Parse Error

draspen2

I am getting a Parse error on this code....any ideas?

<html>
<head>
<title>Input Test</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>

<?php
if ( isset ( $first_name ) && isset ( $state )&& isset ($last_name)) {
//check input here
$dberror= "";
$ret = add_to_database ($first_name, $state, $last_name, $dberror );
if ( ! $ret )
print "Error $dberror<BR>";
else
print "Thanks you very much";
} else {
write_form();
}

function add_to_database ($first_name, $state, $last_name, &$dberror){

$username = "user77";
$password = "pass77";
$hostname = "localhost";
$db = "user77_homes";
$link = mysql_connect($hostname, $username, $password);
// or die("Unable to connect to MySQL");
if ( ! $link ) {
$dberror ="Couldn't connect to the server";
return false;
}
if ( ! mysql_select_db( $db, $link )) {
$db_error = mysql_error();
return false;
}
$query = "INSERT INTO foster_homes ( first_name, state, last_name)
values ( '$first_name', '$state', '$last_name' )";
if ( ! mysql_query( $query, $link ) ) {
$dberror = mysql_error();
return false;
}
return true;
}

function write_form() {
global $PHP_SELF;
print "<form method=\"POST\">\n";
print "<input type=\"text\" name=\"first_name\">";
print "The first name you would like<p>\n";
print "<input TYPE=\"text\" name-\"last_name\">";
print "Your last_name address<p>\n";
print "<select name=\"state\">";
print "\t<option value=\"CA\">CA\n";
print "\t<option value=\"MN\">MN\n";
print "</select>\n";
print "<input type=\"submit\" value=\"submit!\">\n</form>\n";

)
//mysql_close($link);
?>

</body>
</html>

cgraz

That's a lot of code. First of all, edit your post so your code is within [p h p] code here [ / p h p ] tags (minus the spaces). That way it's easier to follow and you'll get syntax highlighting.

Once that is done, post the parse error and the line number so we know where to look.

Cgraz

draspen2

Not sure if this is what you mean...but....here goes...

function add_to_database ($first_name, $state, $last_name, &$dberror){

)
//mysql_close($link);
?>

draspen2

Parse error: parse error in /home2/user77/public_html/insert2.php on line 61

cgraz

i meant actually write out

[ p h p ] php code here [ / p h p ] around your code (without the spacing).

Anways, look at your function write_form()

You end it with a parenthesis when you should use a closing bracket }

Cgraz

draspen2

That was it....thanks.

That cleared up the parse error now I have to find out why it is not writing to the db?

cgraz

use mysql_query($query) or die(mysql_error()

also print out the query string $query to see what it is being processed as.

Those are the two best ways to solve mysql problems.

Cgraz

keith73

Originally posted by draspen2
Not sure if this is what you mean...but....here goes...

read this and you'll know what it means.

http://www.phpbuilder.com/board/misc.php?action=bbcode

keith

draspen2

if ( isset ( $first_name ) && isset ( $state )&& isset ($last_name)) { 
//check input here 
$dberror= ""; 
$ret = add_to_database ($first_name, $state, $last_name, $dberror ); 
if ( ! $ret ) 
print "Error $dberror<BR>"; 
else 
print "Thanks you very much"; 
} else { 
write_form(); 
} 

function add_to_database ($first_name, $state, $last_name, &$dberror){ 

$username = "user77"; 
$password = "pass77"; 
$hostname = "localhost"; 
$db = "user77_homes"; 
$link = mysql_connect($hostname, $username, $password); 
// or die("Unable to connect to MySQL"); 
if ( ! $link ) { 
$dberror ="Couldn't connect to the server"; 
return false; 
} 
if ( ! mysql_select_db( $db, $link )) { 
$db_error = mysql_error(); 
return false; 
} 
$query = "INSERT INTO foster_homes ( first_name, state, last_name) 
values ( '$first_name', '$state', '$last_name' )"; 
if ( ! mysql_query( $query, $link ) ) { 
$dberror = mysql_error(); 
return false; 
} 
return true; 
} 

function write_form() { 
global $PHP_SELF; 
print "<form method=\"POST\">\n"; 
print "<input type=\"text\" name=\"first_name\">"; 
print "The first name you would like<p>\n"; 
print "<input TYPE=\"text\" name-\"last_name\">"; 
print "Your last_name address<p>\n"; 
print "<select name=\"state\">"; 
print "\t<option value=\"CA\">CA\n"; 
print "\t<option value=\"MN\">MN\n"; 
print "</select>\n"; 
print "<input type=\"submit\" value=\"submit!\">\n</form>\n"; 

) 
//mysql_close($link);

draspen2

Sorry but I can't get the to print to the screen..... I tried these two way's

print "mysql_query($query )";

print "$query ";

And nothing happened???

Thanks

draspen2

Ok...so for some reason I am not connecting to the database correctly" I get the following message when I connect:

Connected to MySQL

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/olivia77/public_html/query.php on line 20
There are currently in the table

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/olivia77/public_html/query.php on line 23

but it does not show any info? the database is called: olivia77_homes
the table is:
olivia77_homes

<?php
$username = "olivia77_draspen";
$password = "wolly";
$hostname = "localhost";	
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to MySQL");
print "Connected to MySQL<br>";
// you're going to do lots more here soon

$result = mysql_query ( "SELECT * FROM olivia77_homes" );
//$num_rows = mysql_num_rows ($results);
print "There are currently $num_rows in the table <P>";
print "<table border=1>\n";
//while ( $a_row = mysql_num_rows( $results ) ) {
	//print "<tr>\n";
	//foreach ( $a_row as $field )
	//	print "\t<td>$field</td>\n";

//print "</tr>\n";
//}
print "</table>\n";
mysql_close ($dbh);

//mysql_close($dbh);

Thanks....

Bunkermaster

try this after mysql_query (and replace all or dies) and tell us what the script returns

...or die ("your message ".mysql_error()." line ".__LINE__);

Or die alone is useless unless you know what error you have.

draspen2

I added the code and now I get the following:

Connected to MySQL

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/olivia77/public_html/query.php on line 21
There are currently in the table

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/olivia77/public_html/query.php on line 25

Bunkermaster

Originally posted by draspen2
olivia77_homes

<?php
$username = "olivia77_draspen";
$password = "wolly";
$hostname = "localhost";	
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to MySQL");
print "Connected to MySQL<br>";
// you're going to do lots more here soon

$result = mysql_query ( "SELECT * FROM olivia77_homes" );
//$num_rows = mysql_num_rows ($results);
print "There are currently $num_rows in the table <P>";
print "<table border=1>\n";
//while ( $a_row = mysql_num_rows( $results ) ) {
	//print "<tr>\n";
	//foreach ( $a_row as $field )
	//	print "\t<td>$field</td>\n";

//print "</tr>\n";
//}
print "</table>\n";
mysql_close ($dbh);

//mysql_close($dbh);

[/B]

ok my bad I hadn't seen the mistake in the code 😃

<?php
$username = "olivia77_draspen";
$password = "wolly";
$hostname = "localhost";	
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to MySQL");
print "Connected to MySQL<br>";
// you're going to do lots more here soon

$result = mysql_query ( "SELECT * FROM olivia77_homes" );
$num_rows = mysql_num_rows ($result);
print "There are currently $num_rows in the table <P>";
print "<table border=1>\n";
while ( $a_row = mysql_num_rows( $result ) ) {
	print "<tr>\n";
	foreach ( $a_row as $field ){
		print "\t<td>$field</td>\n";
	}
	print "</tr>\n";
}
print "</table>\n";

mysql_close($dbh);

Try this (difference $result and $results got you in all this)