[PHP] Could some one check my code
i am getting this error returned but i dont know why :(
error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in /home/pferrie/public_html/vinrev/adm/insert2.php on line 6
php file
?php
include(connection.php);
if(!empty($rating)){
$query=SELECT * FROM $tablename WHERE rating = '$rating';
if(mysql_query($query)) {
$myrow = mysql_fetch_array($query);// This line returns an error!
$id = $myrow ['id'];
echoError! There is another row in the database with a rating
value of $rating\n
You must first edit the row with a rating of $rating ! Row id =
$idbrbr;
echoa href='edit2.php?tablename=$tablenameid=$id'Edit
record/abr;
}
}else {
if(!empty($_FILES['userfile']['tmp_name'])) {
$name = strtolower(eregi_replace('#| |\?|!', '',
$_FILES['userfile']['name']));
$destination = 'img/'.$name;
if(move_uploaded_file($_FILES['userfile']['tmp_name'], $destination)) {
$picture=$destination;
$query = INSERT INTO $tablename (artist, title, picture, review,
label, format, price, rating)VALUES ('$artist', '$title', '$picture',
'$review', '$label', '$format', '$price', '$rating');
if(!mysql_query($query)) {
fail(Error updating DB1);
}
Header(Location: main.php?success=New row has been inserted into the
database);
}
}
if(empty($userfile)) {
$query2 = INSERT INTO $tablename (artist, title, picture, review,
label, format, price, rating)VALUES ('$artist', '$title', '$picture',
'$review', '$label', '$format', '$price', '$rating');
if(!mysql_query($query2)) {
fail(Error updating DB2);
}
Header(Location: main.php?success=New row has been inserted into the
database);
}
}
?
I am trying to check for a duplicate entry for rating and if one is found
grab the row ID and then use it to load the edit page for the row.
Anyone
Cheers
Paul
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Re: [PHP] Could some one check my code
PAUL FERRIE wrote:
i am getting this error returned but i dont know why :(
error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in /home/pferrie/public_html/vinrev/adm/insert2.php on line 6
php file
?php
include(connection.php);
if(!empty($rating)){
$query=SELECT * FROM $tablename WHERE rating = '$rating';
if(mysql_query($query)) {
$myrow = mysql_fetch_array($query);// This line returns an error!
you sould write :
if ($result = mysql_query($query)) {
$myrow = mysql_fetch_array($result);
Hope this helps
--
Cordialement,
---
Sophie Mattoug
Développement web dynamique
[EMAIL PROTECTED]
---
$id = $myrow ['id'];
echoError! There is another row in the database with a rating
value of $rating\n
You must first edit the row with a rating of $rating ! Row id =
$idbrbr;
echoa href='edit2.php?tablename=$tablenameid=$id'Edit
record/abr;
}
}else {
if(!empty($_FILES['userfile']['tmp_name'])) {
$name = strtolower(eregi_replace('#| |\?|!', '',
$_FILES['userfile']['name']));
$destination = 'img/'.$name;
if(move_uploaded_file($_FILES['userfile']['tmp_name'], $destination)) {
$picture=$destination;
$query = INSERT INTO $tablename (artist, title, picture, review,
label, format, price, rating)VALUES ('$artist', '$title', '$picture',
'$review', '$label', '$format', '$price', '$rating');
if(!mysql_query($query)) {
fail(Error updating DB1);
}
Header(Location: main.php?success=New row has been inserted into the
database);
}
}
if(empty($userfile)) {
$query2 = INSERT INTO $tablename (artist, title, picture, review,
label, format, price, rating)VALUES ('$artist', '$title', '$picture',
'$review', '$label', '$format', '$price', '$rating');
if(!mysql_query($query2)) {
fail(Error updating DB2);
}
Header(Location: main.php?success=New row has been inserted into the
database);
}
}
?
I am trying to check for a duplicate entry for rating and if one is found
grab the row ID and then use it to load the edit page for the row.
Anyone
Cheers
Paul
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Re: [PHP] Could some one check my code
On Wed, Nov 26, 2003 at 11:23:19AM -, PAUL FERRIE wrote:
:
: i am getting this error returned but i dont know why :(
: error:
:
: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
: resource in /home/pferrie/public_html/vinrev/adm/insert2.php on line 6
:
: php file
: ?php
: include(connection.php);
: if(!empty($rating)){
: $query=SELECT * FROM $tablename WHERE rating = '$rating';
: if(mysql_query($query)) {
: $myrow = mysql_fetch_array($query);// This line returns an error!
mysql_query() might fail, in which case it returns a boolean FALSE.
You should first check whether the query succeeded or failed.
$result = mysql_query($query);
if ($result === false)
{
// query failed, report the error
echo mysql_error();
}
else
{
// query succeeded, do your thing
}
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Re: [PHP] Could some one check my code
PAUL FERRIE wrote:
i am getting this error returned but i dont know why :(
error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in /home/pferrie/public_html/vinrev/adm/insert2.php on line 6
php file
?php
include(connection.php);
if(!empty($rating)){
$query=SELECT * FROM $tablename WHERE rating = '$rating';
if(mysql_query($query)) {
$myrow = mysql_fetch_array($query);// This line returns an error!
You need to capture the return value of mysql_query(). mysql_query() is
going to return a Resource that should be passed to mysql_fetch_array(),
so it knows where to get each row from.
if($result = mysql_query($query))
{ $myrow = mysql_fetch_array($result);
--
---John Holmes...
Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/
php|architect: The Magazine for PHP Professionals www.phparch.com
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