I need a little help for my problem...
Scene: I was trying to search something in my database but the result is "Query was empty" but the one I'm trying to search is already in my database. I'm trying to search the "Atrium Hotel"
Here's my screenshot of my Database:
Here's my screenshot of my result Page:
And Lastly here's my code:
<input type='submit' name='search' value='Search Building' onClick="this.form.action='search_bldg.php'; this.form.submit()">
<input type="text" id="idSearch"name="searchBldg" size="40" value="<?php echo $_POST['searchBldg']; ?>">
<fieldset width= "50px">
<legend>BUILDING/S</legend>
<?php
$search = $_POST["searchBldg"];
$data = mysql_query("SELECT * FROM tbldata WHERE fldBldgName LIKE '%$search%'");
$result = mysql_query($data) or die(mysql_error());
while($row = mysql_fetch_array( $result ))
{
echo $row['fldBldgName'];
}
?>
</fieldset>
I was wondering what is the problem in my query...
Thanks in advance...
You are executing your query twice (Line #7 and #8). That may be the problem. Try something like this:
<input type='submit' name='search' value='Search Building' onClick="this.form.action='search_bldg.php'; this.form.submit()">
<input type="text" id="idSearch"name="searchBldg" size="40" value="<?php echo $_POST['searchBldg']; ?>">
<fieldset width= "50px">
<legend>BUILDING/S</legend>
<?php
$search = $_POST["searchBldg"];
$query= "SELECT * FROM tbldata WHERE fldBldgName LIKE '%$search%'"; //Your sql
$result = mysql_query($query) or die(mysql_error()); //execute your query
while($row = mysql_fetch_array( $result ))
{
echo $row['fldBldgName'];
}
?>
</fieldset>
P.S. use mysqli_* or PDO instead of mysql_* since it is deprecated as of PHP 5.4
You should use mysqli or PDO since mysql_* is depreciated.
Try the below code: It's using mysqli . It should be working...
<?php
$search = $_POST["searchBldg"];
//connecting to db...mysqli_connect("example.com","peter","abc123","my_db")
$con=mysqli_connect(host,username,password,dbname);
//searching...
$query= "SELECT * FROM tbldata WHERE fldBldgName LIKE '%$search%'";
//execute the query
$result = mysqli_query($query) or die(mysqli_error());
while($row = mysqli_fetch_array( $result ))
{
echo $row['fldBldgName'];
}
mysqli_close($con);
?>
There are some issues in your code that I'll demonstrate by commenting your code:
<input type='submit' name='search' value='Search Building' action='search_bldg.php' onClick="this.form.submit();"> <!-- setting action-attribute directly in form-tag, no need to use js for that -->
<input type="text" id="idSearch" name="searchBldg" size="40" value="<?php echo $_POST['searchBldg']; ?>"> <!-- added a space between id="search" and name-attribute -->
<fieldset width= "50px">
<legend>BUILDING/S</legend>
<?php
$search = $_POST["searchBldg"];
$data = "SELECT * FROM tbldata WHERE fldBldgName LIKE '%$search%'"; //mysql_query is removed, because the actual query is executed below
$result = mysql_query($data) or die(mysql_error());
while($row = mysql_fetch_assoc( $result )) //mysql_fetch_array doesn't return associative arrays, therefore it's replaced with mysql_fetch_assoc
{
echo $row['fldBldgName'];
}
?>
</fieldset>
You should of course sanitize data (so not unwanted data gets inserted into db) and use something else besides mysql_* as stated in previous answers.
Another issue (which is not really a problem) is the name of elements, column-names etc. searchBldg is very similar to searchB1dg and it might be easy to do typing-errors which might be hard to find.
Related
I am new to coding and keep getting this error and I am not sure what the reason is... I am trying to search/retrieve data from a Mysql database... Idea is that someone selects a category of search (such as first name) and inputs a first name and then code retrieves all relevant matches from database table Customers.
I am getting the following error:
Fatal error: Call to undefined function query() in ..../search.php on line 36
Can anyone help.
<html>
<head>
<title>pageName</title>
<style type="text/css">
table {
background-color: #ADD8E6;
border: 1px solid black;
font-family: Arial; font-size: 14px;
}
th {
text-align: left;
}
</style>
</head>
<body>
<h1>MEMBERS SEARCH</h1>
<form method="post" action="search.php">
<input type="hidden" name="submitted" value="true"/>
<label> Search Category:
<select name="category">
<option value="firstname">First NAME</option>
<option value="lastname">Last NAME</option>
</select>
</label>
<label> Search Criteria:<input type="text" name="criteria" /></label>
<input type="submit" />
</form>
<?php
if (isset($_POST['submitted'])){
// connect to the DB
include('connect.php');
$category = $_POST['category'];
$criteria = $_POST['criteria'];
$query = "SELECT * FROM Customers WHERE $firstname LIKE '%" . $criteria ."%'";
$result = $query ($con, $query) or die ('error getting data from database');
$num_rows = mysql_num_rows ($result);
echo "$num_rows results found";
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
?>
<table>
<tr>
<td width="300" ><font face="Arial Black" size="2"><?php echo $row['firstname']?> <?php echo $row['lastname']?></font></td>
</tr>
</table>
<table>
<tr>
?>
</body>
</html>
Try changing your code to look like this.
<form method="post" action="search.php">
<input type="hidden" name="submitted" value="true"/>
<label> Search Category:
<select name="category">
<option value="firstname">First NAME</option>
<option value="lastname">Last NAME</option>
</select>
</label>
<label> Search Criteria:<input type="text" name="criteria" /></label>
<input type="submit" />
</form>
<?php
if (isset($_POST['submitted'])){
// connect to the DB
include('connect.php');
$category = mysqli_real_escape_string($con, $_POST['category']);
$criteria = mysqli_real_escape_string($con, $_POST['criteria']);
$query = "SELECT * FROM Customers WHERE firstname LIKE '%" . $criteria ."%'";
$result = mysqli_query($con, $query);
$num_rows = mysqli_num_rows($result);
echo "$num_rows results found";
Brief explanation of changes:
Added 'real_escape_string' function to your user submitted variables
to help prevent sql injection
Changed '$firstname' in your query to 'firstname'
Took out 'or die()' part after query and used mysqli_ functions to
handle the data.
You have $query instead of query, it should be a function not a variable. That may fix your problem, if there is more let me know.
$result = $query ($con, $query) or die ('error getting data from database');
should be:
$result = query($con, $query) or die('error getting data from database');
Also you have query mistake:
$query = "SELECT * FROM Customers WHERE $firstname LIKE '%" . $criteria ."%'";
Should be:
$query = "SELECT * FROM Customers WHERE firstname LIKE '%" . $criteria ."%'";
Should be firstname and not $firstname.
Also
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
While loop doesnt have a closing brackets
It should be like:
<table>
<?php
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
?>
<tr>
<td width="300" ><font face="Arial Black" size="2"><?php echo $row['firstname']?> <?php echo $row['lastname']?></font></td>
</tr>
<table> <---- This is html mistake also remove it
<tr> <----- This is html mistake also remove it
<?php
}
?>
</table>
I moved your table tag around loop because this is logical way to every row from loop generates row in table, but in some examples you can loop also whole tables just you will get lots of tables in your html code.
I think what you are looking for is mysqli_query($con, $query).
Try replacing your query() function call with mysqli_query()
Note that mysql_query() is depricated in favor of mysqli_query() for security reasons.
See the documentation for it here:
http://php.net/manual/en/mysqli.query.php
If that doesn't help then I'd suggest examining the code in your 'connect.php' that you are including. Also it would be helpful to get the exact text of the error, it would help us debug your code.
The line
$result = $query ($con, $query) or die ('error getting data from database');
doesn't make sense at all. $query is a variable containing the SQL instructions, not a function. And the variable $con has not been defined, at least not in the code you've shown.
Also, you should use the function mysqli_query() - http://php.net/manual/en/mysqli.query.php
So the line should be:
$result = mysqli_query($query) or die ('error getting data from database');
If you are using procedural style and $con is the link identifier returned by mysqli_connect() or mysqli_init(), use:
$result = mysqli_query($con,$query) or die ('error getting data from database');
Assuming you are using php mysql extension, below is the corrected solution to your problem.
<?php
if (isset($_POST['submitted'])){
include('connect.php');
$category = $_POST['category'];//I guess it is search category(e.g. firstname)
$criteria = strtolower($_POST['criteria']);
$query = "SELECT * FROM Customers WHERE LOWER({$category}) LIKE '%{$criteria}%'";
$result = mysql_query($query) or die (mysql_error());
$num_rows = mysql_num_rows($result);
echo "{$num_rows} results found";
echo '<table>';
while ($row = mysql_fetch_assoc($result)) {
?>
<tr>
<td width="300" >
<font face="Arial Black" size="2"><?php echo "{$row['firstname']} {$row['lastname']}"?></font>
</td>
</tr>
<?php } echo '</table>'; ?>
I have a table in MySQL where there is a row with this data.
id = 187
friendly name = i don't like mustard
filetype = exe
This first block of code below works perfectly, and echos text i don't like mustard into an HTML form. Similarly, if I change $row['friendlyname'] to $row['filetype'], text exe is echoed. All good, no issues yet.
<?php
$con = mysqli_connect('domain','user','pass','db');
$sql = "select * from installers where id=187";
$result = mysqli_query($con,$sql);
while($row=mysqli_fetch_array($result))
$friendlyname = htmlspecialchars(" ".$row['friendlyname']." ",ENT_QUOTES);
$con->close();
?>
<input type='text' value='<?php echo $friendlyname; ?>'>
The problem I'm having is if I try to echo both $row['friendlyname'] and $row['filetype'], only the variable that is listed first will be echoed. For example, in the below code, $row['friendlyname'] is listed before $row['filetype']. In this example, only $row['friendlyname'] (i don't like mustard) will be echoed. Similarly, if $row['filetype'] is listed before $row['friendlyname'], then only $row['filetype'] (exe) is echoed, and the second other HTML input form is empty.
<?php
$con = mysqli_connect('domain','user','pass','db');
$sql = "select * from installers where id=187";
$result = mysqli_query($con,$sql);
while($row=mysqli_fetch_array($result))
$friendlyname = htmlspecialchars(" ".$row['friendlyname']." ",ENT_QUOTES);
$filetype= htmlspecialchars(" ".$row['filetype']." ",ENT_QUOTES);
$con->close();
?>
<input type='text' value='<?php echo $friendlyname; ?>'>
<input type='text' value='<?php echo $filetype; ?>'>
Note 1: It doesn't matter the order of the input type forms. I ruled that out as the issue.
Note 2: If I were to replace $row['friendlyname'] and $row['filetype'] with the text I'm trying to echo, then it work (the below code). So, this definitely appears to be something with these $row variables.
<?php
$con = mysqli_connect('domain','user','pass','db');
$sql = "select * from installers where id=187";
$result = mysqli_query($con,$sql);
while($row=mysqli_fetch_array($result))
$friendlyname = i don't like mustard;
$filetype= exe;
$con->close();
?>
<input type='text' value='<?php echo $friendlyname; ?>'>
<input type='text' value='<?php echo $filetype; ?>'>
You have not added brackets into while loop so only first record is populated.
This block:
while($row=mysqli_fetch_array($result))
$friendlyname = htmlspecialchars(" ".$row['friendlyname']." ",ENT_QUOTES);
$filetype= htmlspecialchars(" ".$row['filetype']." ",ENT_QUOTES);
Should be:
while($row=mysqli_fetch_array($result)){
$friendlyname = htmlspecialchars(" ".$row['friendlyname']." ",ENT_QUOTES);
$filetype= htmlspecialchars(" ".$row['filetype']." ",ENT_QUOTES);
}
I am trying to create a form which allows the user to search for an event using the Venue and category fields which are scripted as dropdown boxes and the Price and finally by event title, as shown via the code if a keyword is entered which matches the fields on the database it should output all the related information for that event if any matches have been made on either search fields, but it seems to output every single event from the database no matter what I type in the search field.
DATABASE: http://i.imgur.com/d4uoXtE.jpg
HTML FORM
<form name="searchform" action ="PHP/searchfunction.php" method = "post" >
<h2>Event Search:</h2>
Use the Check Boxes to indicate which fields you watch to search with
<br /><br />
<h2>Search by Venue:</h2>
<?php
echo "<select name = 'venueName'>";
$queryresult2 = mysql_query($sql2) or die (mysql_error());
while ($row = mysql_fetch_assoc($queryresult2)) {
echo "\n";
$venueID = $row['venueID'];
$venueName = $row['venueName'];
echo "<option value = '$venueID'";
echo ">$venueName</option>";
}# when the option selected matches the queryresult it will echo this
echo "</select>";
mysql_free_result($queryresult2);
mysql_close($conn);
?>
<input type="checkbox" name="S_venueName">
<br /><br />
<h2>Search by Category:</h2>
<?php
include 'PHP/database_conn.php';
$sql3 ="SELECT catID, catDesc
FROM te_category";
echo "<select name = 'catdesc'>";
$queryresult3 = mysql_query($sql3) or die (mysql_error());
while ($row = mysql_fetch_assoc($queryresult3)) {
echo "\n";
$catID = $row['catID'];
$catDesc = $row['catDesc'];
echo "<option value = '$catID'";
echo ">$catDesc </option>";
}
echo "</select>";
mysql_free_result($queryresult3);
mysql_close($conn);
?>
<input type="checkbox" name="S_catDes">
<br /><br />
<h2>Search By Price</h2>
<input type="text" name="S_price" />
<input type="checkbox" name="S_CheckPrice">
<br /><br />
<h2>Search By Event title</h2>
<input type="text" name="S_EventT" />
<input type="checkbox" name="S_EventTitle">
<br /><br />
<input name="update" type="submit" id="update" value="Search">
searchfunction.php file
<?php
$count = 0;
include 'database_conn.php';
$venuename = $_REQUEST['venueName']; //this is an integer
$catdesc = $_REQUEST['catdesc']; //this is a string
$Price = $_REQUEST['S_price'];
$EventT = $_REQUEST['S_EventT'];
$sql = "select * FROM te_events WHERE venueID LIKE '%$venuename%' OR catID LIKE '%$catdesc%' OR eventPrice LIKE '%Price%' OR eventTitle LIKE '%$EventT%'";
$queryresult = mysql_query($sql) or die (mysql_error());
while ($row = mysql_fetch_assoc($queryresult))
{
echo $row['eventTitle'];
echo $row['eventDescription'];
echo $row['venueID'];
echo $row['catID'];
echo $row['eventStartDate'];
echo $row['eventEndDate'];
echo $row['eventPrice'];
}
mysql_free_result($queryresult);
mysql_close($conn);
?>
The query should be
$sql = "select * FROM te_events
WHERE (venueID LIKE '%$venuename%'
OR catID LIKE '%$catdesc%'
OR eventPrice LIKE '%$Price%'
OR eventTitle LIKE '%$EventT%')
;
To get values from the form submitted with method POST we use $_POST to access form data and not $_REQUEST:
$venuename = $_POST['venueName']; //this is an integer
$catdesc = $_POST['catdesc']; //this is a string
$Price = $_POST['S_price'];
$EventT = $_POST['S_EventT'];
That was about your problem - now some important notes:
Do not use mysql extension as it's deprecated. Read this official documentation.
Use mysqli and prevent SQL injections by using prepared queries and parameters like in official documentation again.
Since you are matching on any fields surrounded by wildcards, if any of the fields are blank, then the MySQL query will match all rows.
Also, you need to prevent MySQL injection. Otherwise, your MySQL table will eventually be hacked.
By the way, the code eventPrice LIKE '%Price%' is invalid and is missing a dollar sign.
Lastly, the mysql extension has been deprecated. I would recommend using mysqli instead as it is fairly similar.
I am currently working on a school project using php and mysql. I have created a form with three drop down boxes where users select types of data they are looking for. However, I am having a lot of trouble displaying the results after the form is submitted. Here is my current code:
<?php
require_once 'connection.php';
?>
<form action="stats.php" method ="post">
<input type="hidden" name="submitted" value="true" />
<fieldset>
<legend>
Specify Date, Month, and County
</legend>
<p>
<label for="year">
Please select a year
</label>
<select name= 'year'>
<?php
$query = "select distinct year from unemployed";
$result = $conn->query($query);
while($row = $result->fetch_object()) {
echo "<option value='".$row->year."'>".$row->year."</option>";
}
?>
</select>
</p>
<p>
<label for="month">
Please select a month
<label>
<select name= 'month'>
<?php
$query = "select distinct month from unemployed";
$result = $conn->query($query);
while($row = $result->fetch_object()) {
echo "<option value='".$row->month."'>".$row->month."</option>";
}
?>
</select>
</p>
<p>
<label for="location">
Please specify a location
</label>
<select name='select'>
<?php
$query = "select * from unemployed";
$result = $conn->query($query);
while ($finfo = $result->fetch_field()) {
echo "<option value='".$finfo->name."'>".$finfo->name."</option>";
}
?>
</select>
</p>
<input type ="submit" />
</fieldset>
</form>
<?php
if (isset($_POST['submitted'])) {
include('connection.php');
$gYear = $_POST["year"];
$gMonth = $_POST["month"];
$gSelect = $_POST["select"];
$query = "select $gSelect from unemployed where year='$gYear' and month='$gMonth'";
$result = $conn->query($query) or die('error getting data');
echo"<table>";
echo "<tr><th>Year</th><th>Time</th><th>$gSelect</th></tr>";
while ($row = $result->fetch_object()){
echo "<tr><td>";
echo $row['Year'];
echo "</td><td>";
echo $row['Month'];
echo "</td><td>";
echo $row['$gSelect'];
echo "</td></tr>";
}
echo "</table";
} // end of main if statement
?>
I am almost certain my problem lies within my while statement. I get the titles of my table columns to show up (Year, Month, $gSelect), but I am not getting my query results to be displayed.
I have tried:
while ($row = $result->fetch_object())
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
Neither of these are working for me. I have looked at php.net for guidance. I am still confused with what to do. If anyone could help me, I would really appreciate it.
Always check your returns:
if( ! $result = $conn->query($query) ) {
die('Error: ' . $conn->error());
} else {
while($row = $result->fetch_object()) {
echo "<option value='".$row->year."'>".$row->year."</option>";
}
}
Also putting error_reporting(E_ALL); at the top of your script while you're developing it will help enormously as well.
You should really look into passing variables as parameters to your query instead of injecting them as variables directly into your query. This can lead to sql injection attacks.
Also, here's a quick example of how to write a query using PDO and mysql:
//Simple Query
$dbh = new PDO('mysql:host=localhost;dbname=testdb;charset=utf8', 'username', 'password');
//Useful during development.
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
//mysql can have prepares depending on the version: http://stackoverflow.com/questions/10113562/pdo-mysql-use-pdoattr-emulate-prepares-or-not
$dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$sth = $dbh->query("SELECT * FROM table");
var_dump($sth->fetchAll(PDO::FETCH_ASSOC));
//Now with pass params and a prepared statement:
$query = "SELECT * FROM table WHERE someCol = ?";
$sth = $dbh->prepare($query);
$sth->bindValue(1,"SomeValue");
$sth->execute();
$results = $sth->fetchAll(PDO::FETCH_ASSOC));
again I'm trying to study php mysql and it seems that I tried everything thing to figure the problem out.. but it seems as a beginner codes in the internet are not helping.. I really can't update the records in the database.
<html>
<body>
<?php
$db = mysql_connect("localhost", "root");
mysql_select_db("dbtry",$db);
$id = isset($_GET['id']) ? $_GET['id'] : null;
$submit = isset($_POST['submit']);
if ($id) {
if ($submit) {
$result = mysql_query("select * from employees where id = " . mysql_real_escape_string($_GET['id']) );
$row = mysql_num_rows($result);
if ($myrow != 0) {
mysql_query ("UPDATE employees SET firstname='$first',lastname='$last',address='$address',position='$position' WHERE id = '$id'");
}
echo "Thank you! Information updated.\n";
} else {
// query the DB
$result = mysql_query("SELECT * FROM `employees` WHERE `id` = " . mysql_real_escape_string($_GET['id']), $db);
$myrow = mysql_fetch_array($result);
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']?>">
<input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
First name:<input type="Text" name="first" value="<?php echo $myrow["firstname"] ?>"><br>
Last name:<input type="Text" name="last" value="<?php echo $myrow["lastname"] ?>"><br>
Address:<input type="Text" name="address" value="<?php echo $myrow["address"]
?>"><br>
Position:<input type="Text" name="position" value="<?php echo $myrow["position"]
?>"><br>
<input type="Submit" name="submit" value="Enter information">
</form>
<?php
}
} else {
// display list of employees
$result = mysql_query("SELECT * FROM employees",$db);
while ($myrow = mysql_fetch_array($result)) {
printf("%s %s<br>\n", $_SERVER['PHP_SELF'], $myrow["id"],
$myrow["firstname"], $myrow["lastname"]);
}
}
?>
</body>
</html>
There are two things potentially causing you a problem: firstly, the values you are trying to set are variables which have not been defined. I'm assuming the begginers code you found assumed you had register globals enabled, you really don't want to do this!
The second problem, is that if you do have register globals enabled, the data isn't being sanitized, so a quotation mark could send the update awry.
Try this instead:
$first = mysql_real_escape_string( $_POST['first'] );
$last = mysql_real_escape_string( $_POST['last'] );
$address= mysql_real_escape_string( $_POST['address'] );
$position = mysql_real_escape_string( $_POST['position'] );
mysql_query ("UPDATE employees SET firstname='$first',lastname='$last',address='$address',position='$position' WHERE id = '$id'");
This should at least get you up and running. I'd strongly advise that you use either the MySQLi library, or PHP PDO, and think about using prepared statements for added security.
mysql_query("UPDATE `employees` SET `firstname`='".$first."', `lastname`='".$last."',
`address`='".$address."', `position`='".$position."' WHERE `id` = '".$id".' ; ", $db) or
die(mysql_error());
I think the problem may lie in your connection to the database. The third parameter of the mysql_connect function is a password. Therefore this:
$db = mysql_connect("localhost", "root");
should be:
$db = mysql_connect("localhost", "root", "yourPassword");
It would also help a lot if you posted what type of error you are getting.
You need to differentiate post and get. Follow the working example below. It will sort you out :D
<html>
<body>
<?php
$db = mysql_connect("localhost", "root","");
mysql_select_db("test",$db);
if($_SERVER['REQUEST_METHOD']=='POST')
{
//SUBMIT FORM
$id=isset($_POST['id'])?$_POST['id']:0;
if ($id) {
$result = mysql_query("select * from parameter where id = " . mysql_real_escape_string($id) );
$rows = mysql_num_rows($result);
if ($rows != 0) {
mysql_query ("UPDATE parameter SET name='".$_POST['name']."',value='".$_POST['value']."' WHERE id = '".$id."'");
echo "Thank you! Information updated.\n";
}
}
}
if($_SERVER['REQUEST_METHOD']=='GET')
{
//SELECT WHERE ID=GER VAR AND DISPLAY
$id = isset($_GET['id']) ? $_GET['id'] :0;//
if ($id) {
// query the DB
$result = mysql_query("SELECT * FROM parameter WHERE `id` = " . mysql_real_escape_string($_GET['id']), $db);
$myrow = mysql_fetch_array($result);
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']?>">
<input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
First name:<input type="Text" name="name" value="<?php echo $myrow["name"] ?>"><br>
Last name:<input type="Text" name="value" value="<?php echo $myrow["value"] ?>"><br>
<input type="Submit" name="submit" value="Enter information">
</form>
<?php
}
else {
// display list of employees
$result = mysql_query("SELECT * FROM parameter",$db);
while ($myrow = mysql_fetch_array($result)) {
echo "<a href='".$_SERVER['PHP_SELF']."?id=".$myrow['id']."'>".$myrow['name'].": ".$myrow['value']."</a><br>";
}
}
}
?>
</body>
</html>
Usually when I run into this problem, it's because auto commit is off and I forgot to tell the connection explicitly to commit.
EDIT: Have you tried this: How can I implement commit/rollback for MySQL in PHP?? Depending on your settings, InnoDB can be set to auto commit off, which means you need to tell MySQL explicitly to commit updates after your done.