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May 6th, 2009

update record error-mysql_numrows(): supplied argument is not a valid MySQL

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Hi!, I am trying to update record but an error is coming. Can any one help.
error is - mysql_numrows(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\xampp\Test\Report\Despatch\update.php on line 10
here is the script

<?
$id=$_GET['id'];
$username="root";
$password="mysql";
$database="service";
mysql_connect(localhost,$username,$password);

$query = "SELECT * FROM contacts WHERE id = '$id'";
$result = mysql_query($query);
$num = mysql_numrows($result);
mysql_close();

$i=0;
while ($i < $num) {
$first=mysql_result($result,$i,"first");
$last=mysql_result($result,$i,"last");
$phone=mysql_result($result,$i,"phone");
$mobile=mysql_result($result,$i,"mobile");
$fax=mysql_result($result,$i,"fax");

++$i;
}
?>

<form action="updated.php" method="post">
<input type="hidden" name="ud_id" value="<? echo $id; ?>">
First Name: <input type="text" name="ud_first" value="<? echo $first; ?>"><br>
Last Name: <input type="text" name="ud_last" value="<? echo $last; ?>"><br>
Phone Number: <input type="text" name="ud_phone" value="<? echo $phone; ?>"><br>
Mobile Number: <input type="text" name="ud_mobile" value="<? echo $mobile; ?>"><br>
Fax Number: <input type="text" name="ud_fax" value="<? echo $fax; ?>"><br>

<input type="Submit" value="Update">
</form>

san_cash

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since May 2009

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May 6th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

just echo this line and execute that query on your database area

PHP Syntax (Toggle Plain Text)

$query = "SELECT * FROM contacts WHERE id = '$id'";

or post your table structure..

Shanti C

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Posting Virtuoso

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May 6th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

hi,
problem is here you have forget to write (_)

PHP Syntax (Toggle Plain Text)
$num = mysql_numrows($result);
$num = mysql_numrows($result);

change it to

PHP Syntax (Toggle Plain Text)
$num = mysql_num_rows($result);
$num = mysql_num_rows($result);

hope will help u..

Aamit

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Solved Threads: 15

Posting Whiz

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May 6th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

ya exactly,, i even don't see this before..

Shanti C

Reputation Points: 137

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Posting Virtuoso

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May 6th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

Click to Expand / Collapse Quote originally posted by Shanti Chepuru ...

ya exactly,, i even don't see this before..

thanks..mark thread as solved

Last edited by Aamit; May 6th, 2009 at 3:53 am.

Aamit

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May 6th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

Click to Expand / Collapse Quote originally posted by Aamit ...

hi,
problem is here you have forget to write (_)
PHP Syntax (Toggle Plain Text)
$num = mysql_numrows($result);
$num = mysql_numrows($result);
change it to
PHP Syntax (Toggle Plain Text)
$num = mysql_num_rows($result);
$num = mysql_num_rows($result);
hope will help u..

I tried but still the following errar is coming

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\xampp\Test\Report\Despatch\update.php on line 10

san_cash

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May 6th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

If you are not setting password then left blank or write your password

PHP Syntax (Toggle Plain Text)
$con = mysql_connect("localhost","root","");
	if (!$con)
	  {
	  die('Could not connect: ' . mysql_error());
	  }
	  else
	  {
	 // echo('Connected with Mysql');
	  }
		@mysql_select_db("service", $con);
 
$id=$_GET['id'];
echo $query = "SELECT * FROM contacts WHERE id = '$id'";
$result = mysql_query($query);
echo "num of rows".$num = mysql_num_rows($result);
$con = mysql_connect("localhost","root","");
	if (!$con)
	  {
	  die('Could not connect: ' . mysql_error());
	  }
	  else
	  {
	 // echo('Connected with Mysql');
	  }
		@mysql_select_db("service", $con);

$id=$_GET['id'];
echo $query = "SELECT * FROM contacts WHERE id = '$id'";
$result = mysql_query($query);
echo "num of rows".$num = mysql_num_rows($result);

check what result you get..

Aamit

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Solved Threads: 15

Posting Whiz

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May 7th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

hi amith, i tried that too but still its telling the error.

Click to Expand / Collapse Quote originally posted by Aamit ...

hi,
problem is here you have forget to write (_)
PHP Syntax (Toggle Plain Text)
$num = mysql_numrows($result);
$num = mysql_numrows($result);
change it to
PHP Syntax (Toggle Plain Text)
$num = mysql_num_rows($result);
$num = mysql_num_rows($result);
hope will help u..

san_cash

Reputation Points: 10

Solved Threads: 0

Newbie Poster

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since May 2009

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May 7th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

Click to Expand / Collapse Quote originally posted by Aamit ...

If you are not setting password then left blank or write your password
PHP Syntax (Toggle Plain Text)
$con = mysql_connect("localhost","root","");
	if (!$con)
	  {
	  die('Could not connect: ' . mysql_error());
	  }
	  else
	  {
	 // echo('Connected with Mysql');
	  }
		@mysql_select_db("service", $con);
 
$id=$_GET['id'];
echo $query = "SELECT * FROM contacts WHERE id = '$id'";
$result = mysql_query($query);
echo "num of rows".$num = mysql_num_rows($result);
$con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } else { // echo('Connected with Mysql'); } @mysql_select_db("service", $con); $id=$_GET['id']; echo $query = "SELECT * FROM contacts WHERE id = '$id'"; $result = mysql_query($query); echo "num of rows".$num = mysql_num_rows($result);
check what result you get..

please check the post and tell me the result...

Aamit

Reputation Points: 3

Solved Threads: 15

Posting Whiz

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341 posts
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May 7th, 2009

Re: update record error-mysql_numrows(): supplied argument is not a valid MySQL

You most likely have a problem with your query.
Replace:

PHP Syntax (Toggle Plain Text)
$result = mysql_query($query);
$result = mysql_query($query);

with

PHP Syntax (Toggle Plain Text)
$result = mysql_query($query) or die(mysql_error());
$result = mysql_query($query) or die(mysql_error());

This will show the error.

Last edited by kkeith29; May 7th, 2009 at 9:26 am.

kkeith29

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