PHP Forum

**hotstar07** · 09-20-2008, 06:16 AM

there is no out put with this code,there is no errors as well, can some one help me, here is the code

<html>
<body>
<form method='post' action='output.php'>
<input type='text' name='f_name' size='12'>
<input type='submit' value='show'>
</form>
</body>
</html>
<?php
$a=$_POST['f_name'];
$user="root";
$host="localhost";
$password="";
$database="diki";
$connection=mysql_connect($host,$user,$password);
$db=mysql_select_db($database,$connection) or die( "Unable to select database");
$query = "select contact from member where f_name='$a'";
$result=mysql_query($query) or die ("couldnot excute query");
if ($a=f_name)
while ($row=mysql_fetch_array($result));
{
echo "$row[f_name]";
echo "$row[address]";

}
mysql_close();
php?>

**ndr** · 07:12 AM

With "</body></html>" you're telling the browser that the page is over before your script has had a chance to output something. Move them after the script.
The line that says "if ($a=f_name)" is an assignment, not a comparison. Likely not what you want.
You're using a nonstandard closing tag for PHP: use ?>

**priti** · 09-20-2008, 08:15 AM

Hi,

<html>
<body>
<form method='post' action='output.php'>
<input type='text' name='f_name' size='12'>
<input type='submit' value='show'>
</form>
</body>
</html>

The following code must be in output.php

<?php

$a=$_POST['f_name'];
$user="root";
$host="localhost";
$password="";
$database="diki";
$connection=mysql_connect($host,$user,$password);
$db=mysql_select_db($database,$connection) or die( "Unable to select database");
$query = "select contact from member where f_name='$a'";
$result=mysql_query($query) or die ("couldnot excute query");

1. IF f_name IS A STRING COMPARISION THEN KEEP IN ''
2. if($a == 'f_name') or if($a == $f_name')

if ($a=f_name) //This doesn't make any sense!!

while ($row=mysql_fetch_array($result));
{
echo "$row[f_name]";
echo "$row[address]";
}
mysql_close();

?>

There i can see some logic problems !! you are compaing and then fetching the result !!

Thanks