PHP function, calling from another file error with variable

Question

I have a functions.php file with the following content.

function maxSavePerHour($ip, $maxSave = 15){
//DB connection and query. Working fine.
$totalRows = mysql_num_rows($res);
    if($totalRows <=$maxSave){
        $safe=true;
    }else{
        $safe=false;
    }

if($safe){
    echo "OK";
}else{
    echo "Not OK";
}
}

I call that function from another page. And I use the function as following.

maxSavePerHour($ip, 5);

If I change the value the function echoes text as it should. The problem is the $safe variable. If I echo $safe i get nothing.

Any clues? The reason I need $safe to be true or false is because I use it in a if statement later.

I put global $safe;` and now its working. Thanks Nile.

i am unable to see any function definition and what do you mean by: "i call that function from another page"? — aurora
– aurora, Commented Aug 7, 2012 at 5:59
you may need to put global $safe at the beginning of your function, but I couldn't know without seeing the whole thing. — jeremy
– jeremy, Commented Aug 7, 2012 at 6:00
Very unclear question, please show more code context and where you output the $safe variable. Spontaneous guess: variable scope! — deceze
– deceze ♦, Commented Aug 7, 2012 at 6:00

Lock · Accepted Answer · 2012-08-07 06:06:35Z

1

If you echo $safe, it is echoing nothing because $safe is a boolean value (and false will not echo a value, whereas true will echo 1). I would return the value of $safe and then change the calling code to handle this. E.g.:

function maxSavesPerHour($ip, $maxSave) {
  $totalRows = mysql_num_rows($res);
  return ( $totalRows <= $maxSave ? TRUE : FALSE );
}

Then call it from your code like this:

if ( maxSavesPerHour($ip, 5) ) {
  //do something when true
} else {
 //do something when false
}

answered Aug 7, 2012 at 6:06

Lock

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1 Comment

deceze Over a year ago

I don't think the boolean value is a problem, but this is the only answer with sensible code. +1

米雪儿 · Accepted Answer · 2012-08-07 06:23:21Z

1

function maxSavePerHour($ip, $maxSave = 15){
//DB connection and query. Working fine.
$totalRows = mysql_num_rows($res);
    if($totalRows <=$maxSave){
        $safe=true;
    }else{
        $safe=false;
    }

    if($safe){
        echo "OK";
    }else{
        echo "Not OK";
    }
    return $safe;
}
$safe = maxSavePerHour($ip, 5);
echo $safe;

answered Aug 7, 2012 at 6:23

米雪儿

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1 Comment

EaterOfCode Over a year ago

boolean are not really well showed with echo better use var_dump

Clayton · Accepted Answer · 2012-08-07 06:53:29Z

1

You should be able to simply return $safe and handle it from the other file. Or better yet, implement the following:

function maxSavePerHour($ip, $maxSave = 15){
//DB connection and query. Working fine.
$totalRows = mysql_num_rows($res);
    if($totalRows <=$maxSave){
        return true;
    }else{
        return false;
    }
}

You do not actually need a $safe variable. Marking your $safe variable as global is a very bad idea and should hardly ever be done.

edited Aug 7, 2012 at 6:53

answered Aug 7, 2012 at 6:03

Clayton

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1 Comment

Clayton Over a year ago

From where you call the method you can do something of this sort: if(maxSavePerHour('valuesfor$ip', 15)){//true} else {//false} or $ok = maxSavePerHour('valuesfor$ip', 15); if($ok){//do something} else {//do someething else}

Mithun Satheesh · Accepted Answer · 2012-08-07 06:12:37Z

0

function maxSavePerHour($ip, $maxSave = 15){ 
    //DB connection and query. Working fine. 
    $totalRows = mysql_num_rows($res); 
    $safe = false;
    if($totalRows <=$maxSave){ 
         $safe=true; 
    }else{ 
         $safe=false; 
    } 

    if($safe){ 
        echo "OK"; 
    }else{ 
        echo "Not OK"; 
    } 
}

safe only exsists in the if not outside it

edited Aug 7, 2012 at 6:12

Mithun Satheesh

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answered Aug 7, 2012 at 6:06

EaterOfCode

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1 Comment

EaterOfCode Over a year ago

@mithunsatheesh I was on a mobile phone so I couldn't format it, thanks :)

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