Data from Database into Text Area?

michaelness · June 23, 2009

Right I have a page set up where a user can edit fields in the database specific to that user (interests, music etc). What i want to be able to do is to have the data from the database that the user has already put in showing in the text area when the user goes to edit them. Because at the moment everytime the user goes to edit a field and send it. it resets the other field to blank if there is nothing in their.

<?php
echo '<link rel="stylesheet" type="text/css" href="original.css" />';
include 'connection.php';
?>

<head>
<title>yourFace</title>
</head>

<body>

<?php 

$var = $_COOKIE['username'];

include 'connection.php';
$query = "SELECT * FROM	 `userinfo` WHERE username='$var'"; 
$result = mysql_query($query);
$row = mysql_fetch_object($result);

if(isset($_COOKIE['username'])){

echo '

<form enctype="multipart/form-data" action="uploader.php" method="POST">
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form>






<form action="updateprofile.php" method="post">

<label>Age: </label><br>
<input type="int" name="age" id="age"><br>


<label>Interests: </label><br>
<TEXTAREA name="interests", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>

<label>Hometown: </label><br>
<TEXTAREA name="hometown", ROWS="1", COLS="45", wrap=virtual></TEXTAREA><br>

<label>Hometown: </label><br>
<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual> </TEXTAREA><br>

<label>:Favourite TV: </label><br>
<TEXTAREA name="tv", ROWS="5", COLS="45", wrap=virtual><"$tv";</TEXTAREA><br>

<label>:Favourite Films: </label><br>
<TEXTAREA name="films", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>


<label>:Relationship Status: </label><br>
<TEXTAREA name="relation", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>


<input type="submit" name="submit" id="submit" value="Submit">
</form>

';


}
//when not logged in
else
{
echo 'You Are Not Logged In';
}
?>

</body>


</html>

Please help me with this problem, I know something should go into the text area bits but i cant figure out what.

elis · June 23, 2009

Put whatever you want shown into a variable

or

echo '<textarea>' . $variable . '</textarea>';

michaelness · June 23, 2009

I cant seem to get it to work

echo '<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>' . $music . '</TEXTAREA>';

>:(

Alex · June 23, 2009

I cant seem to get it to work

echo '<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>' . $music . '</TEXTAREA>';

>:(

That's fine as long as $music has the value you want..

michaelness · June 23, 2009

hmmm well im getting the error

Parse error: parse error, expecting `','' or `';'' in C:\xampp\htdocs\yourFace\editprofile.php on line 50

I am so lost ???

Alex · June 23, 2009

hmmm well im getting the error

Parse error: parse error, expecting `','' or `';'' in C:\xampp\htdocs\yourFace\editprofile.php on line 50

I am so lost ???

Show us more code.

michaelness · June 23, 2009

<!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" />

<?php
echo '<link rel="stylesheet" type="text/css" href="original.css" />';
include 'connection.php';
?>

<head>
<title>yourFace</title>
</head>

<body>

<?php 

$var = $_COOKIE['username'];



include 'connection.php';
$query = "SELECT * FROM	 `userinfo` WHERE username='$var'"; 
$result = mysql_query($query);
$row = mysql_fetch_object($result);

if(isset($_COOKIE['username'])){

echo '

<form enctype="multipart/form-data" action="uploader.php" method="POST">
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form>

<form action="updateprofile.php" method="post">

<label>Age: </label><br>
<input type="int" name="age" id="age"><br>


<label>Interests: </label><br>
<TEXTAREA name="interests", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>

<label>Hometown: </label><br>
<TEXTAREA name="hometown", ROWS="1", COLS="30", wrap=virtual></TEXTAREA><br>

<label>Music: </label><br>
<TEXTAREA name="music", ROWS="1", COLS="30", wrap=virtual>$music</TEXTAREA><br>

echo '<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>' . $music . '</TEXTAREA>';
<br>

<label>:Favourite TV: </label><br>
<TEXTAREA name="tv", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>

<label>:Favourite Films: </label><br>
<TEXTAREA name="films", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>


<label>:Relationship Status: </label><br>
<TEXTAREA name="relation", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>


<input type="submit" name="submit" id="submit" value="Submit">
</form>
';


}
//when not logged in
else
{
echo 'You Are Not Logged In';
}
?>

</body>


</html>

I've been messing around with $music = $_POST['music']; aswell.

the problem is im exceedingly tired. this is literally the last thing I have to do on this site, and im slowly going insane

J.Daniels · June 23, 2009

You added a second echo inside the first echo

<TEXTAREA name="music", ROWS="1", COLS="30", wrap=virtual>$music</TEXTAREA><br>

<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>' . $music . '</TEXTAREA>
<br>

michaelness · June 23, 2009

im sorry, that was a typo and me being tired.

this is the actual code

<!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" />

<?php
echo '<link rel="stylesheet" type="text/css" href="original.css" />';
include 'connection.php';
?>

<head>
<title>yourFace</title>
</head>

<body>

<?php 

$var = $_COOKIE['username'];
$music = $_GET['music'];

include 'connection.php';
$query = "SELECT * FROM	 `userinfo` WHERE username='$var'"; 
$result = mysql_query($query);
$row = mysql_fetch_object($result);

if(isset($_COOKIE['username'])){

echo '

<form enctype="multipart/form-data" action="uploader.php" method="POST">
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form>

<form action="updateprofile.php" method="post">

<label>Age: </label><br>
<input type="int" name="age" id="age"><br>


<label>Interests: </label><br>
<TEXTAREA name="interests", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>

<label>Hometown: </label><br>
<TEXTAREA name="hometown", ROWS="1", COLS="30", wrap=virtual></TEXTAREA><br>

<label>Music: </label><br>
<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>' . $music . '</TEXTAREA>;
<br>

<label>:Favourite TV: </label><br>
<TEXTAREA name="tv", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>

<label>:Favourite Films: </label><br>
<TEXTAREA name="films", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>


<label>:Relationship Status: </label><br>
<TEXTAREA name="relation", ROWS="5", COLS="45", wrap=virtual></TEXTAREA><br>


<input type="submit" name="submit" id="submit" value="Submit">
</form>
';

}
//when not logged in
else
{
echo 'You Are Not Logged In';
}
?>

</body>


</html>

J.Daniels · June 23, 2009

You have a semi-colon at the end of line 50

<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>' . $music . '</TEXTAREA>

Alex · June 23, 2009

$music the a $_GET variable, I thought you wanted it to get it from your query? Change it from $music to $row['music']

michaelness · June 23, 2009

$music the a $_GET variable, I thought you wanted it to get it from your query? Change it from $music to $row['music']

im sorry to sound ignorant but im not sure what you mean mate

Alex · June 23, 2009

$music the a $_GET variable, I thought you wanted it to get it from your query? Change it from $music to $row['music']

im sorry to sound ignorant but im not sure what you mean mate

Don't you want the stuff inside the text area to be taken from the database? If your query is right, and there's a row in the table called 'music' replace $music with $row['music'] inside of your textarea.

michaelness · June 23, 2009

ahh after i posted that i did what you just explained anyway and it didnt work. i got this line

Parse error: parse error, expecting `','' or `';'' in C:\xampp\htdocs\yourFace\editprofile.php on line 49

using this

<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>$row['music']</TEXTAREA>

Alex · June 23, 2009

You still have to have it like..

<TEXTAREA name="music", ROWS="5", COLS="45", wrap=virtual>' . $row['music'] . '</TEXTAREA>

michaelness · June 23, 2009

I'd tried that aswell.

that comes up with

Fatal error: Cannot use object of type stdClass as array in C:\xampp\htdocs\yourFace\editprofile.php on line 49

if it helps on my other page View Profile i am accessing the tables like this $row->music but i havent been able to get that to work either. is my variable at the top of the page right? the $email = $_POST['music']; bit?

Alex · June 23, 2009

Oh, that's because you're using mysql_fetch_object, and not like mysql_fetch_assoc, or mysql_fetch_array.. Still if $row->music isn't working then there's something wrong with your query.

michaelness · June 23, 2009

Oh, that's because you're using mysql_fetch_object, and not like mysql_fetch_assoc, or mysql_fetch_array.. Still if $row->music isn't working then there's something wrong with your query.

hmmm i have no idea what to do.

michaelness · June 23, 2009

It cant be that complex surely? all i need for it to do is to get info from the database.

i cant believe im stuck on this :S

J.Daniels · June 23, 2009

It's possible there is a problem with the query. To check the query:

$result = mysql_query($query) or trigger_error('SQL Failed', E_USER_ERROR);

Also, you might have an empty result set. Try echoing out the number of rows returned:

echo mysql_num_rows($result);

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Data from Database into Text Area?

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