PHP Forum

**hotstar07** · 09-20-2008, 03:45 AM

i want to get contact number as output but i see no data and no error in code generated, plz help. i have written my code below:

<html>
<body>
<center>
<table border="1">
<form method="post" action="out.php">
<tr>
<td>Name:</td><td><input type="text" name="bhai" size="30"></td>
</tr>
<tr>
<td><input type="submit" value="get data" size="20"></td>
</tr>
</form> </table>
</center>
</body></html>
<?
$a="$_POST[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 execute qrery");
$row=mysql_fetch_array($result);
echo "contact:$a";
mysql_close();
?>

**chrishirst** · 09-20-2008, 03:52 AM

is there an entry in the database of "$_POST[name]" ?

**eugen** · 09-20-2008, 04:39 AM

You're trying to print "contact:$a" after end of page </body></html> ?

**maxxximus** · 09-20-2008, 05:39 AM

Try this for size

PHP Code:


			
<html>
<body>
<center>
<table border="1">
<form method="post" action="<?php $_SERVER['PHP_SELF'] ?>">
<tr>
<td>Name:</td><td><input type="text" name="bhai" size="30"></td>
</tr>
<tr>
<td><input type="submit" value="get data" size="20"></td>
</tr>
</form> </table>
<?php
if ($_POST)
{
$a=$_POST['bhai'];
$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 execute qrery");
$row=mysql_fetch_array($result);
echo "contact: ".$row['contact'];
mysql_close();
}
?>
</center>
</body></html>

**stoot98** · 09-20-2008, 09:40 AM

Listen to Chris, your assignment of the $a variable is to a the string "$_POST[name]" rather than the value in the posted 'name' variable.

Should be -

PHP Code:


			
$a = $_POST['name'];

**maxxximus** · 09-20-2008, 10:01 AM

Quote:

Originally Posted by stoot98

Listen to Chris, your assignment of the $a variable is to a the string "$_POST[name]" rather than the value in the posted 'name' variable.

Should be -

PHP Code:


			
$a = $_POST['name'];

The POST array will not contain the key 'name' as none exists. The POST array will contain the key of 'bhai'so it is this that needs to be accessed.

PHP Code:


			
$a = $_POST['bhai'];

**stoot98** · 09-20-2008, 11:49 AM

Neither it will.

Good spot.

**webhostingultra** · 09-20-2008, 04:49 PM

Quote:

Originally Posted by hotstar07

i want to get contact number as output but i see no data and no error in code generated, plz help. i have written my code below:

<html>
<body>
<center>
<table border="1">
<form method="post" action="out.php">
<tr>
<td>Name:</td><td><input type="text" name="bhai" size="30"></td>
</tr>
<tr>
<td><input type="submit" value="get data" size="20"></td>
</tr>
</form> </table>
</center>
</body></html>
<?
$a="$_POST[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 execute qrery");
$row=mysql_fetch_array($result);
echo "contact:$a";
mysql_close();
?>

By running the script exactly as you have it, I don't see the point in the database call. Your not using any data from the database.

You've refereed to a post variable that isn't available. But even with that change, your code doesn't make use of the DB. $a simply = Your post var, you could do that without the db call.