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how to fix this bug


barnes

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please help me.i unable to find the following error in code 

Parse error: parse error in c:\easyphp1-8\www\insert.php on line 13

my column names in mysql(db_user)(person)

firstname

lastname

age

 

<html>

<body>

<form name=frmname" method="post" action="insert.php" enctype="multipart/form-data">

<table border="0" wodth=200 height=200 align="center">

<tr><td>Firstname</td><td><input type="text" name="fname"></td></tr><br>

<tr><td>Lastname</td><td><input type="text" name="lname"></td></tr><br>

<tr><td>age</td><td><input type="text" name="age"></td></tr><br>

 

<tr colspan="2"><td align="center"><input type="submit" value="submit"></td></tr>

</table>

</body>

</html>

 

insert.php

<?php

$con=mysql_connect("localhost","root","");

if(!$con)

{

die("connection error:".mysql_error());

}

else

{

mysql_select_db("db_user",$con);

$qry="INSERT INTO person(firstname,lastname,age)VALUES('$_POST[fname]','$_POST[lname]','$_POST[age]')

if(!mysql_query($qry,$con));

{

die("connection error:".mysql_error());

}

 

echo "record is added";

mysql_close($con);

}

?>

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you're missing a quote and a semicolon at the end of $qry

 

$qry="INSERT INTO person(firstname,lastname,age)VALUES('$_POST[fname]','$_POST[lname]','$_POST[age]');

 

if you still have problems, change it to this

 

$first = $_POST['fname'];
$last = $_POST['lname'];
$age = $_POST['age'];

$qry="INSERT INTO person(firstname,lastname,age)VALUES('$first','$last','$age');

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Also, this will not do what you want it to:

 

if(!mysql_query($qry,$con));
{
die("connection error:".mysql_error());
}

 

You must remove the ";" at the end of the first line I copied and pasted.  Otherwise the die() will execute every time, regardless of if the query succeeds or fails.

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