localhost

Code:

---

<?php
$user = "username";
$password = "password";
$database = "dbname";

mysql_connect(localhost,$user,$password);
@mysql_selectdb($database)or die("Error!");

$query="CREATE TABLE contacts (id int(6) NOT NULL auto_increment,first varchar(15) NOT NULL,last varchar(15) NOT NULL,phone varchar(20) NOT NULL,mobile varchar(20) NOT NULL,fax varchar(20) NOT NULL,email varchar(30) NOT NULL,web varchar(30) NOT NULL,PRIMARY KEY (id),UNIQUE id (id),KEY id_2 (id))";

mysql_query($query);
mysql_close();

?>

---

although it did work the table was created i got this error

Quote:

---

Notice: Use of undefined constant localhost - assumed 'localhost' in *path here* on line 6

---

Yeah, MySQL doesn't know what localhost is, you should have written 'localhost' (with the 's)

arr ok i'm following a tutorial and it didnt have quotes in it but thanks while i'm here

PHP Code:

---

$query="INSERT INTO contacts VALUES('','John','Smith','01234 567890','00112 334455','01234 567891','[email protected]','http://www.gowansnet.com')"; 


---

can you see any problems with that? i get a parse error, again its exact from this guide i'm following

try chaging the table name to `contacts`, with those kind of quotes.

still get the error....

hmm.. this guide is a bit poor really wonder if it was for older php version,

so how do you go about fixing it?

anyone?>

Do this on your query. It will then tell you the extact error:

PHP Code:

---

($result = mysql_query($query)) or 
    die(mysql_error()); 


---

Try switching the quotes, i.e. put " where you have ', and ' where you have ". It shouldn't matter, but it's better to have it that way. If you begin a string with " PHP has to parse it for variables and stuff but if you begin with a ' PHP doesn't parse it, so it's faster.

Its because you're trying to put a blank string into your auto incrementing int field. Take out the first two 's and the comma and it should work.

PHP Code:

---

<?php
$user="";
$password="";
$database="";

mysql_connect('localhost',$user,$password);
@mysql_select_db($database) or die ("error selecting DB");

$query="INSERT INTO contacts VALUES('John','Smith','01234 567890','00112 334455','01234 567891','[email protected]','http://www.gowansnet.com')";

mysql_query($query) or die (mysql_error());
mysql_close();

?>

---

Ok now it outputs the following error

Quote:

---

Column count doesn't match value count at row 1

---

hm... this tuturiol is starting to p me off big time

PHP Code:

---

<?php
$user="";
$password="";
$database="";

mysql_connect('localhost',$user,$password);
@mysql_select_db($database) or die ("error selecting DB");

$query="INSERT INTO contacts (first, last, phone, mobile, fax, email, web) VALUES('John','Smith','01234 567890','00112 334455','01234 567891','[email protected]','http://www.gowansnet.com')";

mysql_query($query) or die (mysql_error());
mysql_close();

?>

---

never mind did it myself :-)