Re: Newbie Question $2
| From: | Ethan Rosenberg, PhD | Date: | Wed, 18 Jun 2014 04:31:43 +0000 |
| Subject: | Re: Newbie Question $2 | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-48865@lists.php.net to get a copy of this message | ||
On 06/17/2014 12:02 PM, onatawahtaw@yahoo.ca 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