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I'm having a little trouble with a default value in a custom function - could someone help?

Posted on 2011-02-20
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Last Modified: 2012-06-27
The function gets a testimonial from a table like this:

type,id,name,location,testimonial
'type1',1,'Mr A','Manchester','They were great!'
'type2',1,'Mrs B','Liverpool','They were smashing!'

NB There are only 3 of type2, but 5 of each of the others, hence the line starting $count.

I want to be able to get a testimonial either of a specific type, or at random:

getTestimonial(type1);     // Should return a testimonial of type1
getTestimonial();              // Should return a testimonial of any type

I've tried using Nulls & empty strings, but if I don't pass a type, then all I get back is:

<br />
~ ,

Can anyone see where I'm going wrong?
<?php
error_reporting(E_ALL);
function getTestimonial($type="")
{
	if (""==$type) {
		$types = array("type1","type2","type3","type4");
		$type = $types[rand(3)];
	}
//	echo "[".$type."]";
	$count = ($type == 'type2') ? 3 : 5;
	$sqltest = "SELECT * FROM testimonials WHERE type='$type' AND id=".rand(1,$count)." LIMIT 1";
	$restest = mysql_query($sqltest);
	$rowtest = mysql_fetch_assoc($restest);
	return $rowtest[testimonial]."<br /> ~ ".$rowtest[name].", ".$rowtest[location];
}
?>

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Question by:PaulCutcliffe
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9 Comments
 
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Accepted Solution

by:
LAMASE earned 1332 total points
ID: 34939290
When you put an array inside a string it will render as the string "Array" (try to output the query).
You can simply remove the condition where no set is passed.

However you complicated your life with your random selection, use

SELECT .... WHERE .... ORDER BY RAND() LIMIT 1

It will return a random element in the selected set.
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LVL 4

Expert Comment

by:LAMASE
ID: 34939298
PS the right way to specity a set of values in the query is

$query = "SELECT ... WHERE field IN ('".implode("','", $array)."')";

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Will generate

... WHERE field IN ('value1','value2','value3')

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Expert Comment

by:LAMASE
ID: 34939333
I forgot to tell you to use quotes in

getTestimonial("type1");

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LVL 4

Expert Comment

by:LAMASE
ID: 34939347
When you put an array inside a string it will render as the string "Array" (try to output the query).
Sorry, I read the code too quick... however you can solve with the previous hints.
0
 

Author Comment

by:PaulCutcliffe
ID: 34939402
I had thought that perhaps I should have used quotes in my function call, but it seemed to work fine as it was.

I'm sorry though, other than that, you've totally lost me. :-(
0
 

Author Comment

by:PaulCutcliffe
ID: 34939425
I've just checked my code, & I have used quotes.
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LVL 4

Expert Comment

by:LAMASE
ID: 34939430
Is the query correct? can you print it before executing?
0
 

Author Comment

by:PaulCutcliffe
ID: 34939482
It works if I supply a parameter, just not if I don't. I think it's something in the if block, or the handling of an empty string or Null, that isn't working.

I think I can use randomisation in the query as you suggest, then I don't have to worry about how many of each type there are. I had thought that could get complicated!

But I just need to able to pull out a random one of any type, & that currently doesn't work.
0
 

Author Comment

by:PaulCutcliffe
ID: 34939586
Aha, got it.

I've changed it now to this:

<?php
error_reporting(E_ALL);
function getTestimonial($type='')
{
	if (''==$type) {
		$sqltest = "SELECT * FROM testimonials ORDER BY RAND() LIMIT 1";
	} else {
		$sqltest = "SELECT * FROM testimonials WHERE type='$type' ORDER BY RAND() LIMIT 1";
	}
	$restest = mysql_query($sqltest);
	$rowtest = mysql_fetch_assoc($restest);
	return "[".$rowtest[type]."]<i>".$rowtest[testimonial]."</i><br /> ~ ".$rowtest[name].", ".$rowtest[location];
}
?>

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Thanks. Works perfectly.
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