Error displaying "text" in php before calculating the operation

Question

I want to print out a text before printing the output of the calculation, without storing the text in a variable. How can I achieve this?

Here is my code:

$x = 5;
$y = 10.5;
  echo "add: 5+6 =" . $x + $y;

You should display all errors and warnings to see what is wrong: https://stackoverflow.com/questions/1053424/how-do-i-get-php-errors-to-display?rq=1 — jeroen, Oct 11 '19 at 07:31

GabrieleMartini · Accepted Answer · 2019-10-11T09:28:58.460

This code calculate addition $x + $y, then convert to string, then concatenate to string "add: 5+6=" (see https://www.php.net/manual/en/function.strval.php)

$x = 5;
$y = 10.5;
echo "add: 5+6 = ". strval($x + $y);

About your code

See https://www.php.net/manual/en/language.operators.precedence.php : concatenation operators have higher precedence than arithmetic operators.

Your original code is the same as:

$x = 5;
$y = 10.5;
echo (int)("add: 5+6 =" . $x) + $y;

That means: first, concatenate string "add: 5+6 =" . $x

add 5+6 =5

then cast to int (that return 0)

(int)("add: 5+6 =5")

then add $y

0 + 10.5

then echo it

echo 10.5;

So this is why your code return 10.5

Please add some explanation to your code such that others can learn from it — Nico Haase, Oct 11 '19 at 08:16

score 2 · Answer 2 · answered Oct 11 '19 at 07:45

2

<?php
$x = 5;
$y = 10.5;

echo "add: 5+6 =" . ($x + $y); // automatic converts to `string`

answered Oct 11 '19 at 07:45

xayer

score 1 · Answer 3 · answered Oct 11 '19 at 07:37

1

You can try this.

<?php
    $x = 5;
    $y = 10.5;
    $z = $x + $y;
    echo "add: 5+6 = " . $z;
?>

answered Oct 11 '19 at 07:37

Krupal Panchal

Please add some explanation to your code such that others can learn from it – Nico Haase Oct 11 '19 at 08:15

score 1 · Answer 4 · answered Oct 11 '19 at 07:40

1

<?php
  $x = 5;
  $y = 10.5;
  printf("add: 5+6 = %f", $x + $y);
?>

answered Oct 11 '19 at 07:40

alprnkeskekoglu

Please add some explanation to your code such that others can learn from it – Nico Haase Oct 11 '19 at 08:15

score 1 · Answer 5 · answered Oct 11 '19 at 07:50

1

try this:

$x = 5;
$y = 10.5;
$z=$x+$y;
echo "add:5+6 =".$z;

answered Oct 11 '19 at 07:50

PHP Geek

score 1 · Answer 6 · answered Oct 11 '19 at 09:01

1

You can do it by concatinating your variable

<?php
$x = 5;
$y =10.5;
$z=$x+$y;

echo "add: ".$x."+".$y." = ". $z;
?>

output

add: 5+10.5 = 15.5

answered Oct 11 '19 at 09:01

Manish Gupta