hi, thanx for replying!
i installed the new php version 4.2 - but this time as isapi-module. and
since this i have no troubles! ( i used not the php.ini.recommend! )
bye - hermann
Oliver Cronk [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Have you tried
The query should read as below, without the = after the LIKE:
$query = SELECT uid, id, image, iname, quantity, type FROM
{$config[prefix]}_shop WHERE iname LIKE '%$shopsearch% ORDER BY
iname';
HTH
Maureen
Jennifer Downey [EMAIL PROTECTED] said:
Hi all,
I have been trying to build a
Sorry about that, I just pasted and took out the =, but I see another thing
that could be a problem:
$query = SELECT uid, id, image, iname, quantity, type FROM
{$config[prefix]}_shop WHERE iname LIKE = '%$shopsearch%' ORDER BY
iname;
The single quote was in the wrong place, it should have
Jennifer,
In the manual it's probably under the SELECT heading. (I've not checked,
but that's a godd spot to start looking for the bits that can amke up the
WHERE clause.)
Try removing the single quotes - you have one before %shopsear and
another following iname, as iname'
Sorry this is
And Maureen is right! We just so *used* to seeing equal signs in SELECT
statements.
Smacking fist on head Miles Thompson
At 02:23 PM 4/25/2002 +, Maureen wrote:
The query should read as below, without the = after the LIKE:
$query = SELECT uid, id, image, iname, quantity, type FROM
Thanks Maureen,
I don't get the errors I was getting so that shows me you are great. Now my
next problem is if an item is found how do I echo or print that to the
screen. I have this as a starting point but I don't know where to go after
this.
if($search)
{
$query = SELECT uid, id, image,
From what I can see, you are not defining the array $row in here. You would
need to define something like:
$row=mysql_fetch_array($ret);
before calling it.
I finally found where the information on LIKE is in the manual (I can
understand your frustration, it took me a while). It is under
Hello!!
Your right!!
don't use = on the like part...I don't know what I was thinking!!!
Anyway here is the code right from my search page...
include(./navbar2.php);
$nav = new navbar;
$nav-numrowsperpage = 30;
$sql = SELECT Print_Name, Art_Job_Number, Size, Item_Number
HI--
* I have the following variables: $c1 thru $c10.
* I also have a FOR loop like this:
for($count = 0; $count sizeof($artist); $count++)
{ echo i$artist[$count]/ibr; }
// $artist is a list of ten (10) artist's names.
* What I'm trying to do is
I think what you don't know is this:
$varName = c1;
$$varName = 3;
that just set $c1 to equal 3.
Does that help?
-Original Message-
From: Julio Cuz, Jr. [mailto:[EMAIL PROTECTED]]
Sent: Thursday, April 25, 2002 9:56 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Variables
HI--
* I
In an attempt to be less vague I think you want to change your line:
$c . $count+1 = $artist[$count];
to be:
$varName = $c.($count+1);
$$varName = $artist[$count];
Got it now?
---
I think what you don't know is this:
$varName = c1;
$$varName = 3;
that just set $c1 to
Thanks all,
You people are awesome. I have it working but I still have 2 bugs
First it is only showing the first record it finds with the search term, not
all records associated with it.
Second it has to be spelled exactly right or I get 0 results.
in other words if Jelly is spelled Jelly and
Hello.
I have recently installed the freetds package and reconfigured PHP to
connect to MSSQL. I am running a simple join query on two tables and I get
the following error.
bWarning/b: Sybase error: 'REQ' is not a recognized OPTIMIZER LOCK
HINTS option. (severity 15) in myFile.php on line
The if/else/echo should all be within the while loop. The way it is now, it
should only show the last record returned, because that is the last value
within the while loop. The way you have it, the rest would be done after it
had looped through all of the records and it would return
I am quite positive that they are not Binary what I am finding is that it
has to be exactly as it is in the table.
I have one item called StarrRacing Guide if I put in starr or Starr it will
return nothing I have to spell it exactly like it is in the db StarrRacing
Guide.
here is a dump of the
Hello,
Jennifer Downey wrote:
Hi all,
I have been trying to build a search script for my site that deals with only
one table in my db.
As Julie Meloni pointed out look in the MySQL manual for LIKE clauses I
can't seem to locate that clause in ether
manual.
Dan Brunner gave me this to go
Jennifer,
Check if there is a set exact setting for MySQL.
Using LIKE should find everything beginning with Starr. You may also have
to convert search term and target values to one of lower or upper case in
the SELECT statement.
Miles
At 10:55 AM 4/25/2002 -0700, Jennifer Downey wrote:
I am
Jennifer
I suggest you check out MySQLs full text indexing capabilities. If you
create a fulltext index on the text fields you want to search you can use
the MATCH function. This will perform a fuzzy matching search that returns
results in order of relevance. For details see
Ok I am looking at the code an I'm thinking it has to be in the code because
I tried it on my box at home and it did the same thing.
So while looking hard at the code I came up with this:
if($searchword == $iname) this is the problem
if $searchword == $iname then it will echo everything I want
Here is some more information I found in the manual about case sensitivity in
searches. The whole page is at
http://www.mysql.com/doc/C/a/Case_sensitivity.html
Here is the text that shows what might be the problem:
In older MySQL versions LIKE comparisons where done on the uppercase value
Yes it seems off-topic but I only joined to this mail-list. So I don't have any other
place to ask.
I sent mail to unsubscribe me from the list but nothing happened. I'm still recieving
mails from the list. What's wrong?
See Dave Matthews Band live or win a signed guitar
hehehe, you recoded PHP to reconnect to MSSQL, but you forgot
to redo your connection stuff. Sybase error: means that you're
connecting to Sybase and not MSSQL. *wink*
'Luck
-Szii
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, April 25, 2002
Hi,
how can I append from one table to another with same structure some selected
records
Thanks,
Rosen
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Help. Help
Please, Please tell me what I am doing wrong here...I have a form to input data but
it is not posting ithere is the code. It seems to work no errors but a query show
no new additions to the table.
$sql = INSERT INTO employees
As many people have told me and I have just now got it through my thick head
try either echoing the query or use
$result = mysql_query($sql) or die(mysql_error());
You will find out what is being passed (or not passed) through your
variables or find out what error you are getting.
HTH
Thanks Maureen,
The == was the problem. I can search with upper and lower case and only
partial words.
Hosting is using 3.23.45 mysql so I don't believe that was the case.
Thanks again!
Jennifer
Maureen [EMAIL PROTECTED] wrote in message
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Here is
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