[minglou]

Hi guys,

I need help.

I want to build a phone card table rates in PHP,

basicly, customer choose coutry name from drop down list, and it will retrieve
information from mysql database, it will show like this

Coutry Name Coutry Code Area/City_Code Us Dollars Toll Free
Albania 355 0.2150 0.2450
Algeria 213 0.1704 0.2004
Algeria-mobile 213 61,98-99 0.2300 0.2600

only one coutry name each time, at the same page.


this is what i did:

Code:

<?
$localhost="localhost";
$username="xxx";
$password="yyy";
$database="zzz";

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

mysql_select_db("rates",$linkid);
$resultid=mysql_query("SELECT * From rates",$linkid);

echo"<table border=1><tr><th>Coutry Name</th>";
echo"<th>Coutry Code</th><th>Area/City_Code</th><th>Us Dollars</th><th>Toll Free</th></tr>";
while ($row = mysql_fetch_row($resultid))
{
echo"<tr>";
foreach ($row as $field)
{
echo"<td>$field</td>";
}
echo"</tr>";
}
echo"</table>";
mysql_close($linkid);
?>

but this code retrieve all the information at the same time.

what i want is drop down list, choose coutry name, and retrieve information from database.

this is database i created:

Code:

CREATE TABLE `rates` (
  `country_name` varchar(20) NOT NULL default '',
  `country_code` varchar(20) default NULL,
  `area/city_code` varchar(20) default NULL,
  `us_dollars` decimal(15,4) default NULL,
  `toll_free` decimal(15,4) default NULL,
  PRIMARY KEY  (`country_name`)
) TYPE=MyISAM;

#
# Dumping data for table `rates`
#

INSERT INTO `rates` VALUES ('Albania', '355', '', '0.2150', '0.2450');
INSERT INTO `rates` VALUES ('Algeria', '213', NULL, '0.1704', '0.2004');
INSERT INTO `rates` VALUES ('Algeria-mobile', '213', '61,98-99', '0.2300', '0.2600');

Thank you guys, anybody kindly help me please?

sean

[mdhall]

Create a form something like this. Change "name of your script" to the name of the script you posted above..

<form action="name of your script" method="post">
<select name=country_name size=1>
<option value=Albania>Albania
<option value=Algeria>Algeria
</select>
<p>
<input type=submit name=submit value="Get Rate">
</form>

In your script, right after the opening <? tag, add this...
$country_name=$_POST['country_name'];

Change your query to this...

$resultid=mysql_query("SELECT * FROM rates WHERE country_name='$country_name", $linkid);

See if that works.

[minglou]

Hi mdhall,

Thank you so much helping me!

This the code( i put in the same page ):

Code:

<form action="rates2.php" method="post">
<select name=country_name size=1>
<option value=Albania>Albania
<option value=Algeria>Algeria
</select>
<p>
<input type=submit name=submit value="Get Rate">
</form>



<?php

$country_name=$_POST['country_name'];

$localhost="localhost";
$username="xxx";
$password="yyy";
$database="zzz";

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

mysql_select_db("rates",$linkid);
$resultid=mysql_query("SELECT * FROM rates WHERE country_name='$country_name", $linkid);

echo"<table border=1><tr><th>Coutry Name</th>";
echo"<th>Coutry Code</th><th>Area/City_Code</th><th>Us Dollars</th><th>Toll Free</th></tr>";
while ($row = mysql_fetch_row($resultid))
{
echo"<tr>";
foreach ($row as $field)
{
echo"<td>$field</td>";
}
echo"</tr>";
}
echo"</table>";
mysql_close($linkid);
?>

However, i got this error:

Code:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/ddd/public_html/rates2.php on line 31

line 31 is while ($row = mysql_fetch_row($resultid)), i think the problem is $resultid, $resultid=mysql_query("SELECT * FROM rates WHERE country_name='$country_name", $linkid);

How can i correct this?

Thank you!

sean

[cj158]

$resultid=mysql_query("SELECT * FROM rates WHERE country_name='$country_name' ", $linkid);


now you try this n c........
there was a small mistake. i corrected it
good luck