Re: AW: [PHP-DB] getting mysql_fetch_row into array

2002-09-23 Thread Johannes Janson


Hi,

 $res = mysql_query (SELECT COUNT(deptid) FROM maintenance GROUP BY
 deptid);
 $deptcount = array(); // reset the array
 while ($row = mysql_fetch_row ($res)) {
   $deptcount[] = $row[0];

this should be $deptcount[] .= $row[0]; shouldn't it?

 }
 

cheers
J


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Re: AW: [PHP-DB] getting mysql_fetch_row into array

2002-09-23 Thread Johannes Janson

Hi,

Micah Stevens wrote:
 Nope. Anytime you set an array with no index equal to something, (i.e. 
 $deptcount[]) it will add another element to the end of the array.
 
 If you use .=, that means append the data, but since you're not 
 specifying an element, I'm not sure what it would do. Perhaps PHP is 
 smart enough to just add another element anyway. Have you tried this?
 

Well it's been some time since I dealt with PHP but as far as I remember
I did it like that and it worked. Well, sorry for the misinformation.

cheers
J


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[PHP-DB] Re: getting a details-page

2001-09-12 Thread Johannes Janson

Hi,


 I have a problem:
 with a mysql_query I get the results of a table field into a list page.
 Is it possible to use this result (as link) to open a new page where the
 detailed results of each row are shown?

So you select one field and list it with a while-loop?
Assuming this you can make a link like this:

while ($row = mysql_fetch_array($result)){
echo $row[field]. a
href=\show_details.php?criteria=$row[field]\Details/a;
}

and on show_details.php something like this:

$query = SELECT * FROM table WHERE criteria='$criteria';
$result = mysql_query($query);

hope this is what you want

cheers
Johannes



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[PHP-DB] Re: Show abbreviated article

2001-08-13 Thread Johannes Janson

hi,

[...]
 My question is this: how can I select only those first 20 words in the
 PHP code?

SELECT LEFT(YourBlobFiekd, 20) FROM YourTable;

hope it helps
Johannes



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[PHP-DB] Re: update mysql table using php

2001-08-13 Thread Johannes Janson

Hi,

 Any ideas as to what the problem is?

What error do you get? Is the query not working?
Put a mysql_error() into your or die(The category...) to get
the exact error message. And try putting quotes around CAtID = '$value'.

  print a href= . '' . display.php . '' . Return to the
print a href=\display.php\Return to CatIndex/a;
looks much nicer, doesn't it

hope it helps
Johannes



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[PHP-DB] Re: semi newbie Q

2001-08-09 Thread Johannes Janson

Hi,

 Here is a quick question. Cane the mail() function handle file
attachments?
 I couldnt find anything like that on the PHP reference but maybe someone
 knows out there.  If not does anyone have any simple scripts that will
allow
 me to do this?

I think it doesn't. Check e.g http://phpclasses.upperdesign.com/ for a
MIME-class.
Look at the links on php.net

hope it helps
Johannes



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[PHP-DB] Re: Array and ID

2001-08-03 Thread Johannes Janson

Hi,

 I have a drop-down menu that has been prepopulated with an array from a
 column name CompanyName. What I need to do is take the CompanyName lookup
 the CompanyID in the company table and insert the CompanyID into a row in
a
 table called contacts with a CompnayID column.
[...]
Is this a multiple or a single select?

A quick and dirty thought:
?php
$sql and $query as they were

echo select name=\company\;

while ($row = mysql_fetch_array($query)) {

echo option value=\$row['CompanyID']\$row['CompanyName']/option;

// I didn't get the if-else statement you had here before. I think it is
better to have
// the id as the value and not the name.

}
?

So if the from (which I assume you are using) is submitted you have $company
with
the CompanyID as its value which then can be inserted. If it is a multiple
select you
have to do it with an array but shouldn't be too tricky either

hope it helps
Johannes





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[PHP-DB] Re: PHP redirect?

2001-08-02 Thread Johannes Janson

Hi,

 I have a form and based off of this form if my user selects option a then
I
 want them to be redirected to another page. If they select option B then I
 want to continue with my php scrips as it does now.

 Does any one have any ideas on how I can accomplish this?

use the header() function and keep in mind: no output before the call of
the function in clear text that means no HTML, no whitespaces or anything
of this kind.

Johannes




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[PHP-DB] Re: SQL Error

2001-08-02 Thread Johannes Janson

Hi,

[...]
 Any ideas?

 $add_contact_sql = INSERT INTO $table_name
 (ContactID, FirstName, LastName, Title, WorkPhone,
 HomePhone,Mobile,EmailName,Birthday,CompanyID)
 VALUES
 ('',
 '$FirstName',
 '$LastName',
 '$Title',
 '$WorkPhone',
 '$HomePhone',
 '$Mobile',
 '$EmailName',
 '$Birthday',
 '$ContactID',

is that comma really there??? if so this might be the problem.

 );

cheers
johannes



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[PHP-DB] Re: problem with form

2001-07-10 Thread Johannes Janson

Hi,

 hope anyone has an idea on how to fix this...

 form
 input type=submit name=Submit value=Submit Request
 OnClick=submit.php class=button
 input type=submit name=Edit value=Edit Request OnClick=edit.php
 class=button
 /form

you could give the form one action and then check on the following page
wether $Submit or $Edit is set.

form.php
form action=doit.php method=post
...
/form

doit.php
?php

if(isset($submit)) {
do this, display that
}
if (isset($edit)) {
alter this, whatever
}

cheers
Johannes



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[PHP-DB] Re: MySQL Error

2001-07-10 Thread Johannes Janson

Hi,

 Dear Experts

 When ever i try to run MySQL under linux it gives me following error

 ERROR 2002 : can't connect to local MYSQL server through socket
'/var/lib/mysql.sock' (111)

this is with PHP isn't it? can you log in the monitor and then type SHOW
VARIABLES. A long list
will be displayed where the socket is listed aswell. Then check if the file
mysql.sock is in the
specified directory. You can also check in the my.cnf what the socket is.
I don't know the exact
path where my.cnf is located but I think on my SuSE box it was /etc Usually
it is in /tmp I think.

hope it helped
Johannes



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[PHP-DB] Re: Login Help

2001-07-09 Thread Johannes Janson

Hi,

[...]
Now, if i hit enter again it will say:
Please fill out all fields to proceed.
Sorry, username is already taken, please choose another.
There seems to be a problem with the database.

So this is the problem?

[...]
Here is my code:

htmlhead
titleRegister/title/head
body
form method=post action=? echo $PHP_SELF; ?
table cellpadding=2 cellspacing=0 border=0
tdUsername:/tdtdinput type=text name=username size=10/tdtr
tdPassword:/tdtdinput type=password name=password
size=10/tdtr
tdEmail:/tdtdinput type=text name=email size=15/tdtr
td /tdtdinput type=submit name=submit value=register/td
/table/form
/body
/html
?php
include'dbcon.inc';
if(isset($submit)) {
dbConnect('login');

// Make sure all fields are filled out
if ($username== or $password== or $email==) {
echo(Please fill out all fields to proceed.br);

--exit;
}

You need to add an exit; to tell PHP to terminate the if-loop.
If oyu don't do so the rest of the script is executed and thus
the empty usrname and password are entered. That's why you
get usrname already taken after the second submit.


// Make sure there is not the same name in the database
$query = SELECT COUNT(*) FROM user WHERE name = '$username';
$result = mysql_query($query);
if (!$result) {
echo(There seems to be a problem with the database.br);
}
if (mysql_result($result,0,0)0) {
echo(Sorry, username is already taken, please choose another.br);

--exit;
}

here again. but if your username is unique entering it to the DB won't
work anyway and you could leave it, but I think it is cleaner to put
it there.

You could also put the whole processing into a seperate script and then use
header(...)s to redirect back to the loginscreen.

hope it helps
Johannes




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Re: [PHP-DB] Header error

2001-06-29 Thread Johannes Janson

Hi,

[...]
 Warning: Cannot add header information - headers already sent by (output
 started at /home/thesis/public_html/registro.php:2) in
 /home/thesis/public_html/registro.php on line 80
[...]

you can't have any output before calling header(). Output is any HTML or
even
whitespaces before the opening ?php-tag. You'll have to move the header
before
line 2.

cheers
Johannes




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Re: [PHP-DB] What does this error mean?

2001-06-29 Thread Johannes Janson

Hi,

[...]
 Warning: fopen(logs/993700800.log,w+) - Permission denied in
 /home/sites/site20/users/guide/web/counter.php on line 28

are you sure you have the rights to access the file?

 Warning: Supplied argument is not a valid File-Handle resource in
 /home/sites/site20/users/guide/web/counter.php on line 29

fopen failed and so fgets, fputs and the other f* - functions (not in terms
of swearing ;-)) fail also cause $hCounter is not vaild.

 Warning: Supplied argument is not a valid File-Handle resource in
 /home/sites/site20/users/guide/web/counter.php on line 30
 1
[...]
 I'm new to this mailing list and have never used it before so I'm not sure
 which one/s to join or post this specific message to.  I apologize if this
 is the wrong forum for this type of question.  Thank you for your time.
[...]

perhaps better in general but hey, it's PHP so it can't be really OT.

cheers
Johannes




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Re: [PHP-DB] PHP and Access

2001-05-24 Thread Johannes Janson

Hi,

 I normally work PHP/MySQL but now I have a customer that is adamant
 about keeping their data in an Access DB.

you could work around it by using an odbc connection from Access to MySQL.
So the administration is still done via Access but actually it is a MySQL
DB.
If you're interested see here:
http://www.devshed.com/Server_Side/MySQL/ODBC/

cheers
johannes



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Re: [PHP-DB] Help With Getting Values Please!

2001-05-23 Thread Johannes Janson

Hi,

$sql2 = SELECT ServiceDes FROM ServiceTypes WHERE ServiceID = $serviceid;

put single quotes around $serviceid.

$service_type = mysql_result($sql2, 1);

I might be perpetuating the spreading of rumors but I've heard and read
somewhere
that mysql_query() is supposed to be better in the performance. Perhaps
someone
else could clear this point for me thus stopping the rumors.

Cheers
Johannes



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Re: [PHP-DB] error

2001-05-23 Thread Johannes Janson

Hi,

 Warning: MySQL Connection Failed: Unknown MySQL Server Host '0f2scom' (2)
in
 /web/sites/139/gamie/www.game-boy-advance.f2s.com/addartikel.php on line
22
 Unable toconnect to database
[...snip...]
 mysql_connect(db.game-boy-advance.f2s.com,*,**) or die(Unable
 toconnect to database);

You need to specify the port 3306 like this
mysql_connect(db.domainname.f2s.com:3306, user, pass);

Johannes



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Re: [PHP-DB] Pick a row, any row.

2001-05-17 Thread Johannes Janson

Hi,

 How can I pick a random row?

SELECT * FROM table ORDER BY RAND() LIMIT 0, 1;
I'm not really sure but it worked with order by rand. Check
the manual on that if it doesn't work

hope it helps
Johannes



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Re: [PHP-DB] do table alias's work in PHP?

2001-05-16 Thread Johannes Janson

Hi,

 The following statement will not work:

 $sql = select * from sometable t1, sometable t2 where t1.field1 =
 t2.field2;

do you get an error? It should work.

Johannes




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Re: [PHP-DB] Detect OS

2001-05-16 Thread Johannes Janson

Hi,


 How can I detect client's OS/platform with PHP? (Win, Mac, etc.)

with PHP_OS. $os = PHP_OS;

See Chapter 8 - Constants of the manual.

Johannes



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Re: [PHP-DB] do table alias's work in PHP?

2001-05-16 Thread Johannes Janson

I'm not sure but couldn't it really be that this is
due to the MySQL version? PHP is just passing
a string which is then executed by MySQL. I think
I recall having no problems with aliases. I don't quite
remember which version it was but I think it was 3.23.*.

 Now I am using a shareware version which, perhaps, is buggy:
 3.22.24-shareware-debug

Probably is a problem of the version...
I simply had to test it again. Here's very simple a script that worked fine.


 $sql = SELECT a.id as ida, b.id as idb FROM alleuebungen a,
allevorlesungen b WHERE a.id=b.id;

 $result = mysql_query($sql) or die(mysql_error());

 while (list($ida, $idb) = mysql_fetch_array($result)){

echo $ida, $idbbr;
}

I'm using 3.23.36.

hope it helps
Johannes






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Re: [PHP-DB] Newbie: IF statement

2001-05-14 Thread Johannes Janson

Hi,

 I have tried these if statements, but without luck:

  if ($Page='') { $Page=homepage }

you use only one = So you assign  or '' or null to $page.
use if ($page == ) instead.

hope it helps
Johannes



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Re: [PHP-DB] MySq lType and PHP??

2001-05-11 Thread Johannes Janson

Hi,

 I'm trying to get MySql type of a field (as string) into a php script.

sometime the answer is easier as you think:
mysql_field_type(link_identifier);

Johannes




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Re: [PHP-DB] How to randomly select data in a field. Please help newbie.

2001-05-02 Thread Johannes Janson

Well the only way to do this properly is a cron job.
have a look at
http://forums.devshed.com/showthread.php?threadid=13642forumid=5

hope it helps

MAX [EMAIL PROTECTED] schrieb im Newsbeitrag
002301c0d2c0$b49b3c00$0300a8c0@cv2">news:002301c0d2c0$b49b3c00$0300a8c0@cv2...
 Hi,
 I'm newbie from Indonesia.
 Your query has helped me to create random text using PHP and MySQL.
 Could you please give me an example script for random text, the text
changes
 in every 24 hours/every 7 o'clock in the morning not after
refresh/reload.

 Thanks for your help.

 Max
 Indonesia



 - Original Message -
 From: Johannes Janson [EMAIL PROTECTED]
 To: [EMAIL PROTECTED]
 Sent: Monday, April 30, 2001 6:56 PM
 Subject: Re: [PHP-DB] How to randomly select data in a field. Please help
 newbie.


  Hi,
 
  you could do the random selection in MysSQL if you use it
  using a query like this.
  SELECT  FieldWithURL FROM YourTable ORDER BY RAND() LIMIT 0,1;
  This will give you a random field. If you want more fields you have to
  set different values for the limit. Like LIMIT 0,5 for the first 5 row.
  Just if you don't know the syntax of LIMIT: limit 'StartingPoint',
  'NumberOfRows'
 
  hope it helps
  Johannes
 
  Greg K [EMAIL PROTECTED] schrieb im Newsbeitrag
  9cfe9n$ggd$[EMAIL PROTECTED]">news:9cfe9n$ggd$[EMAIL PROTECTED]...
   Can someone explain to me the best way to randomly select data from
 field
   that consist of urls. The fields consist  urls that  are images of
 people.
  I
   believe I have to setup that field as a array and turn it into a
 interger.
   But I have no experience doing this can someone sort of give me a
 example
  on
   how I would do this .
  
   Thank you ..
  
   Your Help would be well appreciate it ..
  
  
  
  
  
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[EMAIL PROTECTED]
  
 
 
 
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Re: [PHP-DB] How to randomly select data in a field. Please help newbie.

2001-04-30 Thread Johannes Janson

Hi,

you could do the random selection in MysSQL if you use it
using a query like this.
SELECT  FieldWithURL FROM YourTable ORDER BY RAND() LIMIT 0,1;
This will give you a random field. If you want more fields you have to
set different values for the limit. Like LIMIT 0,5 for the first 5 row.
Just if you don't know the syntax of LIMIT: limit 'StartingPoint',
'NumberOfRows'

hope it helps
Johannes

Greg K [EMAIL PROTECTED] schrieb im Newsbeitrag
9cfe9n$ggd$[EMAIL PROTECTED]">news:9cfe9n$ggd$[EMAIL PROTECTED]...
 Can someone explain to me the best way to randomly select data from field
 that consist of urls. The fields consist  urls that  are images of people.
I
 believe I have to setup that field as a array and turn it into a interger.
 But I have no experience doing this can someone sort of give me a example
on
 how I would do this .

 Thank you ..

 Your Help would be well appreciate it ..





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Re: [PHP-DB] Variable passing

2001-04-24 Thread Johannes Janson

You can output the passed varibles from the previous page
in an input type=hidden  in the form of the following page.

Johannes


Brinzoi Constantin Aurelian [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 Hi !

 Here is the problem: I got 8 pages, shown one after another. Each have 2
 buttons (one for continue and one for abort). When I click to Continue I
 go to the next page. On the last page I have to submit ALL variables from
 these 8 pages to the database.

 My test shown that only the variables from the last page are filled-in and
 otherones are empty.

 I have tried with another php page that includes all 8 -
 include(page...). Worthless!

 What can I do?

 TIA,

 Constantin Brinzoi
 Dept. Sondaje
 IRECSON


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Re: [PHP-DB] getting rows from multiple tables into a single array ? (newbie, mysql)

2001-04-24 Thread Johannes Janson


Hi,

 $sql4=select * from quicktimes, other_images, storyboards, where
 quicktimes.spot_id, other_images.spot_id, storyboards.spot_id =
 \$spot_id\ order by date_posted desc;

this should be:
...WHERE quicktimes.spot_id = '$spot_id', other_images.spot_id = '$spot_id'
storyboards.spot_id =
'$spot_id'order by date_posted desc;

Johannes




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Re: [PHP-DB] calling a column into a popup

2001-04-23 Thread Johannes Janson

So a certain company is previously selected? And then you want to
populate the form with the info of the company?

to do this link the menu with a onChange action to a display page.

But I'm not sure if I get the layout. Could you describe it again?

Cheers
Johannes


Beckie Pack [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 That's almost it. I want to populate the menu automatically from the
 database without having to manually modify the HTML. I believe the below
 will pull the data but it doesn't modify the menu in HTML. I tried a few
 things like using the variable name as the same as the co_name but the
menu
 always comes up blank unless I put an option name in the list. it doesn't
 pass the variable for $co_name in the HTML.

 thanks,
 boo





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Re: [PHP-DB] calling a column into a popup

2001-04-22 Thread Johannes Janson

Hi,

let me get this right. What I understood is the following:

1. Your admin page where you enter the information of a company
2. A Page with a drop-down list where all the company names are listed.
   (These names should then be linked to an information page with more
info?)
3. A page where new companies can be added.

well the drop-down list you can do with a simple while and a select(html)
like
this:
$result = mysql_query("SELECT companyName FROM companyTable ORDER BY
companyName");
echo "select name=company";
while (list($c_name) = mysql_fetch_array($result)) {
echo "option name=$c_name]$c_name/option";
}

If this is what you want (what I'm not sure about) I'm glad I understood
you. If not supply more detailed information.

Cheers
Johannes

"Beckie Pack" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 I'm creating a database where I'd like to relate a company name field to
all
 the field in the database (mysql).  what i'm trying to do is have a field
 pop up with all the current company names in a pop up list on a browser.
so,
 for example, using a form i can enter all the data for a company including
 name, address, etc. what i'm trying to do is have a popup for the company
so
 if the information is already entered it is there for selection or the
 person can enter a new company if it's not there.

 any ideas?

 thanks,
 beckie


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Re: [PHP-DB] nested ifs?

2001-04-22 Thread Johannes Janson

Hi,

 This is the code that I have so far (please note, it is within a
 function)... any ideas of how I can make it work?

this arouses the assumption that is doesn't work. Did you get
errors? And could you supply more information about
your database relations especially about the one of
orders to (probably existing) "ordering persons".

And yes nesting ifs is possible.

Johannes



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Re: [PHP-DB] Loading jpg (etc)

2001-04-21 Thread Johannes Janson

Hi,

why storing the pictures in the db. just store the link to
it.

just a thought
Johannes

"Sharmad Naik" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 Hi,
 I wanted to know how to store and retrive images like jpeg,bmp from
 database.I have a table called member with mem_id and photo.
 Pls instruct how to do it
 I m using postgresql but any reference would do

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Re: [PHP-DB] Help with Apache+PHP4 on Windows ME

2001-04-20 Thread Johannes Janson

Hi,


 Does anybody know a solution? I am willing to show you my httpd.conf and
 php.ini if you like.

httpd.conf would be helpful. only the relevant stuff like AddTpye
application.

Johannes



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Re: [PHP-DB] Variables and DB

2001-04-19 Thread Johannes Janson

how about posting the code? or describe
the problem a bit closer.

Johannes

"Rui Machado" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 Hello

 I'm writting a registration form with connection to
 mysql DB, but I have a problem in the last step, I
 can't access the variables to put them in the DB.
 I'm using the same file to process the form, but the
 last step is confirmation of the data and insertion in
 the DB, here it fails because I can't submit the
 variables.

 Any clue?

 I was thinking of doing hidden types but I think this
 is not the best way.

 Use GLOBALS?

 Thanks

 Rui Machado

 __
 Do You Yahoo!?
 Yahoo! Auctions - buy the things you want at great prices
 http://auctions.yahoo.com/

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Re: [PHP-DB] Sessions w/ Communicator 4.7 Problem

2001-04-19 Thread Johannes Janson

Hi,

 $test = "The session data was transferred.";
 echo "Click this link: a href=\"test2.php?".SID."\"test2.php/a";

change this to a href=\"test2.php?id=$PHPSESSID\"test2.php/a";

you need to specify a variable name first behind the ?. It worked
for me on NS 4.75

Johannes


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Re: [PHP-DB] How to redirect a user to main page once I capture his info in a form?

2001-04-18 Thread Johannes Janson

Hi,

you have output in line 42. any html stuff. You have to send haeder
information
before any output. If you capture the info via a form you can make a
seperate
capturing script (not $PHP_SELF) to enter the info into a DB and at the end
of this
-after successfully inserting the data- redirect the user to the page of
needs.

Johannes

""Hector M Banda"" [EMAIL PROTECTED] schrieb im Newsbeitrag
004801c0c84a$fee799a0$[EMAIL PROTECTED]">news:004801c0c84a$fee799a0$[EMAIL PROTECTED]...
Hi all,
I have a simple form where I capture the information from users. Once the
information is captured, I'd like to send the users the the main page. I
already tried  the header command but I get an error like this:

Warning: Cannot add header information - headers already sent by (output
started at page.php:42) in page.php on line 45



Any ideas?




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Re: [PHP-DB] Supplied argument is not a valid MySQL result resource...

2001-04-17 Thread Johannes Janson

Hi,

you usually get the error if your query is wrong and
you try to call mysql_fetch_array in a while loop.

It's easier if you post the relevant lines of code but
I'd say there is something wrong with your query.

Johannes


""Klaus Haberkorn"" [EMAIL PROTECTED] schrieb im Newsbeitrag
9bh369$d2h$[EMAIL PROTECTED]">news:9bh369$d2h$[EMAIL PROTECTED]...
 Installation Details:
 SuSE Linux 7.1 kernel 2.4 + apache 1.3.14 + php4.0.4pl1-1 + mysql
3.23.30-2

 The following error message appears:
 "Warning: Supplied argument is not a valid MySQL result resource in
 /www/wp/php/login.php on line 29"

 Remarks:
 php4 runs fine and returns "not FALSE" for function mysql_db_query (means
 query is ok) or mysql_query.
 But the result of the query causes the error above. Any other function
using
 the query-result does so also.
 The same php-script runs fine on other systems with the same environment.
 The Access-Rights are completely "open" for the whole development site and
 all of the mysql-users have
 full access, there are only a few test-records in the table (no question
of
 size!).

 Can anybody help us with this problem?
 Thanks in Advance
 Klaus
 ([EMAIL PROTECTED] and/or [EMAIL PROTECTED])




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Re: [PHP-DB] A bit of help with a query needed.

2001-04-17 Thread Johannes Janson

Hi,

which part is not working? Put a "or die(mysql_error())" behind your
queries to see which one produces an error.

like this:  $result = mysql_query("SELECT * FROM ships ORDER BY ship",$db)
or die(mysql_error());

 while ($row = mysql_fetch_array($result)) {

 If ($result) {
 $id = $row["id"];

you should put this the other way round that is the if before the while.
If($result) will always be "true"
within the while-loop.


 While ($result) {

you don't need another while here. Leave it out.

  $result2 = mysql_query("SELECT * FROM convicts WHERE
 convicts.ship = $id ORDER BY convict",$db);

put '' around the $id thus: WHERE convicts.ship='$id'...; // you don't
neccesarily need the
convicts before your columns name. There is no possible ambuguity.

 $row2 = mysql_fetch_array($result2)
  $id = $row2["id"];
 $convict = $row2["convict"];
 $ref = $row2["ref"];
 $contact = $row2["contact"];

In your original code you don'T do anything with $id, $ref or $contact. I
guess
that is just left out ;-)

Keep on asking.
Cheers
Johannes





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Re: [PHP-DB] Supplied argument is not a valid MySQL result resource...

2001-04-17 Thread Johannes Janson

Hi,

i don't get that.
Examlpe: tablename: test , entry: id 001; name: John Doe

$result = mysql_query("SELECT * FROM test");
$row = mysql_fetch_whatever($result); // everything ok


$result = mysql_query("SELECT * FROM tet");
$row = mysql_fetch_wahtever($result); // Warning: Supplied arggument is not
a valid 

$result = mysql_query("SELECT nonexistent FROM test");
$row = mysql_fetch_whatever($result); // Warning: Supplied argument is not


$result = mysql_query("SELECT * FROM test WHERE id='002'");
$row = mysql_fetch_whatever($result); // no errormsg, $result=true but empty

That's what I get. So there must be something wrong with the query.

Johannes


""Klaus Haberkorn"" [EMAIL PROTECTED] schrieb im Newsbeitrag
9bhh1c$2gg$[EMAIL PROTECTED]">news:9bhh1c$2gg$[EMAIL PROTECTED]...
 thank You, Johannes,
 but I think we can skip this,
 we reduced the query statement to the most simple one:
 "SELECT * FROM {tablename}"
 if the tablename is wrong, then another error message is issued.
 the result is not treated by a while-loop but just by fetch_num_rows,
 which at least reports the error.
 (and, by the way, the same script runs on another system)
 so we are more looking for problems with apache versions
 and/or modules.
 we also checked access-rights, but this also causes different errors.
 any other idea?





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Re: [PHP-DB] Error messge regarding session variables

2001-04-17 Thread Johannes Janson

Hi,

cookies are sent as header information. This means you can't send
any output (html-tag, whitespaces before ?php) at all before
sending cookie/header information.
As your error message tells you the output already started in line 4.
Move the cookie-part from line 6 to the beginning and it should
work.

Johannes

"Dan Guja" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 i've got error message when i put some session variables on my pagge:

 Warning: Cannot send session cookie - headers already sent by (output
started at
/home/httpd/vhosts/coliba.freephp.digiro.net/htdocs/user/index.php:4) in
/home/httpd/vhosts/coliba.freephp.digiro.net/htdocs/user/index.php on line 6

  Anyone can expline me whta does it mean?



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Re: [PHP-DB] A bit of help with a query needed.

2001-04-17 Thread Johannes Janson

Hi,

  $result2 = mysql_query("SELECT * FROM convicts WHERE ship =
 '$id' ORDER BY convict",$db);
 $row = mysql_fetch_array($result2) // ! you are missing a ";"
this is where you get the parse error from!

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Re: [PHP-DB] PHP + Interbase

2001-04-17 Thread Johannes Janson

Hi,

check the relevant configure-flag for interbase.
most likely to be something like --with-interbase.
just type ./configure --help |more and you'll get
a lng list of options.

Johannes

[EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...

Quiero instalar PHP con apache en linux y conectarme a un
servidor interbase
corriendo en solaris. Ya tengo php, que mas necesito para
"activar" la conexion a
interbase?.

I have a database in solaris (in interbase) and I need to conect to
that database. Now there is installed PHP in linux with apache.
How I heve to do to configure PHP to conect to that database?.
What more I need? (interbase client software, a new module).

Ruben,


Bienvenido Rubén Núñez Riveros [EMAIL PROTECTED]
Centro Nacional de Computación - CNC
San Lorenzo - Paraguay
Tel.: 585-619, 585-550


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Re: [PHP-DB] Checking database if already exsists

2001-04-16 Thread Johannes Janson

Hi,

assuming you have unique passwords.
$result = mysql_query("SELECT COUNT(*) AS used FROM userTable WHERE
passowrd=$password");
$exists = mysql_fetch_object($result);

if ($exists-used) {
display error that pswd is already in use 
}

If there is an entry in your DB with the password the _fetch_object contains
"used".
You can do the same thing for username and e-mail.

Johannes

""DC"" [EMAIL PROTECTED] schrieb im Newsbeitrag
9bfam9$jlm$[EMAIL PROTECTED]">news:9bfam9$jlm$[EMAIL PROTECTED]...
 Hi again

 How do i check my database when someone trys to become a member to check a
 see if

 username
 password
 email address

 is already in use and show the the bit they need to change for my script
to
 allow them to register.

 Thanks again in advance

 Dave C



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Re: [PHP-DB] Permanent Cookies

2001-04-16 Thread Johannes Janson

Hi,

could it be the setting of cookie_lifetime in your php.ini?
or the coolie_path?

just two thoughts

cheers
Johannes

""Lisa Elita"" [EMAIL PROTECTED] schrieb im Newsbeitrag
004f01c0c4df$1bd0e540$69fc2bca@hadinata">news:004f01c0c4df$1bd0e540$69fc2bca@hadinata...
 I tried to use cookies with this PHP code:

 SetCookie("cookiename", "cookievalue", "Friday, 16-Jan-2037 00:00:00 GMT",
 "/", ".myserver.com", 0);

 but it seems that the cookie is only store in memory (not as file).
 Can someone explain it to me...

 Best regards,
 Lisa Elita


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[PHP-DB] link_identifier in mysql_query

2001-04-10 Thread Johannes Janson

Hi,

just a theoretical question:

is it neccessary/useful/good to specify the
link_identifer for a query if I include an
ypplication file where the link is opened
in every file.

quote:
If link_identifier isn't specified, the last opened link is assumed.
If no link is open, the function tries to establish a link as if
mysql_connect
was called with no arguments, and use it.
quote_end;

So I assume the link opened by the application-include would be used
or could this cause some trouble?

thanks in advance
johannes



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Re: [PHP-DB] headers

2001-04-07 Thread Johannes Janson

Hi,

you can't have ANY output whatsoever before adding a
header. Even whitespaces before the opening ?php-tag
are not possible. You also could turn "output_buffering"
"ON" in your php.ini. But your ISP has to have them same
setting and I don't know about disadvantages. So go
with the first solution.

Johannes

"Sharmad Naik" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 When ever i add headers in my pages i get the following errors
 Can anybody tell me what's wrong?

 Warning: Cannot add header information - headers already sent by (output
started
  at /usr/src/apache_1.3.14/web/server/apache/htdocs/searchdisplay.php:4)
in /usr
 /src/apache_1.3.14/web/server/apache/htdocs/searchdisplay.php on line 6

 -Sharmad
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Re: [PHP-DB] msql_numrows HELP please !!!

2001-04-06 Thread Johannes Janson

Hi,

?php

$dbhost = "yourhost";
$dbuser = "youruser";
$dbpass = "yourpassword";
$dbname = "name_of_DB";

$connection = mysql_connect("$dbhost, $dbuser, $dbpass") or die ("can't
connect");
$database = myslq_select_db($dbname) or die ("can't select DB");

$result = mysql_query("SELECT * FROM yourtable");

while (list($date, $author, $subject, $contents) = mysql_fetch_row($result))
{

code to display the stuff you want referencing to the vars with the
given
vars in the list(...)

}

Johannes

""Bigbosss"" [EMAIL PROTECTED] schrieb im Newsbeitrag
9aket9$i1t$[EMAIL PROTECTED]">news:9aket9$i1t$[EMAIL PROTECTED]...
Hi,

I'm  a beginner in PHP dev, and i have find how to connect a DB and select a
TABLE. But i search a example to ahve the numbre of ligne in a rows on my
table.

my table is :

date
author
subject
contents



i want to print each rows with HTML tag like this :

dateauthor
subject
contents


can u help me please ??   thx




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Re: [PHP-DB] duplicate a table

2001-04-03 Thread Johannes Janson

Hi,

just install phpMyAdmin which has a feature to
duplicate tables structure only or structure and
data.

""McShen"" [EMAIL PROTECTED] schrieb im Newsbeitrag
9aak89$ruv$[EMAIL PROTECTED]">news:9aak89$ruv$[EMAIL PROTECTED]...

 how do i duplicate a table?
 i tried
 create table refer2 select * from refer;
 create table refer2 as select * from refer;

 they didn't work.
 my mysql version is  mysql 3.22.28,



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Re: [PHP-DB] Auto_Increment

2001-03-29 Thread Johannes Janson

Hi,

well usually yuto_increment is used on ID fields.
then depending on how many rows your table has
tinyint (up to 255; 1k), smallint (up to 65535; 2k)
mediumint (up to 16777215; 3k) int (up to 4294967295, 4k).

Johannes

"Ben Cairns" [EMAIL PROTECTED] schrieb im Newsbeitrag
3AC2FA65@tech-01">news:3AC2FA65@tech-01...
 What is the best field type to use for an Auto_Increment field?

 i am using int at the moment, is there a better one?

 -- Ben Cairns - Head Of Technical Operations
 intasept.COM
 Tel: 01332 365333
 Fax: 01332 346010
 E-Mail: [EMAIL PROTECTED]
 Web: http://www.intasept.com

 "MAKING sense of
 the INFORMATION
 TECHNOLOGY age
 @ WORK.."


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Re: [PHP-DB] Take out one Data and show it in page

2001-03-28 Thread Johannes Janson

Hi,

hte first part after WHERE is the column name of your db.
if you want to search with name as a criteria you write:
selelct name, nickname from table where name=$name;

johannes

"Naga Sean" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 I had a question here.
 How to pull up one database, and then view that row
 what I mean here is

 NO  NAME   AGE
 1   JOHN   12
 2   RYAN   23
 3   JIM42

 I want to take JOHN out from database using URL
 www.example.com/template.php?name=JOHN

 here is my coding so far, and I know it's wrong, But will make you guys
had the idea what I want to do.
 Thanks

 ---

 ?php
 $db = mysql_connect("localhost", "user","pass");
 mysql_select_db("media",$db);
 $result = mysql_query("SELECT nickname,name FROM database WHERE
$myrow[2]=$name",$db);
 $myrow = mysql_fetch_row($result);
 ?


 Your name $myrow[name], $myrow[age]

 _
 www.kaskus.com - FREE EMAIL SERVICE

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Re: [PHP-DB] mysql privilege

2001-03-25 Thread Johannes Janson

Hi,

to do this you'd need to be a quite priviledged user.
in order to create users you'll need acces to the
user-table of mysql. not likely to be given to a "normal"
user. developing on your home pc, well...

you need GRANT to grant priviledges to any user.
also not likely to be given to a user.

 Hello php-db,

   what privileges that i need to do this (privileges for any number):
   1 create database
   2 Give privilege to any user
   3 create user
   4
   -
   and what does GRANT and REFERENCE privileges for ?

 --
 Best regards,
  andrie  mailto:[EMAIL PROTECTED]



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Re: [PHP-DB] Yet more strings

2001-03-21 Thread Johannes Janson

Hi,

I'm not quite sure but try to play a bit with the '".$row[...]."' part.
that looks too much. leave out the "" and/or the period.

good luck
Johannes


""Mick Lloyd"" [EMAIL PROTECTED] schrieb im Newsbeitrag
00c001c0b1ea$54a5f5c0$5c0083ca@oemcomputer">news:00c001c0b1ea$54a5f5c0$5c0083ca@oemcomputer...
 CC's suggestion of using mysql_num_rows helped to clarify the problem
rather
 than find a solution! The problem is with mysql not PHP. When I use:

 $resultp = mysql_query("select Primaryid from primarycodes where Code =
 '".$row[Primaryexpertise]."'") or die (mysql_error());

 it doesn't return a resource id#, ie the query fails. mysql_num_rows
returns
 0.

 Using:

 $resultp = mysql_query("select Primaryid from primarycodes where Code LIKE
 '%$row[Primaryexpertise]%'") or die (mysql_error());

 it works, mysql_num_rows returns 1.

 I have also checked the string length of $row[Primaryexpertise] (the value
 of which is Biology for example) and it shows that there are no "hidden"
 characters in the field value (ie strlen returns 7).

 Running all this at the mysql shell (is that the word?) results in the
same
 errors.

 The row values are Primaryid (8) and Code (Biology).

 mysqlselect Primaryid,Code from primarycodes where Code like '%Biology%';
 \\returns the correct values

 mysqlselect Primaryid,Code from primarycodes where Primaryid = 8;
\\returns
 the correct values

 mysqlselect Primaryid,Code from primarycodes where Code = 'Biology';
 \\returns Empty set

 I'm no expert (obviously) but something seems odd here.
 Regards

 Mick Lloyd
 [EMAIL PROTECTED]
 Tel: +44 (0)1684 560224


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