Plugging in Array

[geosite]

The script below displays "North America," where $Name is an array that includes the value "North_America."

$val = array_search('North_America', $Name);
if ($val>-1)
echo 'North America';
else
echo "Value isn't present."; 

Can I replace 'North_America' with an array that includes all the continents? If so, how? I tried the script below, but it isn't working. Thanks.

$Continents = array("North_America", "South_America", "Africa");
$query = "SELECT * FROM gzdistpractice";
 $result = mysql_query($query);
 while($row = mysql_fetch_array($result)) {
      $Name[] = $row['Name'];
echo($row['Name']);
print_r($Name);
}

$val = array_search($Continents, $Name);
if ($val>-1)
echo 'Continent';
else
echo "Value isn't present."; 

[Scheisskopf]

Instead of

$val = array_search($Continents, $Name);
if ($val>-1)
echo 'Continent';
else
echo "Value isn't present."; 

How about

foreach($Continents as $continent)
     if (in_array($continent, $Name)) 
          echo "$continent present"; 

[geosite]

Instead of

$val = array_search($Continents, $Name);
if ($val>-1)
echo 'Continent';
else
echo "Value isn't present."; 

How about

foreach($Continents as $continent)
     if (in_array($continent, $Name)) 
          echo "$continent present"; 

Wow, that's a bonus script. It works, plus it illustrates that the in_array function works on my computer. It appears that in_array doesn't work with dynamically generated arrays, apparently because of some sort of bug in PHP. I thought it wouldn't work at all, so this is a nice surprise. Thanks.

[Sorccu]

It appears that in_array doesn't work with dynamically generated arrays, apparently because of some sort of bug in PHP.

Really? I find it kind of hard to believe..

[Dylan B]

Wow, that's a bonus script. It works, plus it illustrates that the in_array function works on my computer. It appears that in_array doesn't work with dynamically generated arrays, apparently because of some sort of bug in PHP. I thought it wouldn't work at all, so this is a nice surprise. Thanks.

Can you give an example of this bug?

[geosite]

Can you give an example of this bug?

That's a good question! I spent a couple days wrestling with some array questions, then started another thread to discuss the error. I think the first link below is one of the original discussions, and the second link regards the error message.

http://www.sitepoint.com/forums/show...=1#post2251964
http://www.sitepoint.com/forums/show...=1#post2252550

I finally asked about it on another forum, and someone told me the problem was me:

"What are you trying to do? Why are you trying to use an array function
on a string? When you get the $output array, the array element
$output['Name'] is presumably a string, not an array. The second
argument to the in_array() function is supposed to be an array. Are you
simply trying to find out if the Name field contains the word
"Neotropical"?"

He suggested I try this script:

if(preg_match("/Neotropical/",$output['Name']))

It seems to work. So I'm assuming I got the error message because I did something wrong. It was really confusing because one or two people told me they duplicated my scripts and they worked. In fact, I think I tried several different PHP functions, and none of them worked. But I may simply have been doing everything wrong.

Also, I found lots of references to this error, including several on Drupal's website.

So I'm thoroughly confused, but things seem to be working out alright for now. I still want to learn how to use the in_array function, but it looks like it's a little more techincal than what was described to me.

Thanks for your reply.

[Dylan B]

The in_array function is really simple, and it works like this:

<?php
$arr = array('Hello', 'Giraffe', 'Dylan', 'Geosite'); //make a new array with a few sample values.
if (in_array('Dylan', $arr)) // The value you want to look for is the first argument, and the array you're looking IN is the second argument.
{
     echo 'DYLAN IS IN THE ARRAY'; //Hooray!  Victory!
}
?>

Dylan

[Sorccu]

The in_array function is really simple, and it works like this:

<?php
$arr = array('Hello', 'Giraffe', 'Dylan', 'Geosite'); //make a new array with a few sample values.
if (in_array('Dylan', $arr)) // The value you want to look for is the first argument, and the array you're looking IN is the second argument.
{
     echo 'DYLAN IS IN THE ARRAY'; //Hooray!  Victory!
}
?>

Dylan

Yeah, it's just the order of the arguments that confuses many First comes the needle (the thing you're looking for), then the haystack (the array). I always have to check it from php.net, just in case Might be easier to remember if you think of it this way:

in_array('the value that "goes IN"', $array) 

And bool strict as the third, optional argument of course.