Re: [PHP] MySql query

[Chris Kwasneski]

Try:

SELECT gallery_name,  image_id, image_name FROM galleries, Images WHERE 
galleries.gallery_id = images.gallery_id

At 07:57 AM 4/19/2002 -0700, you wrote:
Ok this could be the wrong place to ask this but there seems to be some
people in here that know mysql quite well.
My question is this; I have a database that looks like this.
galleries

gallery_id
gallery_name

snip

Images

image_id
gallery_id
image_name
etc
snip


I am trying to run a SQL query that will return an array of rows from the
images table. But I want the gallery_id number to be replaced with the
actual gallery name from the galleries table.


-Chris


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Re: [PHP] SELECT + PARSE ERROR

[Chris Kwasneski]

At 02:23 PM 4/8/2002 +0200, you wrote:
Hy,

I try to do a SELECT but when I test my code I Have a parse error.
If someone can explain to me why.

Code:

$db = mysql_connect(newsmanga_db);
$req = SELECT * FROM dvds WHERE nomdvd='.$nom..$i' ;  /// PARSE 
ERROR IN THIS LINE

Your missing a '.' after the $i .

-Chris


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RE: [PHP] Parsing error

[Chris Kwasneski]

Is it just me or is the for statement missing a begining curly bracket?

-Chris



At 11:29 AM 4/1/2002 -0600, you wrote:
which line is 45

-Original Message-
From: news.php.net [mailto:[EMAIL PROTECTED]]
Sent: Monday, April 01, 2002 10:45 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Parsing error


Hi,

I'm real new in php and trying to read a txt file

this is my code :

?php
  $row = 1;
  $fp = fopen (lecteurs.txt,r);
   while ($data = fgetcsv ($fp, 1000, ;))
   {
 $num = count ($data);
 $row++;
 for ($c=0; $c$num; $c++)
 switch ($row)
  {
  case 1 :
   {
   $vignette = $data[$c];
   }
  case 2 :
{
$photo= $data[$c];
}
case 3 :
{
$marque= $data[$c];
}
   case 4 :
{
$nom = $data[$c];
}
   case 5 :
{
$pdfproduit= $data[$c];
}
   case 6 :
{
$commprod = $data[$c];
}
 }
   fclose ($fp);
   echo $vignette;
   echo $photo;
   echo $marque;
   echo $nom;
   echo $pdfproduit;
   echo $commprod;
?

and when i run it i get : Parse error: parse error in essai.php on line 45

here' my txt file (just 1 line for probe)

petitimage;photo;marque;Nom;lepdf;commproduit

Can someone help me to go thru

TIA

Hubert





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[PHP] Form Problem

[Chris Kwasneski]

I'm having a problem with a HTML form.  When it gets submitted, I keep 
getting an undefined variable error message.  And a blank page (aside from 
a 'Hi' written on it...).  I don't think its a problem with my install of 
PHP as other PHP code is running fine, but can't get it to print out this 
name.  I'm probably missing something simple, but I just can't figure it out.

Any help would be appreciated.

the form:

html
  head
  titleMy Form/title
  /head
  body
  form action=test2.php method=GET

  My name is:
  br input type=text name=YourName

  input type=submit name=submit value=Submit /
  /form
  /body
  /html


Test2.php file:
html
head
titleForm test.../title
/head
body
Hi ?php echo $YourName; ?
/body
/html

The url that is getting passed to the file:

http://localhost/test2.php?YourName=Chrissubmit=Submit

The error message I'm getting:


[Wed Jan 09 14:24:47 2002] [error] [client 127.0.0.1] PHP 
Warning:  Undefined variable:  YourName in c:\program files\apache 
group\apache\htdocs\test2.php on line 10





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