Basic Query returning error

Blue

A VERY basic newbie problem.

I have a database set up with a table named shiptoinfo. This table has 27 fields. All I want to do at this point is have my query show all the records on a html page.

Below is my query, which returns this error when I call the .php page in my browser.
Parse error: parse error, unexpected '}' in /home/nnnnn/public_html/atest/orders.php on line 19

[PHP}}<?php

include "/connections/common_db.inc";

$link_id = db_connect();

$host_location=$HTTP_HOST;

"SELECT *
FROM shiptoinfo";

}

?>PHP]

Help would be appreciated.
Blue

WinterDragon

Apparently! There is 1 too many brackets somewhere!! Look into it... Besides, when posting PHP code, put it between the PHP tags, anyways.. (use the little ubtton above the message body), also... post more of the code if you want us to look at it... shrugs

Blue

Okay I'll try that again, here is the code for the entire .html page which has the Query on it:

1 <html>
2 <head>
3 <title>Untitled Document</title>
4 <meta http-equiv="Content-Type" content="text/html; 
 charset=iso-8859-1">
5 </head>
6 
7 <body bgcolor="#FFFFFF" text="#000000">
8 <?php
9 
10 include "connections/common_db.inc";
11 
12 $link_id = db_connect();
13
14 $host_location=$HTTP_HOST;
15
16 "SELECT *
17 FROM shiptoinfo";
18
19	}
20
21	?>
22 </body>
23 </html>

Here is the error displayed upon calling the page in the browser:
Parse error: parse error, unexpected '}' in /home/nnnnn/public_html/atest/orders.php on line 19

If I remove the bracket on line 19, then a 'blank' white page appears when calling the page in the browser.

Regards,
Blue

shawon22

ur almost there all u need to do is now query the database with that query string. i'm writing down some code below hope this would help u to solve ur prob


include "/connections/common_db.inc"; 

$link_id = db_connect(); 

$host_location=$HTTP_HOST; 

$query = "SELECT * FROM shiptoinfo"; 

$result = mysql_query($query);

while($row = mysql_fetch_array($result , MYSQL_NUM))
{
   echo $row[0] . $row[1] ;
}

WinterDragon

Thats a good sign! That means no errors are being produced, and the reason its not putting anything on the page, is because you aren't asking it to do anything but Do a DB Query, and some other stuff! you don't do anything with the information you receive! You just store it in variables... You need to learn alot more about php, but here goes...

I hope the connection information is being stored off into the INC file you have there... and if so... it should work... but lets see, you have the select field, you are selecting all the information... And are you going to do anything with it? Maybe a while loop, or a for loop to display it all? Or something... either way, You have to grab the information out of the large array you just pulled. hrm... Hopefully all the other information is stored into .INC, anyways I am going to bed... 🙂

Blue

Thanks for the help Folks, I am getting there. I changed the code on the page (see below) Now I get a display.
You can see that the data is displayed here

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

<body bgcolor="#FFFFFF" text="#000000">
<?php

$dbh=mysql_connect ("localhost", "rofarms_mike", "pwhere") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ("rofarms_orders");

$query = "SELECT * FROM shiptoinfo";  

$result = mysql_query($query); 

while($row = mysql_fetch_array($result , MYSQL_NUM)) 
{ 
   echo $row[0] . $row[1] ; 
}


?>
</body>
</html>

All that is being displayed is 1 field(column) of info from the 2 records in the db.

I would like to now modify the code above to be able to display the data in a format like this:

invnum (note=this is the only info being displayed with the above code)
next column data
next columd data
next columd data
etc...

Regards,
Blue