[SOLVED] insert PHP code into function

[djtozz]

Hi,

I'm having problems when trying following:

 

I have:

createBar("Download: <b>$kwd</b><Br />", "100%")

 

and I'm trying to display following after $kwd:

if(count($info['results']))
print '<div align="center" class="info"><b>'.$info['total_results_count'].'</b> results found, page '.$info['curpage'].' from '.$info['total_pages'].'</div>';

 

I replaced all double " by /" but this doesn't seems to work.

createBar("Download: <b>$kwd</b> if(count($info['results'])) print '<div align=\"center\" class=\"info\"><b>'.$info['total_results_count'].'</b> results found, page '.$info['curpage'].' from '.$info['total_pages'].'</div>'<Br />", "100%")

 

I'm getting following error:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in

 

 

 

[Daniel0]

You can't do it like that. I would just generate the text for the first parameter, assign it to a variable and then pass it to the function.

 

$text = "Download: <b>$kwd</b>";
if (count($info['results'])) {
$text .= '<div align="center" class="info"><b>'.$info['total_results_count'].'</b> results found, page '.$info['curpage'].' from '.$info['total_pages'].'</div>';
}
$text .= '<br />';

createBar($text, '100%');

[purpleshadez]

if(count($info['results']))

 

if what? is $info['results'] meant to equal a certain amount, be less than or more than a certain amount? Also use curly brackets with if statements it makes the code easier to read

 

print '<div align="center" class="info"><b>'.$info['total_results_count'].'</b> results found, page '.$info['curpage'].' from '.$info['total_pages'].'</div>';

 

This seems fine to me but what is the rest of the code doing? what line is the error on? Look at the code from the error UP as the fault will most likely be on the line(s) above it. Have you checked for a missing semi colon? Seeing the full script would help. What is createBar() doing?

 

 

[trq]

if what? is $info['results'] meant to equal a certain amount, be less than or more than a certain amount?

 

if(count($info['results']))

 

Checks to see if the result of the call to count() returns true or false. It is perfectly valid code.

[Daniel0]

if(count($info['results']))

 

if what? is $info['results'] meant to equal a certain amount, be less than or more than a certain amount?

 

0 evaluates to false in a boolean context, any other integer evaluate to true. PHP is weakly typed language.

[purpleshadez]

if(count($info['results']))

 

if what? is $info['results'] meant to equal a certain amount, be less than or more than a certain amount?

 

0 evaluates to false in a boolean context, any other integer evaluate to true. PHP is weakly typed language.

 

Good point, my bad. IMO its a bit lazy though

[Daniel0]

Really? So you explicitly convert between all types always?

 

$var1 = 1;
$var2 = 1.6;

echo (string) ((float) $var1 + $var2);

 

Or do you just rely on PHP's automatic type inference and conversion?

 echo $var1 + $var2;

[JAY6390]

lmao nice example

[purpleshadez]

Really? So you explicitly convert between all types always?

 

$var1 = 1;
$var2 = 1.6;

echo (string) ((float) $var1 + $var2);

 

Or do you just rely on PHP's automatic type inference and conversion?

 echo $var1 + $var2;

 

No I don't explicitly convert each type always I only do it when required. But in the context of the if statement (which was what I was commenting on) I would always write it as

if(count($num) > 0) { 
// do someting
}

just as I would also write

if(isset($var)) {
// do something
}

instead of 

if($var) {
// so something
}

 

My apologies if my light hearted and perhaps vague comment has caused offence as that wasn't my intention. 

[Daniel0]

The statement count($num) evaluates to true if and only if count($num) > 0 evaluates to true, so the two statements are equivalent.

 

In a boolean context, the statements isset($var) and $var are not equivalent. Specifically, if you set $var = false; then the first will evaluate to true, but the second one to false. The same thing will happen if you set $var = 0;. If the variable $var is not even set, then the first one will evaluate to false and so will the second one, but the second one will raise an error of level E_NOTICE.

[djtozz]

Thanks for the excellent info guy's!

Once again, I learned a lot from this post!

 

Keep up the good work

[purpleshadez]

The statement count($num) evaluates to true if and only if count($num) > 0 evaluates to true, so the two statements are equivalent.

 

In a boolean context, the statements isset($var) and $var are not equivalent. Specifically, if you set $var = false; then the first will evaluate to true, but the second one to false. The same thing will happen if you set $var = 0;. If the variable $var is not even set, then the first one will evaluate to false and so will the second one, but the second one will raise an error of level E_NOTICE.

 

Yes. Another good point! I'm not sure what I had in my mind but no, I retract my lazy statement.

 

*Considers himself duely corrected*