Not getting proper data in php mysql

Question

table: request

id          friend_id           candidate_id
==          ==============      ==============
1           20180928115958      20180925112428

2           20181008181142      20180925112428

3           20180928115958      20181008181142

4           20181010181207      20180928115958

5           20181008181142      20181010181207

6           20181010181207      20180925112428

Query

<?php
    $_SESSION['candidate_id']=20181010181207;

    $sql_ac = mysqli_query($con,"select friend_id,candidate_id from request where candidate_id='".$_SESSION['candidate_id']."' or friend_id='".$_SESSION['candidate_id']."'");
    while($row_ac = mysqli_fetch_array($sql_ac))
    {
        if($row_ac['friend_id']==$_SESSION['candidate_id'] && $row_ac['candidate_id']==$_SESSION['candidate_id'])
        {
            echo "";
        }
        else
        {
            echo $row_ac['friend_id']."<br/><br/>";
            echo $row_ac['candidate_id']."<br/><br/>";
        }
    }
?>

Unexpected output:

20181010181207

20180928115958

20181008181142

20181010181207

20181010181207

20180925112428

Expected Output

20180928115958

20181008181142

20180925112428

In this code I have table request which have two key field i.e. friend_id and candidate_id. Now, I $_SESSION['candidate_id']='20181010181207'. So, How can I get expected output as I have mention above? Please help me.

Thank You

Still it showing 20181010181207 20180928115958 20181010181207 20180925112428 @MickaelLeger — Rudra
– Rudra, Commented Oct 11, 2018 at 12:43
It is doing exactly what you're asking it to do. It is returning 3 rows which match candidate or friend. Then, for each one of those rows, you're outputting both the candidate ID and also the friend ID. Maybe, if you tell us what you're trying to do we could help. — waterloomatt
– waterloomatt, Commented Oct 11, 2018 at 12:43
Do you want to output friend_id when candidate_id matches and vice versa when friend_id matches? — waterloomatt
– waterloomatt, Commented Oct 11, 2018 at 12:46
have two ifs: if(($row_ac['friend_id'] != $_SESSION['candidate_id'] { echo $row_ac['friend_id'] } - same for conadidate_id — Jeff
– Jeff, Commented Oct 11, 2018 at 12:46
I want if this id 20181010181207 present in both friend_id and candidate_id. it will not show — Rudra
– Rudra, Commented Oct 11, 2018 at 12:49

Jeff · Accepted Answer · 2018-10-11 12:54:18Z

1

I think you want:

while($row_ac = mysqli_fetch_array($sql_ac))
{
    if($row_ac['friend_id'] != $_SESSION['candidate_id']
        echo $row_ac['friend_id']."<br/><br/>";
    }
    if($row_ac['candidate_id'] != $_SESSION['candidate_id']
        echo $row_ac['candidate_id']."<br/><br/>";
    }
}

This will display the values, that are not the same as the 'input', in other words remove the 20181010181207 entries.

Or as a oneliner:

echo $row_ac['candidate_id']==$_SESSION['candidate_id'] ? $row_ac['friend_id] : $row_ac['candidate_id'];

edited Oct 11, 2018 at 12:54

answered Oct 11, 2018 at 12:48

Jeff

6,9611 gold badge19 silver badges34 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Rudra Over a year ago

exactly I need this. Thanks @Jeff

Jeff Over a year ago

You are welcome! There might be a better way in achieving this already on sql level. But it's hard to tell if it would fit your needs.

Jeff Over a year ago

See @MickaelLeger s answer for a sql version!

Mickaël Leger · Accepted Answer · 2018-10-11 12:55:49Z

Your request is :

select 
  friend_id,
  candidate_id
from request 
where candidate_id=:candidate_id or friend_id=:candidate_id

So it's normal that you will return ALL row where :candidate_id is in column friend_id or candidate_id :

4           20181010181207      20180928115958

5           20181008181142      20181010181207

6           20181010181207      20180925112428

Now what you want is :

20180928115958 // this is candidate_id of row 4

20181008181142 // this is friend_id of row 5

20180925112428 // this is candidate_id of row 6

If you only want the value of the column "XXX_id" that is not equal to :candidate_id, try this :

select 
  IF(friend_id = :candidate_id, '', friend_id) as friend_id,
  IF(candidate_id = :candidate_id, '', candidate_id) as candidate_id
from request 
where candidate_id=:candidate_id or friend_id=:candidate_id

This way you will return empty value for column that match your :candidate_id.

Now just add a test before displaying it to avoid empty row :

if (!empty($row_ac['friend_id'])
    echo $row_ac['friend_id']."<br/><br/>";
if (!empty($row_ac['candidate_id'])
    echo $row_ac['candidate_id']."<br/><br/>";

You can do this too :

select 
  IF(friend_id = :candidate_id, candidate_id, friend_id) as result_id
from request 
where candidate_id=:candidate_id or friend_id=:candidate_id

This way you will return only ONE result : the one that is not equal to :candidate_id.

Then just do echo $row_ac['result_id']."<br/><br/>"; to get what you want

Bhargav Chudasama · Accepted Answer · 2018-10-11 12:51:46Z

0

try this one

i hope it will working

if($row_ac['friend_id']==$_SESSION['candidate_id']){
    echo $row_ac['candidate_id']."<br/><br/>";
else{
    echo $row_ac['friend_id']."<br/><br/>";
}

Because And condition always false in your case and else part will execute so try with if else condition and check

edited Oct 11, 2018 at 12:51

answered Oct 11, 2018 at 12:46

Bhargav Chudasama

7,2256 gold badges24 silver badges41 bronze badges

Collectives™ on Stack Overflow

Not getting proper data in php mysql

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related