[PHP] mysql_num_rows() error

[Phil Powell]

 Anyone know why this is happening? I have mySQL on Win2000 Server with IIS
 and PHP:

 mysql_num_rows(): supplied argument is not a valid MySQL result resource

 Here is my code:

 ?
   $conn = mysql_connect('localhost', xxx, yyy) or die('Could not
connect
 to db');
   $result = mysql_query('select * from blah');
   echo Number of rows:  . mysql_num_rows($result);
   if (mysql_num_rows($result)  0) {
while ($row = mysql_fetch_row($result)) {
 for ($i = 0; $i  sizeOf($row); $i++) {
  echo $i:  . $row[$i] . BR;
 }
}
   }
 ?

 It couldn't get any simpler.  I just installed mySQL on my machine and am
 trying to see what I can do with it giving Win2K and IIS.

 AUGH! Thanx
 Phil



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RE: [PHP] mysql_num_rows() error

[Jay \(PHP List\)]

You query is incorrect.  Try doing the same query from the MySQL clent.

|- -Original Message-
|- From: Phil Powell [mailto:[EMAIL PROTECTED]] 
|- Sent: Thursday, January 09, 2003 7:46 PM
|- To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
|- Subject: [PHP] mysql_num_rows() error
|- 
|- 
|-  Anyone know why this is happening? I have mySQL on 
|- Win2000 Server with 
|-  IIS and PHP:
|- 
|-  mysql_num_rows(): supplied argument is not a valid MySQL result 
|-  resource
|- 
|-  Here is my code:
|- 
|-  ?
|-$conn = mysql_connect('localhost', xxx, yyy) or 
|- die('Could not
|- connect
|-  to db');
|-$result = mysql_query('select * from blah');
|-echo Number of rows:  . mysql_num_rows($result);
|-if (mysql_num_rows($result)  0) {
|- while ($row = mysql_fetch_row($result)) {
|-  for ($i = 0; $i  sizeOf($row); $i++) {
|-   echo $i:  . $row[$i] . BR;
|-  }
|- }
|-}
|-  ?
|- 
|-  It couldn't get any simpler.  I just installed mySQL on 
|- my machine and 
|-  am trying to see what I can do with it giving Win2K and IIS.
|- 
|-  AUGH! Thanx
|-  Phil
|- 
|- 
|- 
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|- 
|- 


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RE: [PHP] mysql_num_rows() error

[Sean Malloy]

Are you actually checking that the query result is valid?

$conn = mysql_connect('localhost', xxx, yyy) or die('Could not connect
to db');
$result = mysql_query('select * from blah', $conn);
if ($result)
{
  echo Number of rows:  . mysql_num_rows($result);
  if (mysql_num_rows($result)  0)
  {
while ($row = mysql_fetch_row($result))
{
  for ($i = 0; $i  sizeOf($row); $i++)
  {
echo $i:  . $row[$i] . BR;
  }
}
  }
}
else
{
  echo mysql_error($conn);
}

-Original Message-
From: Phil Powell [mailto:[EMAIL PROTECTED]]
Sent: Friday, 10 January 2003 1:46 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: [PHP] mysql_num_rows() error


 Anyone know why this is happening? I have mySQL on Win2000 Server with IIS
 and PHP:

 mysql_num_rows(): supplied argument is not a valid MySQL result resource

 Here is my code:

 ?
   $conn = mysql_connect('localhost', xxx, yyy) or die('Could not
connect
 to db');
   $result = mysql_query('select * from blah');
   echo Number of rows:  . mysql_num_rows($result);
   if (mysql_num_rows($result)  0) {
while ($row = mysql_fetch_row($result)) {
 for ($i = 0; $i  sizeOf($row); $i++) {
  echo $i:  . $row[$i] . BR;
 }
}
   }
 ?

 It couldn't get any simpler.  I just installed mySQL on my machine and am
 trying to see what I can do with it giving Win2K and IIS.

 AUGH! Thanx
 Phil



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RE: [PHP] mysql_num_rows() error

[Mark Charette]

And which database are you using?? (hint - mysql_select_db() might be a good
idea ...)

 -Original Message-
 From: Phil Powell [mailto:[EMAIL PROTECTED]]
  Anyone know why this is happening? I have mySQL on Win2000
 Server with IIS
  and PHP:



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[PHP] mysql_num_rows error

[Omar Campos]

I have a file. I'll paste from line 32 to 34

?
.
$sql = select USUARIO from docente where USUARIO = '$usuario';
$result = mysql_query($sql, $link);
if (mysql_num_rows($result) == 1) { // line 34

?

I get the next warning:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /home/olimpiad/public_html/base/alta1.php on line 34

I'm running the script on a linux server.
could someone help me.
Thank you.



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Re: [PHP] mysql_num_rows error

[Stuart Dallas]

Omar Campos wrote:
 I get the next warning:
 Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
 resource in /home/olimpiad/public_html/base/alta1.php on line 34

This usually means that the SQL query failed. You can check the value 
returned by mysql_query to see if this is the case - it will be FALSE if 
it failed. In this case my guess would be that the backticks are not 
valid string delimeters.

-- 
Stuart


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Re: [PHP] mysql_num_rows error

[Wilmar Perez]

Well, those error normaly appear when you've got something wrong in your sql 
statement.  Make sure there is some information stored in your database, 
empty tables are sometimes troublesome.

What do you use $link for?


$sql =  select USUARIO from docente where USUARIO = '$usuario';
$result = mysql_query($sql, $link);
if (mysql_num_rows($result) == 1)

Bye

***
 Wilmar Pérez
 Network Administrator
   Library System
  Tel: ++57(4)2105145
University of Antioquia
   Medellín - Colombia
  2002
***
 
 

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RE: [PHP] mysql_num_rows error

[John Holmes]

 I am new to php and that the folowing error: Warning:
mysql_num_rows():
 supplied argument is not a valid MySQL result resource in
 /home/tbonestu/public_html/smallimages.php
 
 i dont know what i am doing wrong here is the code:
 
   $db = mysql_pconnect(connect info);
 
  mysql_select_db(images);
  $query = select * from images where type =.$type.;
  $result = mysql_query($query);
  $num_results = mysql_num_rows($result);

That generally means your query failed. Try to echo mysql_error() after
your query to see what the problem is. You probably need quotes around
$type or $type has no value.

---John Holmes...


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RE: [PHP] mysql_num_rows error

[M . A . Bond]

 I am new to php and that the folowing error: Warning:
mysql_num_rows():
 supplied argument is not a valid MySQL result resource in 
 /home/tbonestu/public_html/smallimages.php
 
 i dont know what i am doing wrong here is the code:
 
   $db = mysql_pconnect(connect info);
 
  mysql_select_db(images);
  $query = select * from images where type =.$type.;  $result = 
 mysql_query($query);  $num_results = mysql_num_rows($result);

That generally means your query failed. Try to echo mysql_error() after
your query to see what the problem is. You probably need quotes around $type
or $type has no value.

You also have too many quotes on the $query line, try echoing your $query
variable to see if it reads properly.

Mark



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RE: [PHP] mysql_num_rows error

[Jon Haworth]

Hi Nick,

 I am new to php and that the folowing error: 
 Warning: mysql_num_rows(): supplied argument 
 is not a valid MySQL result resource in
 /home/tbonestu/public_html/smallimages.php
 
 i dont know what i am doing wrong here is the code:
 
   $db = mysql_pconnect(connect info);

Along with the other suggestions to check your $query variable, you should
change this line to :

$db = mysql_pconnect($whatever);

The  goes in front of the function call, not the variable, if you want to
suppress any potential error messages generated by the function.


Cheers
Jon

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[PHP] mysql_num_rows error

[Nick Varner]

I am new to php and that the folowing error: Warning: mysql_num_rows():
supplied argument is not a valid MySQL result resource in
/home/tbonestu/public_html/smallimages.php

i dont know what i am doing wrong here is the code:

  $db = mysql_pconnect(connect info);

 mysql_select_db(images);
 $query = select * from images where type =.$type.;
 $result = mysql_query($query);
 $num_results = mysql_num_rows($result);





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Re: [PHP] mysql_num_rows error

[Tom Rogers]

Hi,

Sunday, September 23, 2001, 3:25:02 PM, you wrote:
NV I am new to php and that the folowing error: Warning: mysql_num_rows():
NV supplied argument is not a valid MySQL result resource in
NV /home/tbonestu/public_html/smallimages.php

NV i dont know what i am doing wrong here is the code:

NV   $db = mysql_pconnect(connect info);

NV  mysql_select_db(images);
NV  $query = select * from images where type =.$type.;
NV  $result = mysql_query($query);
NV  $num_results = mysql_num_rows($result);


Do it this way

$query = select * from images where type = '$type';

or if you prefer
$query = select * from images where type = '.$type.';


PS check your comps date its a year out


-- 
regards,
Tom


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