php question

Peter B

I've never quite understood how these if statements work but could someone please explain what going on here.

I dont actually understand what the single "=" (equals) means in an if statment.

Does it mean

if $file1 is not false, then $lines1 = file($file1) therefore "Worked".
else
$lines1 = false; therefore ERROR.


[PHP]
if ($lines1 = file($file1))
{
echo "Worked";
} else {
//failed
echo "ERROR";
}
[/PHP]

Goodshape

Fairly straightforward isn't it?

if $lines1 is the same as file($file1) then echo "worked". If it isn't then eco "ERROR".

What's the problem?

Peter B

I thought that was the case with 2 "=" (equals signs) but not with 1.

I would understand that to be the case if I had written

[PHP]
if ($lines1 == file($file1))
{
echo "Worked";
} else {
//failed
echo "ERROR";
} [/PHP]

Maybe it is the same in PHP, I'm not sure

Goodshape

Hmm, right you are acutally. Two equels are better than one.

I'm not actually sure what the single "=" does.. prehaps nothing. It could be that the two equel signs are used so as to keep in line with "=<" (equel to or less than) and so forth.

marcphisto

= is an assignment operator
== is a comparison operator

Think of it this way

X = Y is placing the contents of Y into X

whereas

X == Y is making a comparison between X and Y

If you put X = Y into and if statement it will always return true.

akari no ryu

It depends on what you're trying to do.
Let's look at an example.

[php]
class test{
var $prop;
function test($prop=null){
$this->prop=$prop;
}

function setProp($prop){
$this->prop=$prop;
}

function getProp(){
return $this->prop;
}
}

$test1=new test();
$test2=new test("blah");

if($test1Prop=$test1->getProp()){
echo "got test 1 prop: $test1Prop";
}

if($test2Prop=$test2->getProp(){
echo "got test 2 prop: $test2Prop";
}
[/php]

Essentially, when an assignation fails ($a=doSomething()) the if statement returns false.

akari no ryu

no it won't
[php]if($result=mysql_query($someSQL)){
echo "query succeeded";
}else{
echo "query failed";
}[/php]

Peter B

Hmm think I have found the solution.

Since you are assigning x = y
x will always be assigned value of y, therefore both values will be the same
So if y is equal to "", or y is equal to 0, or y is equal to FALSE, then it will be the same as

[PHP]if (FALSE)
{

} else{

}[/PHP]

I think by using the single "=" in an if statment just allows you to assign at the same time as checking if it is true or not.

I understand exactly the same functioning code would be

[PHP]
$lines1 = file($file1);
if ($lines1)
{
echo "Worked";
} else {
//failed
echo "ERROR";
} [/PHP]

akari no ryu

No.
With assigning, if the assigning happened, it's true. If, for some reason, the assigning didn't happen, then the assignation didn't happen so it's false.

Take for example the mysql_query.

If there's something wrong with the query, the assignation $result=mysql_query($someSQL) will not have happened.

The if will return false.

If is just

if(whatever's in here is true){
  block
}else
  block
}

Assignations do not always return true.

seamus

akari no ryu wrote:

No.
With assigning, if the assigning happened, it's true. If, for some reason, the assigning didn't happen, then the assignation didn't happen so it's false.

Well, yes and no.

Basically, when you insert an assignment operator into an if() statement, the preprocessor evaluates the statement inside the if() conditional, and then tests the value of the item which is being assigned. If the value of the item assigned = true, then the condition is true.

That is, imagine you have

if($x = someFunction())

PHP will carry out someFunction(), assign the return value into the $x variable, and then test the value of $x for true or false.

Peter B

akari no ryu wrote:

No.
With assigning, if the assigning happened, it's true. If, for some reason, the assigning didn't happen, then the assignation didn't happen so it's false.

Thanks for the help akari no ryu

However since I am the OP and was asking the question in the first place I should not really be disagreeing with your comment but if I run the code below the assigning does happen, $hmmm turnes to 0, however the if statment does read it as FALSE.

[PHP]
$hmmm=0;
if ($lines1 = $hmmm)
{
echo "Worked";
} else {
//failed
echo "ERROR";
}
echo "<br> Lines1 ->".$lines1."<-";
if (!$lines)
{
echo "<br>FALSE";
}
[/PHP]

This code echos
[PHP]ERROR
Lines1 ->0<-
FALSE[/PHP]

pH

[php]$hmmm = 0;[/php]

[php]if ($lines1 = $hmmm)[/php] Will equate to FALSE
if (0) is always false.

[PHP]if (!$lines)[/PHP] Will equate to TRUE

see here:
http://www.zend.com/manual/types.comparisons.php
(Last column of the first table on that page)

Cake Fiend

akari no ryu wrote:

Take for example the mysql_query.

If there's something wrong with the query, the assignation $result=mysql_query($someSQL) will not have happened.

Not quite - mysql_query() returns FALSE on error, so the assignment will still take place, assigning $result the value FALSE. $result is then evaluated as Seamus says, and the if statement fails.

democrates

And while we're at it there's '===' which also compares type, eg
[PHP]'1' == 1 //returns true though one is a string type and the other is numeric
'1' === 1 //returns false
[/PHP]