PHP to database

[lerro]

In a class, i have the following - a function to generate an ID, and a function to insert data into the database

function generate_id(){

	$Date = date("Ymdhis");		
	$randNum = rand(100000,999999);		
	$new_string = $Date.$randNum;

	return $new_string;
}

function insert_app_data($input) {

	$item_id = $this->generate_id();


	$query = "INSERT INTO TableName 
	(item_id,...,...,)
	VALUES  ('$item_id',...,...,)";
         }

But i keep getting an error. When i tried to identify the cause of the error by echoing the returned value of the function generate_id(), i got the following response

Catchable fatal error: Object of class Application could not be converted to string

Could someone help me identify what's going wrong in this whole process?
As the following should happen
- generate unique id
- pass value to the variable "$item_id"
- insert value in the database table

[rulian]

Can you put in your actual class, you need to references are right

Also keep in mind the difference when we use " and ' in our programing
basically, " evaluates the string while ' just prints it out
for example
$Test = "Bob";
echo "Hi $Test"; will put out "Hi Bob"
echo 'Hi $Test'; will put out "Hi $Test";

You only need '' in the values column if you are passing it a string by it self
for example
$Test2 = "Bobby"; you woudl only go ) values ($Test2, 'Smith') Because i'm passing SMITH in without using a variable I would use the quotes.

But also you want to look into sprintf and mysql_real_escape _string functions, you NEVER wanna put unfiltered information into the database even IF you are the one generating the information, such is your case.

[lerro]

Thanks.

I tried applying what you said but still ran into errors

I tried changing the function so that instead of passing the return ed value to another function, it's direct.

$item_id = date("Ymdhis").rand(1000,9999);

and then i try to input the $item_id value into the database.

But the problem is that item_id keeps generating exactly the same number -
"2147483647"
which results in duplicate id for subsequent inputs. This means additional records cannot be input.

I therefore tried to echo out $item_id value and it's different each time.

$item_id = date("Ymdhis").rand(1000,9999);
echo ($item_id);

So, why is it that when i try to insert the value into the database, the $item_id value generated is exactly the same number as it's displayed in the error message

Duplicate entry '2147483647' for key 'PRIMARY'

but when i echo it out, it's different each time?

[rulian]

that is wierd, your key should look somethign like
2010010712341832834 /// or be 18 characters long, so its pulling something else because I doubt you are in the year 2147 on your computer.
Post the actual code you have now so I can see it better than snipplets

[big_nerd]

lerro,

The problem seems to be that you have an integer as the datatype for the column that you are trying to use such a LONG unique key.

I Suggest one of two things:

1) Use the auto_increment field and set an auto_increment value of 10000 or 100000 (or something random in there) if you don't expect a HUGE database of rows. (the higher you go the lower the possible number of entries)

lerro wrote:

Duplicate entry '2147483647' for key 'PRIMARY'

Take a look at that number

A normal-size integer. The signed range is -2147483648 to 2147483647

This is from the MySQL Docs.

You will notice they are the same! That's because as rulian pointed out that the number should be 18 char's long, WELL beyond the scope of a signed int.

When a number is provided that goes beyond the boundry of a numeric type, MySQL will assume the highest possible number, since it was already entered once, it tries again and gets a duplicate entry.

If you want to use your current method of creating a unique/primary key (as it seems you are doing) you must change the datatype to BIGINT

BIGINT[(M)] [UNSIGNED] [ZEROFILL]

A large integer. The signed range is -9223372036854775808 to 9223372036854775807. The unsigned range is 0 to 18446744073709551615.

Again from MySQL Docs, you will notice that there are points to keep in mind when using unsigned bigint's, speficially not to go larger than 63 bits (9223372036854775807)

Please let me know if you have any further questions.

[lerro]

Thanks guys.

It finally worked. The problem was the datatype in mysql being set as Integer yet the id was 18 digits.

BIGINT datatype solved the problem.

Thanks.