Notice: Use of undefined constant one - assumed 'one' inHow can to fix this??
Answer:
Let's see your code
error_reporting(2047);
$a=array(1=>'ONE','one'=>1);
$s.=$a[one]."<br>";
die($s);
As we see. One should be type $a['one'] or this will happen: FYI the right answer is 1 not ONE
the code are like this
<?phpIf you define one (or perhaps another script).. The result is not 1 but ONE. For the next time.. LET's More carefully..
error_reporting(2047);
define('one',1);
$a=array(1=>'ONE','one'=>1);
$s.=$a[one]."<br>";
$s.=$a['one']."<br>";
die($s);
Another way if you can't edit all and don't have time.. edit
error_reporting(2047);into (if there is no reporting.. add it)
error_reporting(0);
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