PHP returns floats instead of strings from a database?

Question

I use this to create a unique code for each product in my store,

echo base_convert(uniqid(),16,10);

For instance,

1403802682572650

My query is simple like this,

        $sql = "
            SELECT * 
            FROM page AS p

            LEFT JOIN stock AS o
            ON o.id = p.page_id

            WHERE p.type = ?
            AND p.hide != ?
            ORDER BY p.backdated_on
        ";

And when I want to get the product code, I do this,

var_dump($product->code);

But PHP returns float(1.4038026825726E+15) instead of 1403802682572650

Why does PHP returns float instead of string? How can I get the number as it is?

Or should I convert/cast the number to string or varchar in my SQL query before requesting result from the db table?

EDIT: Found the bug in my code.

I have a line to convert json to array and that convert the string into float, for instance,

echo json_decode("1403802682572650", true); // 1.4038026825726E+15

`varchar(255) utf8_general_ci` so it should be just fine right? — Run, Jun 04 '12 at 01:07
The number is too big: php > echo 1403804750765990; // 1.403804750766E+15 — Roman Newaza, Jun 04 '12 at 01:17

score 1 · Answer 1 · answered Jun 04 '12 at 01:02

1

No. You should change the column itself to VARCHAR since it's not really being used as a number.

answered Jun 04 '12 at 01:02

Ignacio Vazquez-Abrams

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the db table column is set to be `varchar(255) utf8_general_ci` and I can't see where the error is... – Run Jun 04 '12 at 01:08
Verify that the data wasn't fried on the way in then. – Ignacio Vazquez-Abrams Jun 04 '12 at 01:09

score 1 · Answer 2 · edited Jun 05 '12 at 01:23

1

You can do one of two things without making possibly harmful changes to the database schema.

You can first type cast $product->code:

var_dump( (string) $product->code );

Or you can cast/convert the select in the query:

SELECT CAST(columnName as char) as code ...

I've always been a fan of handling most changes like this via the code rather than the query, in the event that a database engine doesn't support your query.

edited Jun 05 '12 at 01:23

user229044

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answered Jun 04 '12 at 01:14

Mike Mackintosh

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thanks. I tried to cast the select in the query but I still get the number as a float - `1.403804750766E+15`.... – Run Jun 04 '12 at 01:25
I'm using PHP PDO and this line may be causing this issue, - `$this->connection = new PDO($dsn, $username, $password, array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));` – Run Jun 04 '12 at 01:34
found the bug in my script. it is caused by `json_decode`. thanks for the help. – Run Jun 04 '12 at 13:42

Roman Newaza · Accepted Answer · 2012-06-04T01:29:02.167

1

The value is out of range.

PHP Man:

The size of an integer is platform-dependent, although a maximum value of about two billion is the usual value (that's 32 bits signed). 64-bit platforms usually have a maximum value of about 9E18. PHP does not support unsigned integers. Integer size can be determined using the constant PHP_INT_SIZE, and maximum value using the constant PHP_INT_MAX since PHP 4.4.0 and PHP 5.0.5.

If PHP encounters a number beyond the bounds of the integer type, it will be interpreted as a float instead. Also, an operation which results in a number beyond the bounds of the integer type will return a float instead.

edited Jun 04 '12 at 01:29

answered Jun 04 '12 at 01:19

Roman Newaza

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Out of range for a string? Don't think so. – Ignacio Vazquez-Abrams Jun 04 '12 at 01:21
1

Somehow, it was converted to int and then interpreted as float. I guess it happens in your DB object. – Roman Newaza Jun 04 '12 at 01:28
Yea, it must be caused by something else! – Run Jun 04 '12 at 11:12
found the bug, please see my edit above - `echo json_decode("1403802682572650", true); // 1.4038026825726E+15` – Run Jun 04 '12 at 13:40
Obviously, you don't need to decode string which is not JSON. – Roman Newaza Jun 05 '12 at 01:59

PHP returns floats instead of strings from a database?

3 Answers3