[SOLVED] Help with array construction

[EchoFool]

Not sure if i understand this correctly but im getting an error to do with my array syntax....

 

the idea is it creates a list of id's then picks one at random, but i cannot get it to work... any help is much appreciated.

 

Parse error: syntax error, unexpected '[', expecting ')' 
in C:\xampp\htdocs\name.php on line 427

 

Im unsure on what i have done wrong.

 

This is what i got:

 

<?php
$SELECT = mysql_query("SELECT ID FROM users")
   Or die(mysql_error());
$a = 0;
While($row = mysql_fetch_assoc($Find)){
$ID = $row['ID'];

Array
(
    [$a] => $ID;
)
$a = $a + 1;
}


shuffle($a);

$_SESSION['MissionID'] = $a;
?>

 

[rhodesa]

what are you trying to accomplish with:

Array
(
    [$a] => $ID;
)

 

..since it's not valid PHP code

[kenrbnsn]

You have done almost everything wrong. This is what I believe you are trying to do:

<?php
$Find = mysql_query("SELECT ID FROM users")
   Or die(mysql_error());
$a = array();
While($row = mysql_fetch_assoc($Find)){
    $a[] = $row['ID'];
}


shuffle($a);

$_SESSION['MissionID'] = $a;
?>

 

Ken

[EchoFool]

Oh i see thankyou for pointing me in the right direction.

 

 

With your suggestion Ken, i produce this error:

 

Fatal error: Cannot use [] for reading in C:\xampp\htdocs\missioninclude.php on line 426

 

Which relates to  $a[] = $row['ID'];.

[kenrbnsn]

Did you put in

<?php
$a = array();
?>

 

Can you post the code you're now using?

 

Ken

[EchoFool]

Sure here it is:

 

<?php
$Find = mysql_query("SELECT ID FROM users")
   Or die(mysql_error());
If(mysql_num_rows($Find)>0){
$a = array();
While($row = mysql_fetch_assoc($Find)){

$ID  = $row['ID'];
    $a[] = $ID;   // this is where the error is
}

shuffle($a);

$_SESSION['MissionID'] = $a;
}
?>

 

Thank you for replying

 

[EchoFool]

Argh i fixed my error. Thank you!

[kenrbnsn]

What was the error?

 

Ken

[EchoFool]

Changed :

 

$ID  = $row['ID'];
    $a[] = $ID;

 

to

 

$a[] = $row['ID'];

 

 

 

Though when i shuffle($a);

 

Then Echo $a;  I get "Array" as the response rather than the id. Why could that be?

[kenrbnsn]

You can't just echo an array, you need to use something like print_r:

<?php
echo '<pre>' . print_r($a,true) . '</pre>';
?>

 

Ken

[EchoFool]

I see so that produces:

 

Array
(
    [0] => 25
)

 

Is there a way to extract the 25 to process with that ID?

[kenrbnsn]

$a[0] is how you reference it, or are you asking for something else?

 

Ken

[wildteen88]

Reading your first post, $a should hold a collection of ids correct? and you're want to select an id from the $a array at random? Well change

 

$_SESSION['MissionID'] = $a;

 

to

$_SESSION['MissionID'] = array_shift($a);

[EchoFool]

Oh i see. So array's start at [ 0 ] not [1] that would explain why I couldn't grab the value!

 

Thanks Ken.

 

wildteen88 - is array_Shift similar to my shuffle function?

[wildteen88]

No, array_shift will remove the first item from the array and return its value. shuffle randomises the order of the items within the array.

 

Actually dump shuffle and array_shift and just use array_rand instead.