replace not updating in database - php

Question

I have a table with id which is the primary key and user_id which is a foreign key but the session is based on this in my code.
I have tried EVERYTHING, so I will post my full code.
The form should insert if there is not a user_id with the same session_id in the table. If there is, it should update.
At the moment, when the user has not visited the form before (no user_id in the table) and data is inserted in, the page returns to the location page: but the data is not inserted in the table. if the user changes the data once it is updated it doesn't change either.
This is the table structure:
`thesis` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`user_id` int(11) NOT NULL,
`thesis_Name` varchar(200) NOT NULL,
`abstract` varchar(200) NOT NULL,
`complete` int(2) NOT NULL DEFAULT '1',
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`)
)
The code I have been using (and failing):
$err = array();
$user_id = intval($_SESSION['user_id']);
// otherwise
if (isset($_POST['doThesis'])) {
$link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection.");
// check if current user is banned
$the_query = sprintf("SELECT COUNT(*) FROM users WHERE `banned` = '0' AND `id` = '%d'",
$user_id);
$result = mysql_query($the_query, $link);
$user_check = mysql_num_rows($result);
// user is ok
if ($user_check > 0) {
// required field name goes here...
$required_fields = array('thesis_Name','abstract');
// check for empty fields
foreach ($required_fields as $field_name) {
$value = trim($_POST[$field_name]);
if (empty($value)) {
$err[] = "ERROR - The $field_name is a required field" ;
}
} // no errors
if (empty($err)) {
$id = mysql_real_escape_string($_POST['id']);
$thesis_Name = mysql_real_escape_string($_POST['thesis_Name']);
$abstract = mysql_real_escape_string($_POST['abstract']);
//replace query
$query = "REPLACE INTO thesis ( thesis_Name, abstract) VALUES ('$thesis_Name',
'$abstract') where id='$_SESSION[user_id]'";
if (!mysql_query($the_query))
echo "the query failed";
else header ("location:myaccount.php?id=' . $user_id");
}}}
$rs_settings = mysql_query("SELECT * from thesis WHERE user_id = $user_id;");
?>
<br>
<form action="thesis.php" method="post" name="regForm" id="regForm" >
class="forms">
<?php
$num_rows = mysql_num_rows($rs_settings);
if($num_rows > 0) { ?>
<?php while ($row_settings = mysql_fetch_array($rs_settings)) {?>
Title of Proposed Thesis<span class="required">*</span>
<textarea name="thesis_Name" type="text" style="width:500px; height:150px"
id="thesis_Name" size="600"><?php echo $row_settings['thesis_Name']; ?> </textarea>
</tr>
<tr>
<td>Abstract<span class="required">*</span>
</td>
<td><textarea name="abstract" style="width:500px; height:150px"
type="text" id="abstract" size="600"><?php echo $row_settings['abstract']; ?>
</textarea></td>
</tr>
<?php }
} else { ?>
//shows fields again without echo
I've tried var_dum($query) but nothing appears
PS I know the code isn't perfect but I'm not asking about this right now

Answer 1

I can't see how your replace statement will ever insert the initial row, as the where clause is always going to be false (there won't be a row with that user Id).
I think of you want to use replace you need to replace into thesis (id, userid, etc) without a where clause. If id and userid have a unique constraint and a row for userid exists then it will be updated; if it doesn't exist it will be inserted.
However- if you don't know id- which you won't if you are using auto increment, then I'm not sure you can do this with replace. See http://dev.mysql.com/doc/refman/5.0/en/replace.html
Why don't you check for the existence of a row an then use update or insert?
BTW, is the idea that a user can enter multiple theses into a form, or just one? Your table suggests they can have multiple. If this is what you are trying to achieve then I think you should be storing the id of each thesis in a hidden field as part of the form data. You would then be able to use REPLACE INTO thesis (id, user_id, thesis_name, abstract) VALUES ($id, $user_id, $thesis_name, $abstract) where id is the id of the thesis obtained from each hidden field. If this is not present, i.e. the user has entered a new thesis, then use NULL for id in the insert. This will work using the REPLACE INTO as the id column is auto increment.

Answer 2

Perhaps you mean user_id not id:
$query = "REPLACE INTO thesis ( thesis_Name, abstract)
VALUES ('$thesis_Name','$abstract')
WHERE user_id='{$_SESSION['user_id']}'";
Or if you do mean the id from $_POST['id']
$query = "REPLACE INTO thesis ( thesis_Name, abstract)
VALUES ('$thesis_Name','$abstract')
WHERE id='$id'";
Also instead of REPLACE you should use UPDATE. Im pretty sure its faster because REPLACE basically deletes the row then inserts it again, im pretty sure you need all the fields and values else your insert default values. From the manual:
Values for all columns are taken from the values specified in the
REPLACE statement. Any missing columns are set to their default
values, just as happens for INSERT
So you should use:
$query = "UPDATE thesis
SET thesis_Name='$thesis_Name', abstract='$abstract'
WHERE id='$id'";

Answer 3

You are doing everything right just one thing you are doing wrong
Your replace query variable is $query and you executing $the_query.
you wrong here:
$query = "REPLACE INTO thesis ( thesis_Name, abstract) VALUES ('$thesis_Name',
'$abstract') where id='$_SESSION[user_id]'";
if (!mysql_query($the_query)) // this is wrong
echo "the query failed";
replace it with:
$query = "REPLACE INTO thesis ( thesis_Name, abstract) VALUES ('$thesis_Name',
'$abstract') where id='$_SESSION[user_id]'";
if (!mysql_query($query)) // use $query
echo "the query failed";