How to make appear pipe according to the conditions

Question

Here i have the variable $json concatenated with possible three conditions.

According the below conditions $first , $second and $third, the output should get concatenate with pipe. So that the output should be either three given below

Expected

first
or 
first | second
or 
first | second | third

Here is my Code simplified :

<?php
$json = "";
$first = "1";
$second = "1";
$third = "1";
$pipe = "|";
        if ($first == 1)
        {        
        $json .= "first";
        }
        $json .= $pipe;
        if ($second == 1)
        {        
        $json .= "second";
        }
        $json .= $pipe;
        if ($third == 1)
        {
        $json .= "third";
        }    
    echo $json;
?>

How can i make the $pipe variable to get my output expected for the variable $json

Note : As i use some complex flexi grid , implode still makes little complex

I'm getting the same output as your desired output. However you should include pipe condition in your `if` condition — Prateek, Oct 26 '14 at 12:27
@Prateek @ Ancient Geek If the $second was set 0 then the output $json will be first||third, which is expected as first | third — Sulthan Allaudeen, Oct 26 '14 at 12:30
You should include pipe condition in your if condition. http://codepad.org/CL33qXI8 — Prateek, Oct 26 '14 at 12:30
Your question didn't explain what was wrong; hence, your code seemed to work. Edit the question and *explain* what is wrong, you'll get better responses. — Jared Farrish, Oct 26 '14 at 12:31
@Prateek Then $third is set 0 the ouput will be first|second| — Sulthan Allaudeen, Oct 26 '14 at 12:33
possible duplicate of [Stripping last character from a foreach loop](http://stackoverflow.com/questions/19050704/stripping-last-character-from-a-foreach-loop) — Glavić, Oct 26 '14 at 13:03

score 3 · Accepted Answer · edited May 23 '17 at 12:06

3

Try this:

$json = "";
$first = "1";
$second = "1";
$third = "1";
$pipe = "|";
$data = array();
if ($first == 1)  $data[] = "first";
if ($second == 1) $data[] = "second";
if ($third == 1)  $data[] = "third";
$json = implode($pipe, $data);
echo $json;

Or use some other way, like this for example.

edited May 23 '17 at 12:06

Community

1
1

answered Oct 26 '14 at 12:32

Glavić

42,781
13
77
107

I didn't think of it. Nice +1 – Prateek Oct 26 '14 at 12:33
Thanks, i should go for implode – Sulthan Allaudeen Oct 26 '14 at 12:43

score 2 · Answer 2 · answered Oct 26 '14 at 12:33

You can do this:

$array = array(
    'first'  => $first,
    'second' => $second,
    'third'  => $third
);

echo implode($pipe, array_keys(array_filter($array)));

array_filter removes any key/value pair whose value is falsely, array_keys returns an array of their keys, and implode joins the values using $pipe.

score 0 · Answer 3 · answered Oct 26 '14 at 12:32

0

The easiest solution for this is to always concatenate a pipe after a variable concatenation Then truncate the the final string by 1 character. If you're printing straight to stdout, remove this last pipe with a back-space, space combination.

answered Oct 26 '14 at 12:32

phil

561
3
10

score 0 · Answer 4 · answered Oct 26 '14 at 12:40

0

$json;
$first = true;
$second = true;
$third = true;
$pipe = " | ";

if ($first === true) {
    $json .= "first";
}
if ($second === true) {
    $json .= $pipe . "second";
}
if ($third === true) {
    $json .= $pipe . "third";
}
echo $json;

answered Oct 26 '14 at 12:40

MrGrigri

304
4
15

What is the output if $first is false? – Glavić Oct 26 '14 at 12:40
Which is expected as second | third – Sulthan Allaudeen Oct 26 '14 at 12:45
You'd have to have a nested if else statement within the second if statement checking the $first variable again. And if the $first variable is true concatenate the pipe at the beginning, else just concatenate the $second. – MrGrigri Oct 26 '14 at 12:51

How to make appear pipe according to the conditions

4 Answers4