "update.php" Code: [Select] <?php $ID = $_POST['ID']; $FName = $_POST['FName']; $LName = $_POST['LName']; $CNumber = $_POST['CNumber']; mysql_connect ("localhost", "root", "vertrigo") or die ('Error: I Failed To Connect To The Database ' . mysql_error()); mysql_select_db ("test"); $query="INSERT INTO testtable (ID, FName, LName, CNumber)VALUES ('".$ID."','".$FName."', '".$LName."', '".$CNumber."')"; mysql_query($query) or die ('Error Updating Database'); echo "Database Updated Sucsessfully With: ".$ID." ".$FName." ".$LName." ".$CNumber ; ?> ok so the script is working like a charm its sending the data to the database as i want it to. the problem i have is that i want to be able to update the info that is already on the database lets say i want to change a phone/contact number i have typed the ID number into the ID text field and the same first and last name into there correct boxes and then typed in the new phone number i then click submit and i get the error ""Error Updating Database"" i have looked all over the forum and net to see what i have done wrong to not allow this code to update can anyone help me out here please im quite new to the php language and could really do with some pointers thanks Steve This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=348130.0 I have a code where i can edit or delete certain details from the database. Right now, if the user clicks on the edit button it takes him edit page where he can edit the details. But, I am not able to Incorporate a Delete button such that, when the user clicks on a delete button, it should ask for a confirmation box. If the user clicks YES, then do the following: Code: [Select] DELETE from emp WHERE emp_id='$emp_id'; When there are multiple entries and I click on delete it deletes everything from the database. how can i make it to delete only the entry that is besides the delete button? Code: [Select] if(mysql_num_rows($emp_query) > 0){ echo "<table border='1'>"; echo "<th>Employee Id </th>"; echo "<th>Employee Name </th>"; while($get_emp = mysql_fetch_assoc($emp_query)){ $emp_id = $get_emp['emp_id']; $emp_name = $get_emp['first_name']." ".$get_emp['last_name']; echo "<tr>"; echo "<td width='100'>"; echo $emp_id; echo "</td>"; echo "<td width='400'>"; echo $emp_name; $edit_path = 'edit_employee.php?id='.$emp_id; ?> <INPUT TYPE="button" style="display:inline;" value="VIEW / EDIT" onClick="location.href='<?php echo $edit_path; ?>'"> <form style='margin: 0; padding: 0; display:inline;' method="post" action="<?php echo $_SERVER['PHP_SELF'];?>" onSubmit="return confirm('Are you sure this is correct?');"> <input style='display:inline;' name="delbutton" type="submit" value="DELETE"> <?php if(isset($_POST['delbutton'])){ $del_emp = mysql_query("DELETE from employee WHERE emp_id = '$emp_id'") or die(mysql_error()); //header('Location:view_employee.php'); } echo '</form>'; echo "</td>"; echo "</tr>"; } } Hey folks, Sorry for being a pain in the ass. I am trying to submit data to my database via a form and when I click Submit, I get: Duplicate entry '' for key 1 I understand that it means I have a duplicate entry with the ID of 1 or something like that. I can't find where the issue is. Here is the form: <form actin="" id="settings" name="settings"> <table class="listing form" cellpadding="0" cellspacing="0"> <tr> <th class="full" colspan="2"><?php echo $lang_settings; ?></th> </tr> <tr> <th colspan="2"><?php echo $lang_settings_description; ?></th> </tr> <tr> <td><?php echo $lang_sitename; ?>: </td> <td><input type="text" name="sitename" value="<?php echo $site_name; ?>" width="172" /> <em>Site name for logo</em></td> </tr> <tr> <td><?php echo $lang_email; ?>: </td> <td><input type="text" name="email" value="<?php echo $site_email; ?>" width="172" /> <em>Your email address</em></td> </tr> <tr> <td><?php echo $lang_yourname; ?>: </td> <td><input type="text" name="name" value="<?php echo $your_name; ?>" width="172" /> <em>Your own name</em></td> </tr> <tr> <td><?php echo $lang_meta_description; ?>: </td> <td><input type="text" name="meta-description" value="<?php echo $description; ?>" width="172" /> <em>SEO</em></td> </tr> <tr> <td><?php echo $lang_keywords; ?>: </td> <td><input type="text" name="meta-keywords" value="<?php echo $keywords; ?>" width="172" /> <em>Separate with Commas</em></td> </tr> <tr> <td><input type="submit" class="button" name="submit" value="<?php echo $lang_button_savesettings; ?>"></td> </tr> </table> </form> Here is the Insert code: $insert = "INSERT INTO settings (site_name, description, keywords, email, name) VALUES ('$sitename', '$meta_description', '$meta_keywords', '$site_email', '$your_name')"; mysql_query($insert) or die(mysql_error()); Can anyone please tell me where I am going wrong here? Much appreciated. Well this sounds weird, but it happens. I have an itemshop script, and I wanna code a system that automatically delete the database entry once the variable item amount falls to 0. However, this doesnt happen when I check phpmyadmin, and the item remains in the database even after the amount becomes 0. In fact, the amount can become negative at times, which annoys me. The code of deleting sql entry looks like this: if($continue == "yes"){ $query = "SELECT * FROM ".$prefix."user_inventory WHERE item_owner = '$loggedinname' AND item_name = '$item_name'"; $result = mysql_query($query); $num = mysql_numrows($result); $item_data = mysql_fetch_array($result); $item_amount = $item_data['item_amount']; $newquantity = $item_amount - $quantity; $query = "UPDATE ".$prefix."user_inventory SET item_amount = '$newquantity' WHERE item_owner = '$loggedinname' AND item_name = '$item_name'"; $result = mysql_query($query); if($newquantity == "0"){ $query = "DELETE FROM ".$prefix."user_inventory WHERE item_owner = '$loggedinname' AND item_name = '$item_name'"; $result = mysql_query($query); } What part of the codes should I edit to fix this issue? Thanks. i'm trying to get the username of the last person to enter something into a database table,and display it an html table. the problem is, the query is failing, and i get the error "Table 'forum.posts' doesn't exist". if in the error message "forum" means the name of the database, and "posts" the name of the table, there is something wrong, as they both exist. the code i'm using is below $list = "SELECT * FROM section_main ORDER BY section_title"; $result = mysql_query($list) or die ("Query failed"); $numofrows = mysql_num_rows($result); for($j = 0; $j < $numofrows; $j++) { $row = mysql_fetch_array($result); echo "<tr><th>". $row['section_title']."</th></tr>"; $query2 = "SELECT * FROM section_sub WHERE section_id = '".$row['section_id']."' ORDER BY section_sub_title"; $result2 = mysql_query($query2) or die("Select Error :" . mysql_error()); $numofrows2 = mysql_num_rows($result2); for($i = 0; $i<$numofrows2; $i++){ $row2 = mysql_fetch_array($result2); echo "<tr><td><a href='display_forum.php?id={$row2[section_sub_id]}'>".stripslashes($row2[section_sub_title])."</a></td><td>"; $new = mysql_query("Select * FROM posts WHERE section_sub_id = '".$row2['section_sub_id'] . " ' ") or die ("Select Error :" . mysql_error()); //this is where the error message is from $lastpost = mysql_fetch_assoc($new); echo $lastpost['username']; echo"</td></tr>"; } } i've checked that the names of the tables and fields in the queries are correct, but i cant think of anything else that might be wrong. any help would be great. My query is not finding the last recieptnum entry, it is finding the number 9 everytime for some odd reason. Im trying to incrementally increase this each time a reciept is created. $getreceiptnum = mysql_query("SELECT receiptnum FROM accounting WHERE agency = '$agency' ORDER BY receiptnum DESC LIMIT 1") or die(mysql_error()); $recieptarray = mysql_fetch_array($getreceiptnum); $recieptnum = $recieptarray['receiptnum']; echo $recieptnum; $insertCount=0; foreach($results[1] as $curName) { if($insert){$insertCount++;} echo <<< END $curName<BR> END; } Right now the results would show up as... Bill Fred Jessica James John How do you make them show up like... 1 Bill 2 Fred 3 Jessica 4 James 5 John For some reason this only allows one SQL to be added... // SQL Connection $username="monstert_admin"; $password="admin"; $database="monstert_admin"; $connection = mysql_connect("localhost", $username, $password) or die("Connection Failure to Database"); // Select Database mysql_select_db($database, $connection) or die ($database . "No Database" . $username); //Select everything from the the table $MyQuery = "SELECT * FROM photos"; $retrieve = mysql_query($MyQuery) or die(mysql_error()); if(mysql_num_rows($retrieve) != 0): $row = mysql_fetch_assoc($retrieve); else: echo ''; endif; if(isset($_POST['Submit']) && !$errors) { $url = $newname; include('img.php'); $image = new SimpleImage(); $image->load($url); $image->resize(500,315); $image->save($newname); mysql_query("INSERT INTO photos (url) VALUES ('$url')"); echo "File Uploaded Successfully as <i> "; echo $newname; echo "</i>"; } What would the issue be? I only have two columns - ID and url Thanks in advance! Hi.. So im currently working on a script.. My script generates a "oid" based on timestamp. Ive made the "oid" field unique in my db, so if i do a quick refresh i get the message: Duplicate entry '1283195988' for key 'oid' Is there some way i can check if its a dublicate, and if it is + it with 1 or something? Hey guys! I know, I know this problem is EVERYWHERE but i just dont understand! I have a solid knowlage of php but my SQL skills are low, so i dont know too much about Keys and stuff. But my error is: Duplicate entry '' for key 2. The thing that im working on at this section is logging in with facebook. The code that presents my error is: $sql = "SELECT * FROM users WHERE uid=".$uid; $fbid = mysql_query($sql); $num_rows = mysql_num_rows($fbid); if(mysql_num_rows($fbid) < 1) { echo "You are not logged in. "; mysql_query("INSERT INTO `users` (`uid`) VALUES ('".$uid."')") or die(mysql_error()); } else { mysql_query("UPDATE users SET logged = '1' WHERE uid=".$uid); //mysql_query("UPDATE users SET full_name = $me WHERE uid=".$uid); echo "Your Logged in "; echo $me['name']; ?> Continue to <a href="removed :)"> My Settings </a>. <? } Any help is welcome Please i ran into this problem help me : Code: [Select] <p>Posted by <a href=""> <?php if(isset($_GET['id'])) { $tpid=$_GET['id']; } else { $tpid=$_POST['id']; } include"header.php"; $sql="SELECT*FROM topics WHERE topicsid='$tpid'"; $result=mysql_query($sql) or die(mysql_error()); while($row=mysql_fetch_array($result)) { echo"{$row['topics_by']}"; echo"</a></p>"; echo"<p>"; echo"{$row['topics_subject']}"; echo" <p align='left'><img src='speakout_1.png'/>"; echo"{$row['topics_content']}"; echo"<p class='post-footer align-right'> <a href='' class='comments'>"; session_start(); if(isset($_SESSION['views'])) $_SESSION['views']=$_SESSION['views']+1; else $_SESSION['views']=1; echo"Comments:".$_SESSION['views']; echo"</a>"; echo"<span class='date'>"; echo"{$row['topics_date']}"; } ?></span> </p> </div> <h3>Comments:</h3> <table> <tr class="row-a"> <td class="first"></td> <td><?php include"header.php"; $sql="SELECT post_content,post_by FROM post WHERE topicsID='$tpid'"; $result=mysql_query($sql)or die(mysql_error()); while($row=mysql_fetch_array($result)) { echo"<strong>{$row['post_by']}</strong>: {$row['post_content']}"."</br>"; } ?></td> </tr> </table> <?php include"header.php"; if(isset($_POST['submit'])) { $comment=mysql_real_escape_string(trim($_POST['comment'])); $name=mysql_real_escape_string(trim($_POST['name'])); $hidden=$_POST['id']; if($comment!=='' && $name!=='') { $ins="INSERT INTO post(topicsID,post_content,post_by)VALUES('$hidden','$comment','$name')"; mysql_query($ins) or die(mysql_error()); } else { echo"you cannot post an empty field"; } } ?> <h3>Post your comments here</h3> <form action=''method='post'> <textarea name="comment" id="content" style="width:400px;height:50px;background-color:#D0F18F;color:#000000;font:15px/20px cursive;scrollbar-base-color:#638E0D;"></textarea> <br /> Name:<input type="text"name="name"/> <input class="button" type="submit"name="submit"value="submit" /> <input type="hidden"name="id"value='<?php echo "$tpid"; ?>'/> </p> </form> <br /> </div> Dear All, I am developing a website www.computationalphotography.in on which I need to register new users. I have designed an HTML form for this purpose, but will need help with writing the code for checking and validating the data in the respective fields. The code for my web form is below and the address of the website is www.computationalphotography.in/registration.php: Code: [Select] <html> <head> <title> Photography Experiment Registration </title> </head> <body> <h1> Questionnaire </h1> <h3> We would like to know about your background and interests in photography. For each multiple-choice question, please circle all answers that apply.</h3> <form method="post" action="registration.php"> Name <input name="name" type="text"> <br> <br> <input name="email" type="text"> <br> <br> 1.) Roughly How many pictures do you take each year? <br> <br> <input name="radios" type="radio" value="a"> a) a few: <10 in a year <br> <input name="radios" type="radio" value="b"> b) many: probably 100's a photos a year <br> <input name="radios" type="radio" value="c"> c) a lot: probably 1000's of photos a year <br> <br> 2.) Do you share your pictures with anyone? <br> <input name="radios" type="radio" value="a"> a) No <br> <input name="radios" type="radio" value="b"> b) Yes, but only with friends or family <br> <input name="radios" type="radio" value="c"> c) Yes, I share with friends or family and also post on public sites such as Flickr, Picasaweb, or Snapfish for anyone to see. <br> <input name="radios" type="radio" value="d"> d) Yes, I share frequently and I have also published my photos in books or magazines, or exhibitde my photos in shows. <br> <br> 3.) How much training in photography have you had? <br> <input name="radios" type="radio" value="a"> a) None <br> <input name="radios" type="radio" value="b"> b) Some, but it's self-taught from books or websites <br> <input name="radios" type="radio" value="c"> c) Quite a bit: I have taken courses in photography <br> <input name="radios" type="radio" value="d"> d) A lot:I have completed a degreee or certificate in the subject <br> <br> 4.) How do you describe yourself as a photographer? <br> <input name="radios" type="radio" value="a"> a) A novice:someone who takes a few pictures occasionally <br> <input name="radios" type="radio" value="b"> b) An amateur:someone who takes many photos, but doesn't usually share them <br> <input name="radios" type="radio" value="c"> c) An enthusiast:someone who loves to take pictures, possibly shows them with oters on public sites, or who might own a DSLR, or frequently uses a camera in other than "automatic mode" <br> <input name="radios" type="radio" value="d"> d) An expert:skilled photographer who has published their work, or exhibited their work in shows, or has been paid for their work <br> <br> 5.) Do you have artistic training or exposure? <br> <input name="radios" type="radio" value="a"> a) No <br> <input name="radios" type="radio" value="b"> b) I visit museums or art galleries <br> c) I have studied art: please describe <br> <textarea name="data5c" cols="50" rows="5"> </textarea> <br> d) I am a practicing artist: please describe <br> <textarea name="data5d" cols="50" rows="5"> </textarea> <br> <br> 6.) Some optional information, if you don't mind. We respect your privacy. <br> <br> Age <br> <textarea name="6a" cols="1" rows="1"> </textarea> <br> <br> Gender <br> <input name="radios" type="radio" value="a"> a) Male <br> <input name="radios" type="radio" value="b"> b) Female <br> <br> Languages Spoken <br> <textarea name="6b" cols="25" rows="2"> </textarea> <br> <br> Languages Read <br> <textarea name="6c" cols="25" rows="2"> </textarea> <br> <br> <br> <input type="submit" value="Send"> <input type="reset" value"Reset"> </form> </body> </html> Kindly help me out with this issue. Thank you for your help in this regard. |
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