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[SOLVED] array variable help


EchoFool

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For some reason using a variable with the in_array() does not work even though the structure in the variable is correct.... heres pretty much what i've got but it doesn't work:

 

<?php
$Function = '"0000","0010","9000"';
$Array = array($Function);
If(in_array($cord,$Array)){echo $cord;}Else{
	echo '<input type="checkbox" name="list[]" value="'.$cord.'">';
}	
echo '</center>';
?>

 

Why won't it allow me to just put the variable function as part of the array?

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and your $Function is not the array you think it is.

 

use print_r(); to test your arrays :

 

print_r($Function);

 

you'll see that key[0] has a value of : "0000","0010","9000" .. easiest way to add values (when on the fly), to the array is get your items and add them to a variable that will later be turned into an array, ie.

 

$arr[] = $cord;

and loop that for however many items you have .. then your array will set to go.

and your $Function is not the array you think it is.

 

use print_r(); to test your arrays :

 

print_r($Function);

 

you'll see that key[0] has a value of : "0000","0010","9000" .. easiest way to add values (when on the fly), to the array is get your items and add them to a variable that will later be turned into an array, ie.

 

$arr[] = $cord;

and loop that for however many items you have .. then your array will set to go.

 

Not quite cos that will guarentee the $cord is in the array when it might not be, only certain coordinates are to be in this array the 3 i gave in the example above are just to show you what im trying to do.. which is build array list in a variable then use that variable as the array to check if $cord is in the array($Function).

The code I posted is correct - you must be implementing it in an erroneous manner.

 

I know, thats why im kinda confused why it error'd cos its correct..... but heres the script with the error :

 

 

<?php
//above here is html so unlikely to be affecting the error

$cord = "0000";
$Collect = mysql_query("SELECT Function FROM usercrimes_blocked WHERE FloorID='$LocationID'")
Or die(mysql_error());
$row = mysql_fetch_assoc($Collect);
$Function = $row['Function']; echo $Function;
//$Function shows  "1000", "0900", "0800", "0700", "0600", "0500"
$Array = explode("," $Function);
If(in_array($cord,$Array)){echo $cord;}Else{     //THIS IS LINE 298
	echo '<input type="checkbox" name="list[]" value="'.$cord.'">';
}	
echo '</center>';
?>

Error:

Parse error: parse error on line 298

explode returns and array so you don't need to declare it one.

 

Just use what jack said:

 

$Array = explode(',' $Function);

 

I am but i get a parse error from it for some reason :S

 

You are still missing a comma ,have to give it like

 

$Array = explode(',',$Function);

and your $Function is not the array you think it is.

 

use print_r(); to test your arrays :

 

print_r($Function);

 

you'll see that key[0] has a value of : "0000","0010","9000" .. easiest way to add values (when on the fly), to the array is get your items and add them to a variable that will later be turned into an array, ie.

 

$arr[] = $cord;

and loop that for however many items you have .. then your array will set to go.

 

Not quite cos that will guarentee the $cord is in the array when it might not be, only certain coordinates are to be in this array the 3 i gave in the example above are just to show you what im trying to do.. which is build array list in a variable then use that variable as the array to check if $cord is in the array($Function).

i just threw $cord in there as an example for adding values to an array using [] .. $cord represents the string you are adding .. that's all.  i post what i know, you play around with it.

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