Message array to string conversion

Question

That might be a fool question, but I need to know how to solve this: Notice: Array to string conversion in C:\xampp\htdocs\search_view.php on line 248 Why am I getting this message, what can I do to solve it?

echo'<div id="thumb"> 

'.$ids = array();
$ids[] = $results['idGames'];  
for ($i = 0; $i < count($ids); $i++) {

$id = $ids[$i];




$v  = $results['total_votes'];
$tv = $results['total_value'];
if ($v)
    $rat = $tv / $v;
else
    $rat = 0;



$j  = $ids[$i];
$id = $ids[$i];
echo '<div class="topcontentstar">

    <div id="' . $id . '" class="">';
for ($k = 1; $k < 6; $k++) {
 if ($rat + 1 > $k)
    $class = "" . $k . "  ratings_stars_index ratings_vote";
    else
    $class = "" . $k . " ratings_stars_index ratings_blank";
     echo '<div class="' . $class . '"></div>';
}
echo ' 
</div>
    </div></div>;

foreach more suit array looping.

Lawrence Cherone
– Lawrence Cherone

2013-07-21 22:38:18 +00:00
Commented Jul 21, 2013 at 22:38 — Lawrence Cherone
– Lawrence Cherone, Commented Jul 21, 2013 at 22:38

Amal · Accepted Answer · 2013-09-21 14:06:54Z

3

Because in this part of code you tried to convert an array to string via concatenation

echo'<div id="thumb"> 
(line 248) '.$ids = array();

Separate them: $ids = array()

echo'<div id="thumb"> 
(line 248) ';
$ids = array();

edited Sep 21, 2013 at 14:06

Amal

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answered Jul 21, 2013 at 22:39

user1646111

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Comments

Hidde · Accepted Answer · 2013-07-21 22:38:18Z

2

echo'<div id="thumb"> 


(line 248) '.$ids = array();

You are concatenating a string and an array, just as the errors says. You are echoing the string, and appending the array $ids to that. Because assigning a value gets a higher precedence than concatenating things, $ids is already an array.

answered Jul 21, 2013 at 22:38

Hidde

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Comments

Amal · Accepted Answer · 2013-07-21 22:41:54Z

1

You're doing this:

echo'<div id="thumb"> 
(line 248) '.$ids = array();

Basically, you can't concatenate array with a string and that's why the error appears.

To fix the error, you can separate the array declaration into a separate line:

echo'<div id="thumb">';
$ids = array();

Hope this helps!

answered Jul 21, 2013 at 22:41

Amal

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3 Comments

Mariangela Vidotto Over a year ago

Great it works, but know is echoing a ";" along the content, was that suppose to happen?

Amal Over a year ago

@MariangelaVidotto: not really. See the last line of your code: </div></div>; -- that might be the issue. :-)

Mariangela Vidotto Over a year ago

Alright now it's perfect, thank you for the help, I really appreciate it, thank you all.

Ben Harris · Accepted Answer · 2013-07-21 22:39:00Z

-1

As a side note, I can see a problem in your last few lines:

echo ' 
</div>
    </div></div>;

Should be:

echo '</div></div></div>';

answered Jul 21, 2013 at 22:39

Ben Harris

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2 Comments

Hidde Over a year ago

-1. Having newlines in a single quoted string is not illegal PHP syntax.

MERT DOĞAN Over a year ago

But you forgot ' character

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