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update record error-mysql_numrows(): supplied argument is not a valid MySQL
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Join Date: May 2009
Posts: 4
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Solved Threads: 0
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>
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>
just echo this line and execute that query on your database area
or post your table structure..
PHP Syntax (Toggle Plain Text)
$query = "SELECT * FROM contacts WHERE id = '$id'";
or post your table structure..
Be intelligent, But Don't try to cheat.. Be innocent But Don't get cheated..
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Join Date: Apr 2008
Posts: 293
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Solved Threads: 11
hi,
problem is here you have forget to write (_)
change it to
hope will help u..
problem is here you have forget to write (_)
PHP Syntax (Toggle Plain Text)
$num = mysql_numrows($result);
change it to
PHP Syntax (Toggle Plain Text)
$num = mysql_num_rows($result);
hope will help u..
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Join Date: May 2009
Posts: 4
Reputation:
Solved Threads: 0
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hi,
problem is here you have forget to write (_)
PHP Syntax (Toggle Plain Text)
$num = mysql_numrows($result);
change it to
PHP Syntax (Toggle Plain Text)
$num = mysql_num_rows($result);
hope will help u..
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
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Join Date: Apr 2008
Posts: 293
Reputation:
Solved Threads: 11
If you are not setting password then left blank or write your password
check what result you get..
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);
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Join Date: May 2009
Posts: 4
Reputation:
Solved Threads: 0
hi amith, i tried that too but still its telling the error.
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hi,
problem is here you have forget to write (_)
PHP Syntax (Toggle Plain Text)
$num = mysql_numrows($result);
change it to
PHP Syntax (Toggle Plain Text)
$num = mysql_num_rows($result);
hope will help u..
•
•
Join Date: Apr 2008
Posts: 293
Reputation:
Solved Threads: 11
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•
•
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If you are not setting password then left blank or write your password
check what result you get..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);
You most likely have a problem with your query.
Replace:
with
This will show the error.
Replace:
PHP Syntax (Toggle Plain Text)
$result = mysql_query($query);
PHP Syntax (Toggle Plain Text)
$result = mysql_query($query) or die(mysql_error());
Last edited by kkeith29; May 7th, 2009 at 9:26 am.
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