Hi Aaron,
I noticed several things, first you are fetching your database row into
$rows but you are referencing the data in $row, and second you do not
have a $ in front of your $row variable name, in order to get the
contents of a variable you must use $variable.
I've included modified code to demonstrate.
Jason
while ($rows=mysql_fetch_array($mysql_result)) {
$ID=row['ID'];
# this is line 30
$first_name=row['first_name'];
$surname=row['surname'];
$email=row['email'];
$comments=row['comments'];
needs to be
while ($row = mysql_fetch_array($mysql_result)) {
$ID=$row['ID']; # this is
line 30
$first_name=$row['first_name'];
$surname=$row['surname'];
$email=$row['email'];
$comments=$row['comments'];
On Tue, 2003-02-04 at 21:44, Aaron Downes wrote:
> Dear Sir/Madam,
>
> trying to displaying records by php get error :
> Parse error: parse error in
>/hsphere/local/home/ozhomene/ardownes.com/ViewContacts.php on line 30
>
> see below for code for ViewContacts.php :
>
> <html>
> <body>
>
> <?
>
> $connection=mysql_connect("localhost","",""); #am entering userID &password
>instead of ,"","")
>
> IF (!$connection) {
> echo "Could not connect to mySQL server";
> exit;
> }
>
> $db=mysql_select_db("test_holding",$connection);
>
> if (!$db) {
> echo "Could not change the database";
> exit;
> }
>
> #sql="SELECT * FROM contacts"; # name of table "contacts" with below
>fields
>
> #mysql_result=mysql_query($sql,$connection);
> $num_rows=mysql_num_rows($mysql_result);
>
> if ( $num_rows == 0 ) {
> echo "Sorry, there is no information.";
> } else {
> #we have results
> while ($rows=mysql_fetch_array($mysql_result)) {
> $ID=row['ID']; #
>this is line 30
> $first_name=row['first_name'];
> $surname=row['surname'];
> $email=row['email'];
> $comments=row['comments'];
> # display results
> echo "$fist_name : $surname $email $comments<br>";
> }
> } # end else
>
> mysql_close($connection);
>
> ?>
>
> </body>
> </html>
>
> Any help will be appreciated.
>
> Best regards,
> Aaron Downes
>
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