• Status: Solved
  • Priority: Medium
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  • Last Modified:

duplication insert into SQL

Hay im getting error with line 40 can i get some help its to stop dupication incert into SQL

<?php 

// Read values from form using $_POST (safest) 


$fname=$_POST["vname"]; 


// Connect to server 
// Replace username and password by your details  

$db = @mysql_connect("localhost","root","iqonr301"); 
if (!$db) 
{ 
        do_error("Could not connect to the server"); 
} 


// Connect to the database 
// Note that your database will be called username 

@mysql_select_db("test",$db)or do_error("Could not connect to the database"); 

// Run query 

$sql="INSERT INTO teacher (name) values ('$fname')"; 
$dupesql = "SELECT * FROM table where (name = '$fname')";

$duperaw = mysql_query($dupesql);

if (mysql_num_rows($duberaw) > 0) {
  //your code ...
}
if (mysql_query($sql,$db)) 
{ 
        echo "The following Teacher"; 
        echo "<br>Record: $fname  has been added to          
        database.<p>"; 
} 
else 
{ 
        do_error("Failed to add record"); 
} 

function  do_error($error) 
{ 
        echo  $error; 
        die; 
} 

?>

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Notice: Undefined variable: duberaw in C:\xampp\htdocs\insert.php on line 40
 
Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\xampp\htdocs\insert.php on line 40
 Failed to add record
0
paddy086
Asked:
paddy086
  • 2
3 Solutions
 
Chris StanyonCommented:
Line 29 says $duperaw
line 31 says $duberaw

check your spelling :)
0
 
Meir RivkinFull stack Software EngineerCommented:
change line 29 to:
$duperaw = mysql_query($dupesql, $db);

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0
 
Chris StanyonCommented:
Line 26 is INSERTing from a table called 'teacher'
Line 27 is SELECTing from a table caled 'table'
0

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