[SOLVED] if field empty, submits empty field?

[tronicsmasta]

Anyone have any idea why this doesn't work?

 

when $_POST['name'] is empty, it submits nothing to the database for name...

 

$name = isset($_POST['name']) ? $_POST['name'] : 'Anonymous';

 

thanks,

 

Quinton

[joecooper]

personally i would use IF's

 

IF(isset($_POST['name'])){

$name=$_POST['name'];

}else{

$name="Anonymous";

}

[Barand]

it's always going to be set unless it's an unchecked checkbox or radio button.

 

$name = !empty($_POST['name']) ? $_POST['name'] : 'Anonymous';

[tronicsmasta]

awesome! good call on the isset... haha... I don't mind using if else but it adds more lines to the code therefore increasing processing by .23 milliseconds haha

 

thanks for your guys' help!

 

Quinton

[joecooper]

lol. thats how i would do it anyway. just imo looks neater and easier to follow. but its down to your preference

[tronicsmasta]

it is very easy to follow you 2 guys posted at the same time...

 

help much appriciated!

[Barand]

personally i would use IF's

 

IF(isset($_POST['name'])){

$name=$_POST['name'];

}else{

$name="Anonymous";

}

 

But it still gives the same wrong result as the original problem code.

[tronicsmasta]

personally i would use IF's

 

IF(isset($_POST['name'])){

$name=$_POST['name'];

}else{

$name="Anonymous";

}

 

now that i look at it,

this:

IF(isset($_POST['name'])){
$name=$_POST['name'];
}else{
$name="Anonymous";
}

 

should be this:

 

if (!$_POST['name']) {
$name="Anonymous";
} else {
$name=$_POST['name'];
}

[Barand]

I'd still go for the more efficient ternary

 

$name = $_POST['name'] ? $_POST['name'] : 'Anonymous';

// or

$name = !$_POST['name'] ? 'Anonymous' : $_POST['name'];