Switch error

shalomajay

hey all,

any idea why my code's not working? Something to do with the Switch if/else bit.
Im getting the following error:

Parse error: parse error in c:\program files\easyphp1-8\www\mine\www\aa1.php3 on line 34

regards,
jo

?>
<html>
<head><title>Member Address or Phone Number</title></head>
<body>
<?php
include("dbinfo.inc.php");
mysql_connect($localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

switch ($_POST['display_button']) {
case 'Show Address':
{
$query = "SELECT country_name,currency_code FROM country";
$result = mysql_query($query)
or die ("Couldn’t execute query.");
$row = mysql_fetch_array($result);
extract($row);
echo "$country_name<br>$currency_code<br>";
break;
}

  case 'Show Currency Code':
 {

$query = "SELECT currency_code FROM country ";
$result = mysql_query($query)
or die ("Couldn’t execute query.");
$row = mysql_fetch_array($result);
echo "Phone: {$row['currency_code']}<br>";
break;
}

?>
</body></html>

Weedpacket

You're missing the } that ends the block started on the switch(...) { line.

shalomajay

In the second part of this switch case ie:case 'Show Phone Number':, I'm trying to display more than 1 record but im getting some error. The first switch case is working though.

Any ideas
Many thanks
regards,
Jo

<html>
<head><title>Member Address or Phone Number</title></head>
<body>
<?php
include("dbinfo.inc.php");
mysql_connect($localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

switch ($_POST['display_button']) {
case 'Show Address':
{
$query = "SELECT * FROM country";
$result = mysql_query($query)
or die ("Couldn’t execute query.");
$row = mysql_fetch_array($result);
extract($row);
echo "$country_name<br>$currency_code<br>$comments<br>";
break;
}

 case 'Show Phone Number':

<html>

<TABLE width="90%" border="1">
<TR bgcolor="Gold">
	<TD>Country</TD>
	<TD>Currency Code</TD>
	<TD>Comments</TD>
</TR>

<?
$query ="select * from country";

$result = mysql_query($query);

?>
<?
while($data = mysql_fetch_array($result,MYSQL_BOTH))
{

<TR>
	<TR bgcolor="Red">
	<TD><A HREF="#"><?echo $data[0];?></A></TD>
	<TD><?echo $data[1];?></TD>
	<TD><INPUT TYPE="text" VALUE="<?echo $data[2];?>" NAME="TEXT_BOX"></TD>
	<TD><INPUT TYPE="submit"></TD>
</TR>

<?
}
?>

}
?>
</body></html>

rincewind456

Check your opening and closing php tags you have to many of one and not enough of the other.

BTW can you use

 tags to enclose your code.

shalomajay

No luck so far. I'll keep trying...in the meanwhile if at all anyone get ideas, give me a shout.
In the above code , i'm getting the error near the html line as below

case 'Show Phone Number':

<html>

regards,
jo

Weedpacket

Yes, because you're missing the ?> to separate the PHP from the following HTML.

And like rincewind456 suggests, use

 tags when posting PHP so that it looks like[code=php]<html>
<head><title>Member Address or Phone Number</title></head>
<body>
<?php
include("dbinfo.inc.php");
mysql_connect($localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

switch ($_POST['display_button']) {
case 'Show Address':
{
$query = "SELECT * FROM country";
$result = mysql_query($query)
or die ("Couldn’t execute query.");
$row = mysql_fetch_array($result);
extract($row);
echo "$country_name<br>$currency_code<br>$comments<br>";
break;
}

case 'Show Phone Number':

<html>

<FORM METHOD="POST" ACTION="view.php" name="formname">

<TABLE width="90%" border="1">
<TR bgcolor="Gold">
<TD>Country</TD>
<TD>Currency Code</TD>
<TD>Comments</TD>
</TR>

<?
$query ="select * from country";

$result = mysql_query($query);

?>
<?
while($data = mysql_fetch_array($result,MYSQL_BOTH))
{

?>

<TR>
<TR bgcolor="Red">
<TD><A HREF="#"><?echo $data[0];?></A></TD>
<TD><?echo $data[1];?></TD>
<TD><INPUT TYPE="text" VALUE="<?echo $data[2];?>" NAME="TEXT_BOX"></TD>
<TD><INPUT TYPE="submit"></TD>
</TR>


<?
}
?>


}
?>
</body></html>

You'll notice that your current problem becomes a lot more visible.