mySql query problem !!

Anon

I have a script im having a slight problem with

The page can be seen at www.rpa-solutions.com/stockist
Source code at www.rpa-solutions.com/stockist/index.phps

The script displays stockist in a county - If Lancashire is type into the text box all stockist in Lancashire will be displayed.

The problem Im having is if I type Lancashire into the text box only 5 results are displayed when there are actually 6 row in the database.

It seems to always miss out the fist row in the database. Does the problem lie with the condition statements.

Your help would be most appreciated

[deleted]

We need to see some code to help you with this, but a common mistake is that people check to see if there are results by actually reading one result:

$result = mysql_query(bla);
if ($row = mysql_fetch_array($result))
{
while ($row = mysql_fetch_array($result))
{
// print
}
}
else
{
// no results
}

The IF reads one row that never get's printed. Always use mysql_num_rows($result) instead.

A forum, a FAQ, what else do you need?

Time to YAPF!: http://www.hvt-automation.nl/yapf/

fitzbean

Why not just run the while loop -- if there are no results, it just won't execute it. No need for If or mysql_num_rows.

Anon

Here is the source code. You seem to understand my exact problem however, I am already using the method you are sugesting

<?php

require "./config.inc.php";

<head>
<title>RPA - Test stockist list</title>
<link href="stockist.css" rel="stylesheet" type="text/css">
</head><body>

<font color="#000099" face="verdana" size="2"><b>Stockist Search</b></font>
<form method="post" action=" <?php echo $PHP_SELF; ?>">
<br>

<p>Please enter a county to search: <input type="text" name="search" <?php echo "value='$search'"?>>
<input type="submit" name="Search" value="Find"></p>
<p>
<?php

$db = mysql_connect($host, $user, $pass);
mysql_select_db($database) or die(mysql_error());
$error = "<p>Sorry there is no match for this county</p>";

// echo"search=". $search; // debug!
if (isset($search))
{

$result = mysql_query( "SELECT * FROM stockist WHERE county like '%$search%'",$db) or die(mysql_error());

echo  "<table border='0' cellpadding='3' width='500'>";

if ($myrow = mysql_fetch_row($result)) {

while ($myrow = mysql_fetch_row($result)) {

printf( "<tr>

<td><b>%s</b><br>%s<br>%s, %s<br>%s<p>&nbsp;</p></td>


<td valign='top' align='right'><b>Telephone:<br>
Email:<br>
Website:</td>



<td valign='top'>
<font color='#00009'> %s</font><br>
<font color='#000099'><a href='mailto:%s'>%s</a></font><br>
<font color='#000099'><a href='%s'>%s</a></font></td>




</tr>",
$myrow[1], 
$myrow[2], 
$myrow[3], 
$myrow[4], 
$myrow[5], 
$myrow[6],
$myrow[7],
$myrow[7],
$myrow[8], 
$myrow[8]);

}echo  "</table>";

} else {
echo $error;
}
}

?>
<p> </p>
</form>
</body>