[anchitjindal07]

$selct="SELECT * FROM stu_info";

$result=mysql_query($selct,$status);
while($row = mysql_fetch_array('$result',MYSQL_BOTH))
{
echo $row['Name'],$row['Father s Name'],$row['Address'],$row['Date of birth'],$row['Course'],$row['Branch'],$row['Batch'],$row['Roll No.'],$row['Univ. Reg. No.'],$row['Gender'];
echo "<br />";
}

Above code is giving following Warning:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\EasyPHP 3.0\www\process.php on line 44

Plz help

[nimesh]

I think here's the mistake:
$result will go without quotes in mysql_fetch_array()

while($row = mysql_fetch_array($result,MYSQL_BOTH))

http://in2.php.net/manual/en/functio...etch-array.php

[anchitjindal07]

But the same warning is still there

[nimesh]

Sorry, I don't know php, so can't help much

Maybe some expert would be able to help you.

Shabbir: Can you check this one?

[shabbir]

$result is a variable which is passed in quotes for the function. Try removing the single quotes and see what happens.