Finding result in a PHP Array

Question

I'm trying to find out all of my "Likes" from a mysql table and to put them into an array but I'm really not sure if I'm doing it right as I keep getting a "Wrong Datatype" error. Here is my code:

<?php
$check_like_sql = "SELECT * FROM posts WHERE type = 'Like' && poster = '$yourid'";
$check_like_res = mysqli_query($con, $check_like_sql) or die (mysqli_error());
if(mysqli_affected_rows($con)>0){
    while($likes = mysqli_fetch_assoc($check_like_res)){
        $yourlike = $likes['media'];    
    }
    $likearray = mysqli_fetch_array($con, $yourlike); 
}
?>
<?php
if(in_array($likearray, $postid)) {
            $likethis = "<a href=\"php/unlike.php?poster=$yourid&post=$postid\">Unlike</a> . ";
        }
        else if($posttype == "Like"){
            $likethis = "";
        }
        else{
            $likethis = "<a href=\"php/like.php?poster=$yourid&lat=$yourlat&lon=$yourlon&like=$postid&user=$postusername\">Like</a> . ";
        }
?>

Could anyone please explain where there might be an error? I'm very new to this sort of php coding. Thanks

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in --- on line 8 Warning: in_array() [function.in-array]: Wrong datatype for second argument in --- on line 47 — user2516546
– user2516546, Commented Jun 25, 2013 at 9:19
What are you trying to do with mysqli_fetch_array($con, $yourlike); ? see mysqli_fetch_array — Brewal
– Brewal, Commented Jun 25, 2013 at 9:20
what is in $postid ?? and from where the value is assigned — M Khalid Junaid
– M Khalid Junaid, Commented Jun 25, 2013 at 9:21
@Brewal I want to get all of the posts that I have "Liked" so that when they are shown I can choose to have either a Like or Unlike button depending on what I have already chosen — user2516546
– user2516546, Commented Jun 25, 2013 at 9:23
You have looped around all the returned records from the SQL. Then once you have finished you are trying to use mysqli_fetch_array to return another record — Kickstart
– Kickstart, Commented Jun 25, 2013 at 9:24

Yogesh Suthar · Accepted Answer · 2013-06-25 09:39:57Z

1

You are overriding $yourlike so first initialise it. Also there is no need for mysqli_fetch_array remove that part.

$yourlike = array();
if(mysqli_affected_rows($con)>0){
    while($likes = mysqli_fetch_assoc($check_like_res)){
        $yourlike[] = $likes['media'];    
    }
    //$likearray = mysqli_fetch_array($con, $yourlike);  // remove this part
}

Edit

Change this

if(in_array($likearray, $postid))

to

if(in_array($postid , $yourlike))

edited Jun 25, 2013 at 9:39

answered Jun 25, 2013 at 9:23

Yogesh Suthar

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5 Comments

user2516546 Over a year ago

Thanks, this has cleared up the first error but what do I now do about if(in_array($likearray, $postid)) {?

RDK Over a year ago

I think that if(in_array($yourlike, $postid, true)) is batter

Yogesh Suthar Over a year ago

@user2516546 Post the error. Without knowing that we can't help you out.

user2516546 Over a year ago

I did - It's Warning: in_array() [function.in-array]: Wrong datatype for second argument in C:\--- on line 47

Yogesh Suthar Over a year ago

@user2516546 Ohh yeah it should be if(in_array($postid , $yourlike)). See edited answer also.

user2361529 · Accepted Answer · 2013-06-25 09:22:36Z

0

you are using mysqli_fetch_array incorrecty. The first argument is the result the second is the type to return

$result = mysqli_fetch_array($query_result,MYSQLI_ASSOC);

however it looks like you are using it incorrectlyu as you have already fetched an assoc, can you not just delete the line $likearray = mysqli_fetch_array($con, $yourlike);

and change

$yourlike = $likes['media'];

to

$yourlike[] = $likes['media'];

and

if(in_array($likearray, $postid)) {

to

if(in_array($yourlike, $postid)) {

answered Jun 25, 2013 at 9:22

user2361529

11

Comments

Jahanzeb · Accepted Answer · 2013-06-25 09:28:25Z

0

$yourlike is not mysqli_result.

Replace this line

$likearray = mysqli_fetch_array($con, $yourlike);

with this one

$likearray = mysqli_fetch_array($con, $check_like_res);

edited Jun 25, 2013 at 9:28

answered Jun 25, 2013 at 9:21

Jahanzeb

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