msqli_result could not be converted to string

Question

I'm trying to query the DB with mysqli and then fetch the result, but it's throwing an Object of class could not be converted to a string error.

Here's what I am trying to accomplish:

<?php
include ('conn.php');

$query= "select value from the_table where item = 'url'";
$result = mysqli_query($conn, $query);
?>

And then I am trying to populate a link with the $result:

<a href="<?php echo $result; ?>">This is the Link to the URL</a>

I saw this post: PHP and MySQL error: Object of class mysqli_result could not be converted to string

So, I tried to format the echo with echo $result->fetch_object()->url;, but that didn't work.

I'm not sure if I have to fetch the result and then throw it into a look with a mysqli_fetch_array() and if so, how do I get it to populate that the value outside of the loop?

You need to `$row = mysqli_fetch($result)` to get a row, first. — Marc B, Mar 10 '14 at 21:44
you always first have to fetch the actual result using the fetch_array or fetch_object method. That is correct. However your `->url` part indicates that you have a selected column called `url` in your query. While in the query posted, you do not. Instead, what you wanted is echo `$result->fetch_object->value` — Tularis, Mar 10 '14 at 21:44

score 0 · Answer 1 · answered Mar 10 '14 at 21:52

Assuming that $db is your database connection link you can perform the query:

$result = mysqli_query($db, 'select name from fruits');

Then you can get name using procedural style:

echo mysqli_fetch_object($result)->name;

Or object style:

echo $result->fetch_object()->name;

In your code you're trying to output url property which is not defined - as we can see in your query.

score 0 · Answer 2 · answered Mar 10 '14 at 21:54

Try some thing like this:-

    $query= "select value from the_table where item = 'url'";
    $result = mysqli_query($conn, $query) or die ("Couldn't execute query.");

    // use returned data
    while($row = mysqli_fetch_assoc($result))
    {
        echo $row['value'];
    }

OR

while ( $row = mysqli_fetch_assoc( $result ) )           
   {
      foreach ($row as $key => $value)
      {
     print ($row . " = " . $value . "\n");
      }
      print("================");
   }

msqli_result could not be converted to string

2 Answers2