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What is wrong with my SQL syntax here?

I'm trying to create a IT asset database with a web front end.

I've gathered some data from forms using POST as well as one variable that had already written to a cookie.

This is the first time I have tried to enter the data into the database.

Here is the code:

<?php

//get data
$id = $_POST['id'];
$company = $_POST['company'];
$location = $_POST['location'];
$purchase_date = $_POST['purchase_date'];
$purchase_order = $_POST['purchase_order'];
$value = $_POST['value'];
$type = $_COOKIE["type"];
$notes = $_POST['notes'];

$manufacturer = $_POST['manufacturer'];
$model = $_POST['model'];
$warranty = $_POST['warranty'];

//set cookies
setcookie('id', $id);
setcookie('company', $company);
setcookie('location', $location);
setcookie('purchase_date', $purchase_date);
setcookie('purchase_order', $purchase_order);
setcookie('value', $value);
setcookie('type', $type);
setcookie('notes', $notes);

setcookie('manufacturer', $manufacturer);
setcookie('model', $model);
setcookie('warranty', $warranty);

//checkdata

//start database interactions

// connect to mysql server and database "asset_db"
mysql_connect("localhost", "asset_db", "asset_db") or die(mysql_error());
mysql_select_db("asset_db") or die(mysql_error());

// Insert a row of information into the table "asset"
mysql_query("INSERT INTO asset 
(id, company, location, purchase_date, purchase_order, value, type, notes) VALUES('$id', '$company', '$location', '$purchase_date', $purchase_order', '$value', '$type', '$notes') ") 
or die(mysql_error());
echo "Asset Added";

// Insert a row of information into the table "server"
mysql_query("INSERT INTO server 
(id, manufacturer, model, warranty) VALUES('$id', '$manufacturer', '$model', '$warranty') ") 
or die(mysql_error());
echo "Server Added";


//destination url
//header("Location: verify_submit_server.php");

?>

The error I get is: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '', '678 ', 'Server', '789')' at line 2

That data is just test data I was trying to throw in there, but it looks to be the at the $value, $type, $notes.

Here are the table create statements if they help:

<?php

// connect to mysql server and database "asset_db"
mysql_connect("localhost", "asset_db", "asset_db") or die(mysql_error());
mysql_select_db("asset_db") or die(mysql_error());

// create asset table
mysql_query("CREATE TABLE asset(
id VARCHAR(50) PRIMARY KEY, 
company VARCHAR(50), 
location VARCHAR(50),
purchase_date VARCHAR(50),
purchase_order VARCHAR(50),
value VARCHAR(50),
type VARCHAR(50),
notes VARCHAR(200))")
or die(mysql_error());  
echo "Asset Table Created.</br />";

// create software table
mysql_query("CREATE TABLE software(
id VARCHAR(50) PRIMARY KEY, 
software VARCHAR(50),
license VARCHAR(50))")
or die(mysql_error());  
echo "Software Table Created.</br />";

// create laptop table
mysql_query("CREATE TABLE laptop(
id VARCHAR(50) PRIMARY KEY, 
manufacturer VARCHAR(50),
model VARCHAR(50),
serial_number VARCHAR(50),
esc VARCHAR(50),
user VARCHAR(50),
prev_user VARCHAR(50),
warranty VARCHAR(50))")
or die(mysql_error());  
echo "Laptop Table Created.</br />";

// create desktop table
mysql_query("CREATE TABLE desktop(
id VARCHAR(50) PRIMARY KEY, 
manufacturer VARCHAR(50),
model VARCHAR(50),
serial_number VARCHAR(50),
esc VARCHAR(50),
user VARCHAR(50),
prev_user VARCHAR(50),
warranty VARCHAR(50))")
or die(mysql_error());  
echo "Desktop Table Created.</br />";

// create server table
mysql_query("CREATE TABLE server(
id VARCHAR(50) PRIMARY KEY, 
manufacturer VARCHAR(50), 
model VARCHAR(50),
warranty VARCHAR(50))")
or die(mysql_error());
echo "Server Table Created.</br />";

?>

Running a standard LAMP stack on Ubuntu 10.04.

Thank you.

like image 859
CT. Avatar asked May 17 '10 02:05

CT.


3 Answers

you need a beginning quote ' before $purchase_order

like image 59
Galen Avatar answered Sep 29 '22 16:09

Galen


You're missing a ' before $purchase_order.

like image 26
Keith Randall Avatar answered Sep 29 '22 16:09

Keith Randall


What Galen said. Plus your code is vulnerable to SQL Injection - and it will generally have problems with any user-input data with embedded single-quotes.

like image 23
Cade Roux Avatar answered Sep 29 '22 16:09

Cade Roux