RE: [PHP] R: [PHP] SQL Query Not Kosher?
On 03 February 2004 13:45, Alessandro Vitale contributed these pearls of
wisdom:
> try removing curly braces as follows:
>
> $query = mysql_query("UPDATE stories SET status='approved'
>WHERE story_id={$id}"); |
Nothing wrong with the above, it's perfectly valid -- just a slightly
different way of writing:
> $query = mysql_query("UPDATE stories SET status='approved'
> WHERE story_id=${id}");
Cheers!
Mike
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[PHP] R: [PHP] SQL Query Not Kosher?
try removing curly braces as follows:
$query = mysql_query("UPDATE stories SET status='approved' WHERE
story_id={$id}");
|
|
|
\/
$query = mysql_query("UPDATE stories SET status='approved' WHERE
story_id=$id");
or
$query = mysql_query("UPDATE stories SET status='approved' WHERE
story_id=${id}");
this applies if story_id is of type int in mysql table definition, or you
should enclose it among '' if is of type char, varchar or similar.
cheers
alessandro
-Messaggio originale-
Da: Mr. Austin [mailto:[EMAIL PROTECTED]
Inviato: martedì 3 febbraio 2004 5.35
A: [EMAIL PROTECTED]
Oggetto: [PHP] SQL Query Not Kosher?
Hi all:
I am trying to get this to work, but always get the same error: that the
resource in mysql_affected_rows() is not valid. Anyone see why this would
be? All variables have been tested for accuracy.
$query = mysql_query("UPDATE stories SET status='approved' WHERE
story_id={$id}");
if(mysql_affected_rows($query) == 1) {
print("Your approval of \"$title\" was successful. If this user entered
an email address, they have been sent a notice of its approval and
publication on the site.");
} else {
print("The approval of \"$title\" was not successful. Please check with
the site administrator for assistance.");
}
The above SQL statement works perfectly with phpMyAdmin (and, oddly enough,
works with the above script, yet the Warning is produced and the 'not
successful' message is displayed) Any thoughts are appreciated!
Mr. Austin
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Re: [PHP] SQL Query Not Kosher?
The argument to mysql_affected_rows must be mysql CONNECTION identifier
(from mysql_connect), not RESULT identifier (from mysql_query). Removing
the argument will work for you.
Mr. Austin wrote:
Hi all:
I am trying to get this to work, but always get the same error: that the resource in mysql_affected_rows() is not valid. Anyone see why this would be? All variables have been tested for accuracy.
$query = mysql_query("UPDATE stories SET status='approved' WHERE story_id={$id}");
if(mysql_affected_rows($query) == 1) {
print("Your approval of \"$title\" was successful. If this user entered an email
address, they have been sent a notice of its approval and publication on the site.");
} else {
print("The approval of \"$title\" was not successful. Please check with the site
administrator for assistance.");
}
The above SQL statement works perfectly with phpMyAdmin (and, oddly enough, works with the above script, yet the Warning is produced and the 'not successful' message is displayed) Any thoughts are appreciated!
Mr. Austin
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Re: [PHP] SQL Query Not Kosher?
That would explain why it displays the second message saying it was not
successful. But why then does it still display this warning?
Warning: mysql_affected_rows(): supplied argument is not a valid MySQL-Link
resource in c:\apache\htdocs\admin\stories.php on line 112
Mr. Austin
- Original Message -
From: "Jochem Maas" <[EMAIL PROTECTED]>
To: "Mr. Austin" <[EMAIL PROTECTED]>
Sent: Monday, February 02, 2004 10:41 PM
Subject: Re: [PHP] SQL Query Not Kosher?
> from the php function manual:
>
>
> Note: When using UPDATE, MySQL will not update columns where the new
> value is the same as the old value. This creates the possibility that
> mysql_affected_rows() may not actually equal the number of rows matched,
> only the number of rows that were literally affected by the query.
>
>
> http://nl2.php.net/mysql_affected_rows
>
> Mr. Austin wrote:
>
> > Hi all:
> >
> > I am trying to get this to work, but always get the same error: that the
resource in mysql_affected_rows() is not valid. Anyone see why this would
be? All variables have been tested for accuracy.
> >
> > $query = mysql_query("UPDATE stories SET status='approved' WHERE
story_id={$id}");
> > if(mysql_affected_rows($query) == 1) {
> > print("Your approval of \"$title\" was successful. If this user
entered an email address, they have been sent a notice of its approval and
publication on the site.");
> > } else {
> > print("The approval of \"$title\" was not successful. Please check
with the site administrator for assistance.");
> > }
> >
> > The above SQL statement works perfectly with phpMyAdmin (and, oddly
enough, works with the above script, yet the Warning is produced and the
'not successful' message is displayed) Any thoughts are appreciated!
> >
> > Mr. Austin
> >
>
>
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[PHP] SQL Query Not Kosher?
Hi all:
I am trying to get this to work, but always get the same error: that the resource in
mysql_affected_rows() is not valid. Anyone see why this would be? All variables have
been tested for accuracy.
$query = mysql_query("UPDATE stories SET status='approved' WHERE story_id={$id}");
if(mysql_affected_rows($query) == 1) {
print("Your approval of \"$title\" was successful. If this user entered an email
address, they have been sent a notice of its approval and publication on the site.");
} else {
print("The approval of \"$title\" was not successful. Please check with the site
administrator for assistance.");
}
The above SQL statement works perfectly with phpMyAdmin (and, oddly enough, works with
the above script, yet the Warning is produced and the 'not successful' message is
displayed) Any thoughts are appreciated!
Mr. Austin
