Look at this snippet :
> $word = current($wordsarray);
> next($wordsarray);
> $sql=$sql."$word)";
if you have 2 words the sql would extend to
word1)word2) .. this is invalid sql syntax.
Change the
while ($i < count($wordsarray))
{
$word = current($wordsarray);
next($word
the problem is that your query is not returning a result i would start by
changing
"$Link = mysql_connect ($Host, $User, $Password);"
AND
"$Result = mysql_db_query ($DBName, $Query, $Link);"
To
"$Link = mysql_connect ($Host, $User, $Password) OR die("could not
connect"
I think that the following code looks fine, but I get an error (Warning:
Supplied argument is not a valid MySQL result resource) where it says "while
($Row = mysql_fetch_array ($Result)) {" any ideas?
$category returned the following
results:";
} else { if ($title) {
$Query = "SELECT * from $
Can anyone recommend PHPLIB? for authentication and session management? or
something else?
thanks Ken
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[EMAIL PROTECTED] (Fabian Pichler) wrote:
>Hello together
>
>I'm a real newbie on PHP/Mysql.
>I'm trying to get PHP/mysql to work together on my MOSXS 1.0-2 Server.
>Has anyone done this before?
>
>My problem is the mysql support (--with-mysql=/usr/local/mysql) in PHP.
>If I compile the PHP with
I would like to begin learning ODBC. Will I be able to set this up on my
single local machine?
Also, What about connecting to MS SQL server remotely? Will i simply need an
IP Port and user/pass?
Thank you
olinux
_
Do You Yahoo!?
Get
It works for me... except that you need a semi-colon after 'return
$returned_string' and it sticks the word OR at the end of the whole
string, which you may not want.
Does that help, or did I miss the point?
Ben
-Original Message-
From: Brad Lipovsky [mailto:[EMAIL PROTECTED]]
Sent: Su
- Original Message -
From: "Ken Sommers" <[EMAIL PROTECTED]>
To: "PHP DB Mailing List" <[EMAIL PROTECTED]>; "Hugh Bothwell"
<[EMAIL PROTECTED]>
Sent: Sunday, July 08, 2001 5:30 PM
Subject: Re: [PHP-DB] Re: Homesite Server Mapping
> Thanks alot,
> that mapping works for me too: I use
>
Here is the code that I am trying to use:
//code start
function search_terms($title) {
$array = explode (" ", $title);
for($i=0,$n=count($array); $i<$n; $i++) {
$returned_string = $returned_string . $array[$i] . " OR ";
}
return $returned_string
}
//code end
I want it to turn the stri
Hello all,
I seem to be missing something with my login script,
I have a couple of issues:
1) When you do not enter any informaton, it will say:
"Please fill out all fields to proceed.
There seems to be a problem with the database."
Now, if i hit enter again it will say:
"Please fill out all fie
Ha ha... I should have meantioned "Here's what it -should- be" or
something of the like. :)
I suspect that was the only problem with his code, but perhaps there's
something else there.
-Original Message-
From: Ben Bleything [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 8:13 P
Guess I'm just a big dumbass then, aren't I =P
Oops.
I suppose that would cause it to fail then, wouldn't it?
=> Ben
-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 5:10 PM
To: 'Ben Bleything'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE:
The code you're referencing is my modification of his original post. :)
-Original Message-
From: Ben Bleything [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 8:04 PM
To: 'Matthew Loff'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error
Sure he is. Right here:
$queryResult = mysql_query($sql);
what exact error is occurring?
-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error
You aren't calling mysql_query() to execute the query.
//$find is text box input
$wordsarray = explode(" ",$find);
$sql = "SELECT bandname FROM bands WHERE (bandname
LIKE ";
$i = 0;
while ($i < count($wordsarray))
{
$word = current($wordsarray);
next($wordsarray);
$sql=$sql."$wor
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$find is text box input
$wordsarray = explode(" ",$find);
$sql = "SELECT bandname FROM bands WHERE (bandname
LIKE ";
$i = 0;
while ($i < count($wordsarray))
{
$word = current($wordsar
Thanks alot,
that mapping works for me too: I use
From: C:\Program Files\Apache Group\Apache\htdocs\
To: http://localhost
for local mapping
Have you ever tried mapping to your REMOTE host?
I have tried and failed too many times, what works for Remote mapping?
any one ever done it?
Thanks,
K
"Ken Sommers" <[EMAIL PROTECTED]> wrote in message
001001c107e2$b84ad5c0$aa42500c@zeospantera">news:001001c107e2$b84ad5c0$aa42500c@zeospantera...
> Has anyone successfully mapped their browser to a Remote Server?
>
> Please show your:
> map from:
C:\Program Files\Apache Group\Apache\htdocs\
> Ma
"Brad Lipovsky" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I get the following error:
>
> $Query = "SELECT * from $TableName WHERE title LIKE "%$title%")";
If you tried 'echo $Query' right here, you wouldn't see anything.
Try 'echo "305" % "300";'
You
Hello,
any Homesite users out there??
Has anyone successfully mapped their browser to a Remote Server?
Please show your:
map from:
&
Map to:
thanks
ken
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In my oppinion the error is in the Line:
WHERE title LIKE "%$title%")";
because you have a ")" too much and write the search String a follow:
WHERE title LIKE '%$title%';
important is the symbol ' not " .
I hope this helps you.
bye
Hoth
"Brad Lipovsky" <[EMAIL PROTECTED]> schrieb im Newsbe
I get the following error:
//error
Warning: Supplied argument is not a valid MySQL result resource in
/web/sites/184/syzme/www.syzme.f2s.com/Commerce_System/search.php on line 22
//error
when I try to use this code:
$category returned the following
results:";
while ($Row = mysql_fetch_array
Hello,
When will array processing for Oracle fetch be out ?
regards
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This Problem I already had.
under my mail is a cut of my httpd.conf witch runs under my win2k. Your
Problem is that your apache can't understand ".php" so you had to tell him
what should happen with a file where at the end ".php" is. Therefor add the
lines with "AddType application ...".
I hope th
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