RE: [PHP-WIN] Re: PHP Search DB Table

Okay, thank you all again for your input, I have tried a number of suggestions to work out what is happening. Just to reiterate here is the search form which is on a different page. <form action='datasearch.php' method='POST'> <div id ="options"> <p><label>Search by Client ID<br /><input type="text" name="searchx" /></label></p> <p><input type="submit" value="Search" /></p> </div> shown below is the datasearch.php page <?php $term = $_REQUEST['searchx']; mysql_connect("localhost", "root", "christmas") or die (mysql_error()); mysql_select_db("moneyl") or die(mysql_error()); $query = "SELECT * FROM clients WHERE clientid = '%".$term."%'"; echo $query . '<br />'; echo $term . '<br />'; $result = mysql_query($query); echo"<table border='1'>"; echo "<tr><th>Client ID</th> <th>Feeearner</th> <th>First Name</th> <th>Middle Name</th> <th>Last Name</th> <th>House/Flat Number</th> <th>Address</th> <th>Postcode</th> <th>Gender</th> <th>Date of Birth</th> <th>Landline</th>"; while($row = mysql_fetch_array( $result )) { echo "<tr><td>"; echo$row['clientid']; echo"</td><td>"; echo$row['feeearner']; echo"</td><td>"; echo$row['firstname']; echo"</td><td>"; echo$row['middlename']; echo"</td><td>"; echo$row['lastname']; echo"</td><td>"; echo$row['hfnumber']; echo"</td><td>"; echo$row['address']; echo"</td><td>"; echo$row['postcode']; echo"</td><td>"; echo$row['gender']; echo"</td><td>"; echo$row['dob']; echo"</td><td>"; echo$row['landline']; echo"</td></tr>"; } echo"</table>"; ?> The printout of the query reveals this SELECT * FROM clients WHERE clientid = '%%' the echo of the value $term just showed a blank line. Essentially the numbers entered into the search box which I am retrieving using $term = $_REQUEST['searchx']; are not being pulled through, am I using the correct funtion? I tried GET and POST just to see what would happen an no joy. So why does the above just display a blank value even when there are numbers in the search box? I am very much a novice at this so it is not impossible that I am making a very very basic mistake here but from the tutorials I have seen this should work. Many thanks for your help, this is the first time I have used this method to obtain help and I have been very impressed with both the speed and quality of the repsonse. Kind Regards Oliver > From: aristotlekhan@hotmail.co.uk > To: harlequin2@gmx.de; php-windows@lists.php.net; php-db@lists.php.net; robleyd@aapt.net.au > Date: Tue, 14 Dec 2010 12:55:31 +0000 > Subject: RE: [PHP-WIN] Re: PHP Search DB Table > > > David, Sascha thank you both for your help. Using the query > > $query = "SELECT * FROM clients WHERE clientid = '$term'"; > echo $query . '<br />'; > $result = mysql_query($query); > > as suggested printed out the below > > SELECT * FROM clients WHERE clientid = '' > > This seems to indicate that it is not seeing the value within the single quotes. However a > slight caveats to that > If I change what is contained within the quotes to a specific number then it does read it, for > example > > SELECT * FROM clients WHERE clientid = '123456'; > > So it is only when I enclose a value that it doesn't compute, I tried putting the $term > within %% like so > > $query = "SELECT * FROM clients WHERE clientid = '%".$term"%'"; > > however this just returns we page not found, possibly because I need to use the LIKE parameter > in there somewhere? > > Many thanks for all your help on this > > Kind Regards > > Oliver > > > >