php float number error

Question

I am trying to assign float value in php to variable I tried following,

$_web_lat=‎18.501059;
$_web_long=73.862686;

echo $_web_lat .'='. $_web_long;

[Parse error: syntax error, unexpected '.501059' (T_DNUMBER)]

OR

$_web_lat=floatval('‎18.501059');
$_web_long=floatval('‎‎73.862686');

echo $_web_lat .'='. $_web_long;

Both shows 0 as output? Anyone guide me on this?

Possible duplicate of PHP parse/syntax errors; and how to solve them? — JJJ
– JJJ, Commented Mar 19, 2018 at 8:05

Karlo Kokkak · Accepted Answer · 2018-09-17 10:58:42Z

11

Your code seems to have a hidden character ?

Try copy and use this:

<?php
$_web_lat=18.501059;
$_web_long=73.862686;

echo $_web_lat .'='. $_web_long;

?>

edited Sep 17, 2018 at 10:58

answered Mar 19, 2018 at 8:01

Karlo Kokkak

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10 Comments

Ayaz Ali Shah Over a year ago

@urban Can you let me know how can I check with vi and what is vi?

Adder Over a year ago

The hidden character is before the 18, I see this in php fiddle.

SK IRT Over a year ago

can you please elaborate on this? in my code there is no space/char but it shows error

Ayaz Ali Shah Over a year ago

@SKIRT You need to copy past the script from this answer, it will be functional

Adder Over a year ago

Copy your codes into phpfiddle.org and its editor will show extra characters for both of your code snippets. What editor are you using?

|

Xay · Accepted Answer · 2018-03-19 07:59:46Z

0

Try to write floating values like

<?php

    $_web_lat=floatval('‎18.501059f');
    $_web_long=floatval('‎‎73.862686f');


$float_value_of_var1 = floatval($_web_lat);
$float_value_of_var2 = floatval($_web_long);

echo $float_value_of_var1; // 18.501059
echo $float_value_of_var2; // ‎‎73.862686

?>

answered Mar 19, 2018 at 7:59

Xay

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Comments

ScaisEdge · Accepted Answer · 2018-03-19 08:07:22Z

0

the sintax error is caused by automatic string conversion do the string operator . (dot concat) if you want avoid this you could use or cast float value ast string

  $_web_lat=‎18.501059;
  $_web_long=73.862686;

  echo (string) $_web_lat .'='. (string>$_web_long;

answered Mar 19, 2018 at 8:07

ScaisEdge

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