Why doesn't the html br break line tag doesn't work in this code?

Question

Can some one tell why my php line break not working ( echoing ) ?

I know i can write the code in a different way to make the line break work, but i want to know the reason behind this ?

<?php

    $var1 = 3;

    echo "Addition = "       . $var1 += 3 . "<br>";
    echo "Subtraction = "    . $var1 -= 3 . "<br>";
    echo "Multiplication = " . $var1 *= 3 . "<br>";
    echo "Division = "       . $var1 /= 3 . "<br>";

?>

What do you mean by "does not work"? Try to be more specific. — arkascha, May 25 '15 at 09:49
If you are using CLI try '\r\n' otherwise in web
should work — chandresh_cool, May 25 '15 at 09:52
Read up in operator precedence.(Seems like a war area for vampires.) — Rizier123, May 25 '15 at 10:03

score 8 · Answer 1 · edited Jun 20 '20 at 09:12

Well seems like I have to clean some things up here.

Let's take a look at the operator precedence, which says:

. has a higher precedence, than +=, -=, *=, /=
. is left associative
=, +=, -=, *=, /= is right associative
We also take a look at the note at the bottom of the manual:

Note: Although = has a lower precedence than most other operators, PHP will still allow expressions similar to the following: if (!$a = foo()), in which case the return value of foo() is put into $a.

Means that even tough = has a lower precedence than . it gets evaluated first. You can also see this if you do something like this:

$xy = "HERE";
echo "I am " . $xy = "NOT HERE";

Now you would think that . has a higher precedence than = and will get evaluated first, but as from the note in the manual, the assignment is first and you end up with this:

echo "I am " . ($xy = "NOT HERE");

output:

I am NOT HERE

So if we put all these information's together, we can say, that the assignment gets evaluated first, but it's right assocative. Means this:

$var1 = 3;

echo "Addition = "    . ($var1 += 3 . "<br>");
echo "Subtraction = " . ($var1 -= 3 . "<br>");
echo "Addition = "    . ($var1 *= 3 . "<br>");
echo "Addition = "    . ($var1 /= 3 . "<br>");

So this code will end up in this:

echo "Addition = "    . ($var1 += "3<br>");
echo "Subtraction = " . ($var1 -= "3<br>");
echo "Addition = "    . ($var1 *= "3<br>");
echo "Addition = "    . ($var1 /= "3<br>");

Which then through the arithmetic operator gets convert to an integer we end up with this:

echo "Addition = "    . ($var1 += 3);
echo "Subtraction = " . ($var1 -= 3);
echo "Addition = "    . ($var1 *= 3);
echo "Addition = "    . ($var1 /= 3);

And after the assignment is done the concatenation gets evaluated, which looks like this:

echo "Addition = "    . 6;
echo "Subtraction = " . 3;
echo "Addition = "    . 9;
echo "Addition = "    . 3;

With this you end up in this output:

Addition = 6Subtraction = 3Addition = 9Addition = 3

And now how to solve this? Simply wrap your assignment in parentheses, so that the   tag doesn't get into the assignment. E.g.

echo "Addition = "       . ($var1 += 3) . "<br>";
echo "Subtraction = "    . ($var1 -= 3) . "<br>";
echo "Multiplication = " . ($var1 *= 3) . "<br>";
echo "Division = "       . ($var1 /= 3) . "<br>";
                         //^          ^ So the br tag doesn't get in the assignment of the variable.

This answer doesn't seem to contain false information like some of the others and clarifies what happens and why. +1 — Peter, May 25 '15 at 10:36
@Peter Thanks, yes I though I break everything in single steps to show what happens. — Rizier123, May 25 '15 at 10:41
@Uchiha Thanks I tried to make everything as clear as possible. — Rizier123, May 25 '15 at 10:54

score 4 · Answer 2 · answered May 25 '15 at 09:59

4

This is happening because of the type casting issues. 3 . " " will be converted to number while the operation will be performed. Wrap the inside () so that the operations are performed first then the concatenation.

echo "Addition = " . ($var1 += 3) . "<br>";
echo "Subtraction = " . ($var1 -= 3) ."<br>";
echo "Addition = " . ($var1 *= 3) . "<br>";
echo "Addition = " . ($var1 /= 3) ."<br>";

answered May 25 '15 at 09:59

Sougata Bose

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What difference between my answer and yours? – Deenadhayalan Manoharan May 25 '15 at 10:01
@Deena Why should OP try your answer? What is the issue? – Sougata Bose May 25 '15 at 10:02
1

@Deena might be an explanation – Narendrasingh Sisodia May 25 '15 at 10:02
1

@Uchiha you are right. – Sougata Bose May 25 '15 at 10:03

n-dru · Accepted Answer · 2015-05-25T09:54:33.347

3

You can use commas,

echo "Addition = " . $var1 += 3 , "<br>";
echo "Subtraction = " . $var1 -= 3 ,"<br>";
echo "Addition = " . $var1 *= 3 , "<br>";
echo "Addition = " . $var1 /= 3 ,"<br>";

Or wrap it in brackets:

echo "Addition = " . ($var1 += 3) . "<br>";
echo "Subtraction = " . ($var1 -= 3) ."<br>";
echo "Addition = " . ($var1 *= 3) . "<br>";
echo "Addition = " . ($var1 /= 3) ."<br>";

Otherwise the 3 number is concatenated with  .

edited May 25 '15 at 09:54

answered May 25 '15 at 09:51

n-dru

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Interesting, but lacks any form of explanation. – arkascha May 25 '15 at 09:53
http://stackoverflow.com/questions/1466408/difference-between-and-in-php – n-dru May 25 '15 at 09:53

score 3 · Answer 4 · edited May 25 '15 at 10:01

3

Your PHP means:

echo "Addition = " . $var1 += (3 . "<br>");
echo "Subtraction = " . $var1 -= (3 ."<br>");
echo "Addition = " . $var1 *= (3 . "<br>");
echo "Addition = " . $var1 /= (3 ."<br>");

And number + 3 . ' ' is number + (int)(3 . ' ') which is number + 3. No   exists now due to retyping to number(converting to number).

Use brackets around equations.

echo "Addition = " . ($var1 += 3) . "<br>";
echo "Subtraction = " . ($var1 -= 3) ."<br>";
echo "Addition = " . ($var1 *= 3) . "<br>";
echo "Addition = " . ($var1 /= 3) ."<br>";

edited May 25 '15 at 10:01

Sougata Bose

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answered May 25 '15 at 09:52

pavel

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The concatenation dot is left associative! NOT right! – Rizier123 May 25 '15 at 09:53
@Rizier123: are you sure? It works me well as I wrote above. Or... is there another reason why the behavior is as I wrote? You can insert own answer if you know st. more. – pavel May 25 '15 at 09:55
From the [manual](http://php.net/manual/en/language.operators.precedence.php) : **left** `* / % arithmetic` – Rizier123 May 25 '15 at 09:56

score 1 · Answer 5 · answered May 25 '15 at 09:54

1

Try this..

"." is used for php variable to concate not for numbers

<?php


$var1 = 3;

echo "Addition = ". ($var1 += 3) ."</br>";
echo "Subtraction = ". ($var1 -= 3) ."</br>";
echo "Addition = ". ($var1 *= 3) ."</br>";
echo "Addition = ". ($var1 /= 3) ."</br>";

?>

answered May 25 '15 at 09:54

Deenadhayalan Manoharan

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Sorry, but that explanation above the code is total nonsense. – arkascha May 25 '15 at 09:56
ok can you explain this – Deenadhayalan Manoharan May 25 '15 at 09:56
The dot operator concatenates whatever string operands it is given, regardless whether it is variables or literals or language constructs. Numeric values would get interpreted as strings too by conversion. Out of interest: where is that statement from that the dot operator "is only" for variables? – arkascha May 25 '15 at 09:58
Yours also work but need some explanation Mate@Deena.Your explanation doesn't ,make any sense why OPs code is not working – Narendrasingh Sisodia May 25 '15 at 10:18

Mahadeva Prasad · Answer 6 · 2015-05-25T10:04:01.110

1

Try this way.

 <?php
$var1 = 3;

echo "Addition =" . ($var1 += 3 ).'<br>';
echo "Subtraction =" . ($var1 -= 3).'<br>';
echo "Addition =" . ($var1 *= 3 ).'<br>';
echo "Addition =" . ($var1 /= 3 ).'<br>';

?>

edited May 25 '15 at 10:04

answered May 25 '15 at 09:58

Mahadeva Prasad

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Extremely complicated change to achieve only a minor difference. Why? – arkascha May 25 '15 at 10:01

Why doesn't the html br break line tag doesn't work in this code?

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