Parse Error

nogard

Help me find my code mistake please?

<body>
<?php
$db = mysql_connect("localhost", "user", "pw");
CREATE TABLE event (
name VARCHAR(25),
date DATE,
type VARCHAR(15),
remark VARCHAR(255)
)
;

INSERT INTO event VALUES
(Fluffy, 1995-05-15, litter, 4 kittens, 3 female, 1 male) ;
(Buffy, 1993-06-23, litter, 5 puppies, 2 female, 3 male);
(Buffy, 1994-06-19, litter, 3 puppies, 3 female);
(Chirpy, 1999-03-21, vet, needed beak straightened);
(Slim, 1997-08-0,3 vet, broken rib);
mysql_select_db("coachin3",$db);
$result = mysql_query("SELECT * FROM event",$db);
echo "<table border=1>\n";
echo "<tr><td>Name</td><td>Type</tr>\n";
while ($myrow = mysql_fetch_row($result)) {
printf("<tr><td>%s %s</td><td>%s</td></tr>\n", $myrow[1], $myrow[2], $myrow[4]);
}
echo "</table>\n";
?>
</body>

nogard

I believe it is a simple puncuation mistake of a bracket { or a parenthese ( which both look way too similar for me to distinguish between OR
it is the use of , rather than ; in creating the table. My parse error begins at the CREATE TABLE line of code

[deleted]

Uhm... you are putting raw SQL statements into a PHP script....

nogard

OK
Still learning how to combine mysql code into php document. So I'll take this 1 step at a time ...
Tried this code to create table and still have SQL error so have not attempted to insert info yet.

<body>
<?php
$db = mysql_connect("localhost", "user", "pw");
$query = "CREATE TABLE event (";
$query = "name VARCHAR(25),";
$query = "date DATE,";
$query = "type VARCHAR(15),";
$query = "remark VARCHAR(255)";
$query = ")";
echo "result: $query";
$result = mysql_db_query ("coachin3", $query);
if (mysql_error()) {
echo "< HR/> error: ". mysql_error(). "< HR/>";
}
echo "< HR/> Query sent $result < HR/>";
mysql_close ($db);
?>
</body>

Anon

You are reassigning $query on every one of those lines. Perhaps you want to use the .= operator which will concatenate.

$query = "123";
$query .= "abc";
If you were to look at $query now it would be "123abc".

Secondly, you do not have to split it your $query string into multiple pieces. If you are doing that so that your code is more readable, you can just use whitespace to do that. Neither PHP nor MySQL care much about whitespace.

For example:
$query = "CREATE TABLE event (
name VARCHAR(25),
date DATE,
type VARCHAR(15),
remark VARCHAR(255)
)";

Anon

Also remember to use mysql_select_db. Otherwise, mysql won't know which database you want to use.

nogard

Ok Thanks

still seem be having trouble with insert into

<body>
<?php
$db = mysql_connect("localhost", "user", "pw");
$query = "CREATE TABLE event (
name VARCHAR(25),
date DATE NOT NULL,
type VARCHAR(15),
remark VARCHAR(255)
)";
$row = "INSERT INTO event VALUES (
(Fluffy, 1995-05-15, litter, 4 kittens, 3 female, 1 male) ,
(Buffy, 1993-06-23, litter, 5 puppies, 2 female, 3 male),
(Buffy, 1994-06-19, litter, 3 puppies, 3 female),
(Chirpy, 1999-03-21, vet, needed beak straightened),
(Slim, 1997-08-03, vet, broken rib)
)";
mysql_select_db("coachin3",$db);
$result = mysql_query("SELECT * FROM event",$db);
echo "<table border=1>\n";
echo "<tr><td>Name</td><td>Type</tr>\n";
while ($myrow = mysql_fetch_row($result)) {
printf("<tr><td> %s</td><td>%s</td></tr>\n", $myrow[1], $myrow[2], $myrow[4]);
}
echo "</table>\n";
mysql_close ($db);
?>
</body>

matto

$row = "INSERT INTO event VALUES (
(Fluffy, 1995-05-15, litter, 4 kittens, 3 female, 1 male) ,
(Buffy, 1993-06-23, litter, 5 puppies, 2 female, 3 male),
(Buffy, 1994-06-19, litter, 3 puppies, 3 female),
(Chirpy, 1999-03-21, vet, needed beak straightened),
(Slim, 1997-08-03, vet, broken rib)
)";

within sql, literals have to be single-quoted like this:

insert into thetable values(1, 'text', 'even more text')

your table does not seem normalized. where's the primary key etc.? you should read a sql tutorial first; have a look at the articles section of phpbuilder; there you'll find what you need.

nogard

Thanks Matto

Building a DB Table I found on www.mysql.com.
So that is why there is no primary ID key.

Have not found a simple tutorial yet on how to build a DB and Insert information using php/mysql. All tutorials so far show how to use phpmyadmin or mysql via telenet or SSH which uses straight SQL.

I'll take a look at the articles section here as you recommended.

nogard

Ok sem to have most of this yet do not know why the }else{ line shows a Parse Error.

mysql_select_db("coachin3",$db);
$result = mysql_query("SELECT * FROM event",$db);
echo "<table border=1>\n";
echo "<tr><td>Name</td><td>Type</tr>\n";
do {
printf("<tr><td> %s</td><td>%s</tr>\n", $myrow["name"], $myrow["type"], );
} while ($myrow = mysql_fetch_array($result));
echo "</table>\n";
} else {
echo "Sorry, no records were found!";
}
mysql_close ($db);
?>

[deleted]

Simple: ELSE belongs to IF, there can be no ELSE without an IF, and you have no IF statement....

if (foo==bar)
{
// foo is bar
}
else
{
// foo is not bar
}
}

nogard

Would like to know why code has this result.

Warning: Supplied argument is not a valid MySQL result resource in /host/c/o/a/p/o/r/coaching4u.port5.com/mysql_php/event.php on line 20
Sorry, no records were found!

Does this mean that the CREATE TABLE event or INSERT INTO are still not working correctly?

<body>
<?php
$db = mysql_connect("localhost", "user", "pw");
$query = "CREATE TABLE event (
name VARCHAR(25),
date DATE NOT NULL,
type VARCHAR(15),
remark VARCHAR(255)
)";
$row = "INSERT INTO event VALUES (
('Fluffy', '1995-05-15', 'litter', '4 kittens', '3 female, 1 male') ,
('Buffy', '1993-06-23', 'litter', '5 puppies', '2 female, 3 male'),
('Buffy', '1994-06-19', 'litter', '3 puppies', '3 female'),
('Chirpy', '1999-03-21', 'vet', 'needed beak straightened'),
('Slim', '1997-08-03', 'vet', 'broken rib')
)";
mysql_select_db("coachin3",$db);
$result = mysql_query("SELECT * FROM event",$db);
if ($myrow = mysql_fetch_array($result)) {
echo "<table border=1>\n";
echo "<tr><td>Name</td><td>Type</tr>\n";
do {
printf("<tr><td>%s</td><td>%s</tr>\n", $myrow["name"], $myrow["type"]);
} while ($myrow = mysql_fetch_array($result));
echo "</table>\n";
} else {
echo "Sorry, no records were found!";
}
mysql_close ($db);
?>
</body>

nogard

I am just creating this TABLE so I can learn to run some php querys.

http://www.mysql.com/doc/M/u/Multiple_tables.html

I have created $query for the TABLE and $row for the VALUES. After I create the TABLE and INSERT INTO event VALUES .... How do I pass this information to the server and check return values?

jsobell

Your loop should START with
while ($myrow = mysql_fetch_array($result))
{
do whatever
}

otherwise the 'do whatever' part will be called even if there are no results.

Please please please look at the tutorials and buy yourself a good book on PHP. You are obviously keen to get this working, but these questions are fundamental programming issues that are answered by very easy queries on Google.com.
It's nice to be able to help people, but it also good to help yourself 🙂
Look at http://www.phpbuilder.com/columns/
Also, the PHP help doco has examples of all of the database access code, and specifically the stuff you are asking about in these messages. If you are running under Windows then I would strongly suggest that you download the Windows Help-file version of the doco because it allows very easy searching for words so you can quickly locate stuff.

Cheers,
Jason