On 6/18/2014 12:31 AM, Ethan Rosenberg, PhD wrote:
On 06/17/2014 12:02 PM, onatawahtaw@xxxxxxxx wrote:
Hi Ethan,
Here are some things to clean up your code:
Your line:
$phn = $_POST[phone];
should use quotations as follows:
$phn = $_POST['phone'];
Your line:
$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
Should use double quotes if you need the variable to be interpreted:
$sql1 ="select Lname, Fname from Customers where Phone = $Phn ";
Lastly, as people have mentioned PDO is probably the best way to go.
Try connecting to your database with PDO. Look on Google for "PDO
prepared statements" and use those instead of the mysql escape string
method.
Hope this helps,
-Kevin
Sent from Yahoo Mail on Android
IT WORKS!!!
Here is the code -
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
<html xmlns="http://www.w3.org/1999/xhtml";>
<html>
<?php
$bla = 1;
?>
<head>
</head>
<body>
<div align="center">
<form method="post">
<input type='text' name=phone></input>
<input type='submit'>
<br /><br /><br />
</form>
</div>
<?php
error_reporting(-1);
require '/home/ethan/PHP/ethan.inc';
$db = "Store";
$cxn = mysqli_connect($host,$user,$password,$db);
$phn = $_POST[phone];
$phn = (string)$phn;
$dsh = '-';
$Phn =
$phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];
$sql1 ="select Lname, Fname from Customers where Phone =
'$Phn' ";
$result1 = mysqli_query($cxn, $sql1);
if(!$result)
{
?>
<div align="center">
<strong>No Match Found</strong>
<br /><br />
</div>
<?php
}
?>
<div align="center">
<table border="4" cellpadding="5" cellspacing="55"
rules="all" frame="box">
<tr class='heading'>
<th>Last Name</th>
<th>First Name</th>
<?php
while($row1 = mysqli_fetch_row($result1))
{
$Lname = $row1[0];
$Fname = $row1[1];
?> <tr>
<td> <?php echo $Lname; ?> </td>
<td> <?php echo $Fname; ?> </td>
</tr>
<?php
}
?>
</table>
</div>>
</body>
</html>
As you [those that replied] accurately noted, the problem was with the
quoting.
I appreciate all your comments, take them seriously and will use the
information contained in them for future programming.
No matter how much skill in programming I have, I will remain a NEWBIE;
ie, someone who wishes to grrow in knowledge and acknowledges that there
are many programmers much more skilled than I.
Thanks again.
Ethan
happy to hear you got it working. Sad to see that you didn't heed the
tips provided to you and alter your code, and that you still have errors
in it. oh, well....
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