$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
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/function.mysql-fetch-array.php
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?
$result is a variable which is passed in quotes for the function. Try removing the single quotes and see what happens.