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pass parameter

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ttuser4

ttuser4

MIS
Jun 19, 2008
147
CA
hi, i am trying to select recordset from mysql based on parameter passed from html (vzip is view)

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<meta name="GENERATOR" content="Microsoft FrontPage 4.0">
<meta name="ProgId" content="FrontPage.Editor.Document">
<title>ZIP Code</title>
</head>
<body>
<form action="zip.php" method="post">
ZIP Code: <input type="text" name="zip">
<p><input type="Submit">
<br>
<br>
</p>
</form>
</body>
</html>


zip.php

<?php
$username="root";
$password="jp";
$database="db2";

$zip=$_POST['zip'];

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query = "SELECT * FROM vzip WHERE zip='$zip' ";
mysql_query($query);

$result=mysql_query($query);
$num=mysql_numrows($result);

mysql_close();

echo "<b><center>Database Output</center></b><br><br>";

$i=0;
while ($i < $num) {

$zip=mysql_result($result,$i,"zip");
$state=mysql_result($result,$i,"state");
$city=mysql_result($result,$i,"city");
$county=mysql_result($result,$i,"ziplist_county");
$snowload=mysql_result($result,$i,"s");

echo "<br>ZIP: $zip<br>State: $state<br>City: $city<br>County: $county<br>Snow load: $snowload<br><hr><br>";

$i++;
}

?>

but instead of expected result

ZIP CITY STATE ZIPLIST_COUNTY S
54501 Rhinelander WI Oneida 60


i am getting:

Database Output

"; $zip=mysql_result($result,"zip"); $state=mysql_result($result,"state"); $city=mysql_result($result,"city"); $county=mysql_result($result,"ziplist_county"); $snowload=mysql_result($result,"s"); echo "
ZIP: $zip
State: $state
City: $city
County: $county
Snow load: $snowload

"; ?>

can anyone help me, please? this is my first project with php/mysql.
 
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First off I'd remove the @ from the mysql_select_db, if you get an error you'll never see it and secondly why are you using mysql_result rather than one of the mysql_fetch functions?
 
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thanks, i removed it so it looks now like this:

<?php
$username="root";
$password="jp";
$database="db2";

$zip=$_POST['zip'];

mysql_connect(localhost,$username,$password);
mysql_select_db($database) or die( "Unable to select database");

$query = "SELECT * FROM vzip WHERE zip='$zip'";
echo $query
mysql_query($query);

$result=mysql_query($query);
$num=mysql_numrows($result);

mysql_close();

echo "<b><center>Database Output</center></b><br><br>";

$zip=mysql_result($result,"zip");
$state=mysql_result($result,"state");
$city=mysql_result($result,"city");
$county=mysql_result($result,"ziplist_county");
$snowload=mysql_result($result,"s");

echo "<br>ZIP: $zip<br>State: $state<br>City: $city<br>County: $county<br>Snow load: $snowload<br><hr><br>";

?>


but still not getting right results, my screen says:

Database Output


"; $zip=mysql_result($result,"zip"); $state=mysql_result($result,"state"); $city=mysql_result($result,"city"); $county=mysql_result($result,"ziplist_county"); $snowload=mysql_result($result,"s"); echo "
ZIP: $zip
State: $state
City: $city
County: $county
Snow load: $snowload

"; ?>
 
Upvote 0 Downvote
i did more updates:

<?php

$username="root";
$password="jp";
$database="db2";

$zip=$_POST['zip'];

$con =mysql_connect(localhost,$username,$password);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db($database) or die( "Unable to select database");

$query = "SELECT * FROM vzip WHERE zip='$zip'";
mysql_query($query, $con);

$zip=mysql_result($result,0,"zip");
$state=mysql_result($result,0,"state");
$city=mysql_result($result,0,"city");
$county=mysql_result($result,0,"ziplist_county");
$snowload=mysql_result($result,0,"s");

mysql_close($con);

echo "<br>ZIP: $zip<br>State: $state<br>City: $city<br>County: $county<br>Snow load: $snowload<br><hr><br>";

?>


but still no result :-(
 
Upvote 0 Downvote
Well mysql_result() takes a parameter that is supposed to be the the handle to the result set from the query. $result, however your variable $result does not get set anywhere.

Try this:

Code: 
[red]$result=[/red]mysql_query($query,$con);

I'm more surprised however that you aren't getting any errors from that.


You could of course use the mysql_fetch array method as per Ingresman's suggestion:

Code: 
$query = "SELECT * FROM vzip WHERE zip='$zip'";
$result=mysql_query($query, $con);
while ($row=mysql_fetch_array($result)
{

$zip=$row['zip'];
$state=$row['state'];
$city=$row['city']
$county=$row['county'];
$snowload=$row['county'];
echo "<br>ZIP: $zip<br>State: $state<br>City: $city<br>County: $county<br>Snow load: $snowload<br><hr><br>";
}
mysql_close($con);

----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.
 
Upvote 0 Downvote
I wonder why php didn't report that $result is not a valid handle?
 
Upvote 0 Downvote
i am sorry, it was my mistake how i used it; i am running XAMPP. the major problem wasn;t php.

my zip.php worked fine if i called it directly -
<!DOCTYPE HTML PUBLIC
"-//W3C//DTD HTML 4.01 Transitional//EN"
"<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Wines</title>
</head>
<body>
<pre>
<?php

$zip=$_POST['zip'];

$connection = mysql_connect("localhost","root","jp");

mysql_select_db("db2", $connection);

$result = mysql_query ("SELECT * FROM
vzip WHERE zip='54501'", $connection);

//$result = mysql_query ("SELECT * FROM
// vzip WHERE zip='$zip'", $connection);

while ($row = mysql_fetch_array($result, MYSQL_NUM))
{

foreach ($row as $attribute)
print "{$attribute} ";

print "\n";
}
?>
</pre>
</body>
</html>

when open my zip.html - file:///C:/xampp/htdocs/xampp/zip.html - it didn't work

i have to call it using localhost - it and it works fine.

my apologies and thanks for your help.
 
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