Question for Pro

abnfire

func_name( ($s == 0) ? 'True' : 'False', ($i == 3) ? '1' : '0');

can you tell me what question mark '?' and colon ':' stand here for?

because i tryed to search in a book i can't find it.

shawon22

($s == 0) ---> is the condition, so this condition ($i == 3) ? '1' : '0'); goes like this

if($i==0)
{
1;
}
else
{
0;
}

abnfire

func_name( ($s == 0) ? 'True' : 'False', ($i == 3) ? '1' : '0');
if the string '$s' is empty then how can we call the default value of the function?

shawon22

do it inside the function
example:

function add($a, $b)
{
 if(empty($a) || empty($b))
 $a = 2; //default value
 $b = 2; //default value

return $a + $b;

}

abnfire

a person show me how to call a default value if a string is empty, but i tryed and some how is not working, can you fix it and tell me whats going on, because i been with this problem more then 2 weeks now and hardly ppl help me.

source for another way from another person:

if( $s == 0 )
$foo[0] = 'localhost';

if( $i == 0 )
$foo[1] = 'root';

if( $i == 0 )
$foo[2] = '';

$stuff = implode( ',', $foo );

mysql_connect($stuff);

some how dont pass the value for the agr2 and agr 3. any idea!

shawon22

i don't see any problem!

abnfire

its giving me error. its calling mysql_connect("localhost,root,"), but its not calling like mysql_connect(localhost, root,). and i tryed again and again , but it give me error, can you try it and let me know what comes up and if you find any solution to this.

shawon22

what u trying to do?

abnfire

first check if string is empty then display the default value of the function else display the current value of the string. follow is the example, but it wont work, it's only for my idea that i wanted to do for almost 3 weeks now.
example:

function_name((empty($string)) ? : 'value');

got the idea!

abnfire

we create a function check_s now and it as default values, if the value is not written down then it will give the default values to strings and we have if statment to check if the string is empty then display the default value of the argument.

code:

function check_s($arg1 = 'no', $arg2 = 'argument 2', $arg3 = 'nas'){
return $arg1.$arg2.$arg3;
}

if( $s == 0 )
$foo[0] = ' localhost ';

if( $i == 0 )
$foo[1] = ' arg2 here ';

if( $i == 0 )
$foo[2] = ''; // call default value here

$stuff = implode( ",", $foo );

echo check_s($stuff);

ouput:
localhost,arg2 here,argument 2nas

it should display this outout:
localhost arg2 here nas

so if you can c is not working any idea how to make it work?

abnfire

Is there any pro at all to solve my question? because its been long time to stuck in one question, dont you think? 😕 :mad:

tomhath

After you call the "implode" function $stuff will contain this, because you told it to put a comma between each array element:

"localhost,arg2 here,"

So when you call check_s it will receive one argument (the value of $stuff, that is "localhost,arg2 here,").

The second and third arguments to check_s will get the defaults of "argument 2" and "nas" because you did not supply anything.

Function check_s will the return the three values concatenated together:

"localhost,arg2 here," . "argument 2" . "nas"

That's what you printed. It worked!

tomhath

Try this (what you intended for $arg1, an empty string for $arg2, and let it take the default for $arg3):

function check_s($arg1 = 'no', $arg2 = 'argument 2', $arg3 = 'nas'){ 
return $arg1.$arg2.$arg3; 
} 

if( $s == 0 ) 
$foo[0] = ' localhost '; 

if( $i == 0 ) 
$foo[1] = ' arg2 here '; 

if( $i == 0 ) 
$foo[2] = ''; // call default value here 

$stuff = implode( " ", $foo ); 

echo check_s($stuff, "");

abnfire

:mad:

abnfire

this is having the same problem again, can you really solve it. any pro out there for php? 😕

function check_s($arg1 = 'no', $arg2 = 'argument 2', $arg3 = 'nas'){
return $arg1.$arg2.$arg3;
}
if( $s == 0 )
$foo[0] = ' localhost ';

if( $i == 0 )
$foo[1] = '';// call default value here

if( $i == 0 )
$foo[2] = ' arg3 here ';

$stuff = implode( "", $foo );
echo check_s($stuff);

Weedpacket

Don't understand what you're trying to do any more.

function check_s($arg1='foo', $arg2='bar', $arg3='baz')
{ if(empty($arg1)) $arg1='foo';
  if(empty($arg2)) $arg2='bar';
  if(empty($arg3)) $arg3='baz';
return $arg1.$arg2.$arg3;
}

check_s(); // returns 'foobarbaz'
check_s('wibble'); // returns 'wibblebarbaz'
check_s('wibble','womble'); // returns 'wibblewomblebaz'
check_s('wibble','womble','ork'); // returns 'wibblewombleork'
check_s('wibble','','ork');  // returns wibblebarork

Is that the behaviour you want? Are you sure? What behaviour do you want? Do I undestand your problem? Do you?

tomhath

You don't understand what the implode is doing. It takes the three array elements $foo[0], $foo[1], and $foo[2] and makes them into one big string:

$stuff = implode( "", $foo );

It doesn't matter that you imploded three array elements into $stuff. It is now one string, and it all goes into $arg1.

No value is given here for $arg2 and $arg3, so they take the default.

check_s($stuff);

Is exactly the same as:

check_s($stuff, 'argument 2',  'nas');

abnfire

im calling lots of function same time, but im also getting string from outside for function, so i still want to know how to get default value if the string is empty and i dont want to write anything inside the function.

exmple:
$string1 = '';
$string2 = 'ok';
$string3 = 'bye';

function($string1,$string2,$string3);

any idea how to call default value from outside without going inside?

diego25

any idea how to call default value from outside without going inside?

That's impossible, as many people have told you in this thread and in the other one. Try another approach to what you're doing.

Diego

abnfire

All of you Thank You very much, i really feel really good to be in part of this community and i dont think there is way to call default value from outside without going inside a function if string is empty. thanks all you again. diego25 you are a man who know what he is doing and thanks again.