PHP operators problems

Question

I am building a PHP script. From one form I get min_salary and max_salary. I am trying to get both of them as a string in a salary variable like

$salary = "min_salary TO max_salary";

If the min salary is not specified it should be set to 0, similarly if max salary is not specified it should be set to *.

I have written the below code but am not getting anything when either of the fields are not provided.

$salary = (isset($data['min_salary']) ? $data['min_salary'] : 0) .' TO '. (isset($data['max_salary']) ? $data['max_salary']:'*');

Also I don't want to use if then else statements.

The code looks fine for me, what do you mean with "I have written the below code but not getting anything when any of the field is not provided." - what outputs var_dump($salary) — Flixer
– Flixer, Commented Feb 5, 2014 at 23:19
suppose that the min_salary input box is left blank and max_salary is given 10 then the output should be like "0 TO 10" but I am getting only "TO 10" :( — Jay Chakra
– Jay Chakra, Commented Feb 5, 2014 at 23:21
There is a difference between the methods "isset" an "empty" - be aware of it. If you want to check whether is empty replace "isset(.." with "!empty(..." — Flixer
– Flixer, Commented Feb 5, 2014 at 23:23
You're already using if-else statements. Did you try casting the min_salary variable to integer? — mehmetseckin
– mehmetseckin, Commented Feb 5, 2014 at 23:25
Thanks Flixer your answer works fine just replaced isset with !empty — Jay Chakra
– Jay Chakra, Commented Feb 5, 2014 at 23:35

Josh Crozier · Accepted Answer · 2015-12-10 03:10:47Z

2

This line would produce that code without using if/else statements (even though ternery operator are really syntactic sugar around if/else statements)

$salary = max(0,$data['min_salary']) . ' TO ' . ($data['max_salary'] > 0 ? $data['max_salary'] : '*');

You don't really want the same scripting for both values as one should fallback to 0 and the other to *. The problem with isset():

(isset($data['min_salary']) ? $data['min_salary'] : 0)

is that a variable can be set to an empty string $data = '' which would return true. I'd hazard you do not want this to happen.

edited Dec 10, 2015 at 3:10

Josh Crozier

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answered Feb 5, 2014 at 23:30

Félix Adriyel Gagnon-Grenier

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1 Comment

Jay Chakra Over a year ago

Right, this may be an other alternative to right the same. But I got my answer thanks for all your help.

Jay Chakra · Accepted Answer · 2014-02-06 19:54:44Z

2

Thanks Flixer and Felix for your answer it works fine after I replaced isset with !empty

$salary = (!empty($data['min_salary']) ? $data['min_salary'] : 0).' TO '. (!empty($data['max_salary']) ? $data['max_salary']:'*');

answered Feb 6, 2014 at 19:54

Jay Chakra

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Comments

Roxxorfreak · Accepted Answer · 2014-02-05 23:24:18Z

1

Cast the values to int:

$salary = (isset($data['min_salary']) ? (int)$data['min_salary'] : 0).' TO '...

Same for the second part...

answered Feb 5, 2014 at 23:24

Roxxorfreak

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1 Comment

Jay Chakra Over a year ago

Thanks but not work. Anyways I got answer from Flixer.

Nicolas Defranoux · Accepted Answer · 2014-02-05 23:26:46Z

-1

Your statement is valid and provides a string, it works as-is on my php server (php 5.3.7) with $data not defined or defined and containing only one value.

Maybe it's the way you use $salary that has a concern.

answered Feb 5, 2014 at 23:26

Nicolas Defranoux

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