Weird behaviour when using PHP mysqli row

Question

I am currently translating a PHP script from a PHP server of version 5.4 to a PHP server with version 5.3

I immediately noticed a wierd behaviour in our login script.

After analysing the script for a bit, I found the source of the problem.

A call to a member of the first row in $result->fetch_row was invaild.

The declaration of $result is shown below:

$username = $mysqli->real_escape_string(strtolower($_POST["USERNAME"]));
$password = $mysqli->real_escape_string(md5(strtolower($_POST["PASSWORD"])));

$result = $mysqli->query("SELECT * FROM $table WHERE username='$username' AND password='$password'");

By instinct I checked if I had called the data_seek properly, however; I had.

Printing out the fetch_row() function gave me the following result

Array ( [0] => 3 [1] => admin [2] => d76362b0f640fbcf869033b5e4d1c4bf [3] => Mr [4] => Rasmus [5] => 4 [6] => [7] => 0 )

The array was therefore working.

But the following declaration did not work.

$Title = $result->fetch_row()[3];

Therefore I tried to store the entire row in a single array object, and then call all sub members individually.

Like so:

$row = $result->fetch_row();
$Title = $row[3];

And it worked perfectly.

How come? What is wrong with this declaration:

$Title = $result->fetch_row()[3];

Wrong with which statement?

AndreKR
– AndreKR

2013-01-02 22:01:28 +00:00
Commented Jan 2, 2013 at 22:01 — AndreKR
– AndreKR, Commented Jan 2, 2013 at 22:01
Sorry it was a bad formulation. Just corrected it.

Rasmus Søborg
– Rasmus Søborg

2013-01-02 22:03:43 +00:00
Commented Jan 2, 2013 at 22:03 — Rasmus Søborg
– Rasmus Søborg, Commented Jan 2, 2013 at 22:03

user1736703 · Accepted Answer · 2013-01-02 22:10:29Z

4

The ability to reference the elements of a returned array directly from the calling method is introduced in PHP 5.4 - which is why it's not working in 5.3.

From Example #7 on http://php.net/manual/en/language.types.array.php

As of PHP 5.4 it is possible to array dereference the result of a function or method call directly. Before it was only possible using a temporary variable.

So your temporary solution looks like it will become a long-term one :)

answered Jan 2, 2013 at 22:10

user1736703

Sign up to request clarification or add additional context in comments.

Comments

L84 · Accepted Answer · 2013-01-03 19:04:27Z

3

You said you were translating the script from PHP 5.4 to PHP 5.3. Apparently you forgot that line because

$Title = $result->fetch_row()[3];

is not valid PHP 5.3. You should get an error reading:

Parse error: syntax error, unexpected '[', expecting ',' or ';' in ...

Notes

edited Jan 3, 2013 at 19:04

L84

46.5k59 gold badges183 silver badges260 bronze badges

answered Jan 2, 2013 at 22:04

AndreKR

34k21 gold badges120 silver badges181 bronze badges

2 Comments

Rasmus Søborg Over a year ago

You sir are correct. I just started doing PHP scripting whereas I previously programmed in C#. This is the first time writing in PHP 5.3 - I apologize the trouble.

AndreKR Over a year ago

No trouble at all. Such problems are not uncommon. I would have provided a link to a changelog if I had found one.

Jordi Kroon · Accepted Answer · 2013-01-02 22:05:29Z

0

As you can see in the release notes of PHP.net

functionname()[1] is something what has been added in PHP 5.4.

answered Jan 2, 2013 at 22:05

Jordi Kroon

2,5873 gold badges32 silver badges56 bronze badges

Comments

wonde · Accepted Answer · 2013-01-02 22:09:00Z

0

it is not possible to dereference the result of a function call directly in PHP 5.3 and later versions but as of PHP 5.4 it is possible

answered Jan 2, 2013 at 22:09

wonde

1,18110 silver badges20 bronze badges

Collectives™ on Stack Overflow

Weird behaviour when using PHP mysqli row

4 Answers 4

Comments

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related