Re: [PHP-DB] PHP_SELF and parse error

[Chris Wright]

Try using $_SERVER[PHP_SELF] instead of $PHP_SELF.

I believe $PHP_SELF is only set on servers with register variables turned
on.


ChrisPW

- Original Message - 
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, February 10, 2004 12:49 AM
Subject: [PHP-DB] PHP_SELF and parse error


 Dear friends,

 I tried to modify submit if elseif statement. That didn't work.I have
pasted
 code of form and output with error.
 Any comments on erros, please.

 Asif



 --

 ---
 Notice: Undefined variable: submit in C:\webroot\display.php on line 16

 Notice: Undefined variable: PHP_SELF in C:\webroot\display.php on line 19


 1
 Treatment of Hypertension

 a)loop diuritics
 b)clonazepam
 c)benzodizap
 d)lorazepam

 2
 Thrombolytic treatment of Myocardial infarction

 a)streptokinase
 b)diltiazm
 c)aspirin
 d)heparin


 --

 -
 //code for above output and error


 HTMLbody bgcolor=?php
 // open the connection
 $conn = mysql_connect(localhost, , );

 // pick the database to use
 mysql_select_db(testDB,$conn);


 // create the SQL statement
 $sql = SELECT * FROM testTable;

 // execute the SQL statement
 $result = mysql_query($sql, $conn) or die(mysql_error());
 //go through each row in the result set and display data

 if (!$submit) {


  echo form method=post action=$PHP_SELF;
  echo table border=0;



 while ($newArray = mysql_fetch_array($result)) {
  // give a name to the fields
 $id  = $newArray['id'];
 $testField = $newArray['testField'];
   $testFielda = $newArray['testFielda'];
   $testFieldb = $newArray['testFieldb'];
   $testFieldc = $newArray['testFieldc'];
   $testFieldd = $newArray['testFieldd'];

 //echo the results onscreen


 echo
 trtd colspan=4brb$id/b/td/tr;
 echo
 TD$testFieldinput type=hidden name=$testField value=\$testField\
br
 /td/tr
 TDa)input type=radio name=$testFielda
 value=\$testFielda\$testFielda/td/tr
 TDb)input type=radio name=$testFieldb
 value=\$testFieldb\$testFieldb/td/tr
 TDc)input type=radio name=$testFieldc
 value=\$testFieldc\$testFieldc/td/tr
 TDd)input type=radio name=$testFieldd
 value=\$testFieldd\$testFieldd/td/tr;
 }
 echo /table;


 echo input type='submit' value='See how you did' name='submit';

 echo /form;

 }

 elseif ($submit)

 ?
 /html


-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

Re: [PHP-DB] Cookies ?

[Chris Wright]

Try using $_COOKIE['cookie_name']


- Original Message - 
From: Omelin Morelos [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, February 10, 2004 12:55 AM
Subject: [PHP-DB] Cookies ?


 Friends
 
 Is there a way to read a cookie's with php , that were generated with
 javascript ?
 
 I am now usign :
 
 $xidp=$_COOKIE;
 print Xidp $xidp;
 
 Thank you for your help in advanced.
 
 -- 
 PHP Database Mailing List (http://www.php.net/)
 To unsubscribe, visit: http://www.php.net/unsub.php
 

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

[PHP-DB] mssql_bind problems

[Chris Wright]

I am running PHP 4.3.2 and MS-SQL 2000.   I am trying to run some sample
code to make sure all works well.
The code below works fine.   It simply inserts a record into a table.  This
is done in the stored procedure code on the SQL server.

Code that works:
$myServer = 192.168.200.28;
$myUser = sa;
$myPass = xx;
$myDB = Northwind;

$s = @mssql_connect($myServer, $myUser, $myPass)
or die(Couldn't connect to SQL Server on $myServer);

$d = @mssql_select_db($myDB, $s)
or die(Couldn't open database $myDB);

$query = mssql_init(sp_addnewshipper, $s);
$result = mssql_execute($query);

However, when I try to pass a variable via the following code, I get an
error from my proxy server saying Zero sized reply.  The problem seems to
be with the mssql_bind line.

Code that doesn't work:

$myServer = 192.168.200.28;
$myUser = sa;
$myPass = xx;
$myDB = Northwind;

$s = @mssql_connect($myServer, $myUser, $myPass)
or die(Couldn't connect to SQL Server on $myServer);

$d = @mssql_select_db($myDB, $s)
or die(Couldn't open database $myDB);

$query = mssql_init(sp_getprod, $s);
$supid = 2;
mssql_bind($query, @supplierid, $supid, SQLINT2,true);
$result = mssql_execute($query);
$numProds = mssql_num_rows($result);


echo h1 . $numProds .  Product;

while($row = mssql_fetch_array($result))
{
echo li.$row[0].li;
}

?



-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

[PHP-DB] Problem loading images into MySQL

[Chris Wright]

I'm having big problems getting PHP to load images into a MySQL database. I
can and have loaded images perfectly via other routes but when I try to do
it via PHP the data always ends up corrupted. I don't get any errors but the
binary data is always knackered.

I'm using PHP 4.3.0/Apache 2.0.43 on Windows 2000 and MySQL 4.0.7-gamma
(also tested with MySQL 3.X) running on a linux box. The upload and post max
settings in PHP are both set at 20mb and the size limit to MySQL is also at
20mb. As I said I can load the same files into the database via other routes
without problems.

The code I'm using is as follows, any ideas what the problem is?

session_start();

MYSQL_CONNECT(192.168.1.10,loginname,password) or die('Unable to open
database');
 mysql_select_db(ImageDB) or die('Unable to select database');
$data = addslashes(fread(fopen($_FILES['UpImg']['tmp_name'], r),
filesize($_FILES['UpImg']['tmp_name'])));

 $Qry=INSERT INTO Images (ID,Image,mime,size,OrgName)
VALUES
(.$_SESSION['SessID'].,\$data\,\.$_FILES['UpImg']['type'].\,.$_FILE
S['UpImg']['size'].,\.$_FILES['UpImg']['name'].\);

$result=MYSQL_QUERY($Qry) or die(PDB Write ErrorP.mysql_error());


-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php