PHP & MySQL Error - An error occurred on line 176: Object of class mysqli_result could not be co


I was wondering what is causing this error and how can I correct this problem?

Line 176 is below.

$q = "INSERT INTO users (first_name, middle_name, ". "last_name, rn, username, month, day, year, se, pic, ". "salt, password, rb, aa, ul, rd, ll, ui) ". "VALUES ('" . $fn . "', NULL, '" . $ln . "', NULL, '" . $u . "', NULL, NULL, NULL, NULL, NULL, '" . $salt . "', '" . $p . "', '" . $r . "', '" . $ag . "', 0, NOW(), NULL, '" . $unique_id . "')"; $r = mysqli_query ($dbc, $q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));


One of the variables you are injecting into that query string must be a result from an earlier query.

It's probably the $r.


Does the code following line 176 resemble this posters code? If so, try changing it to resemble more of what comes beneath it.

<a href="http://www.phpfreaks.com/forums/php-coding-help/(solved)-catchable-fatal-error-object-of-class-mysqli_result-could-not-be-conver/" rel="nofollow">http://www.phpfreaks.com/forums/php-coding-help/(solved)-catchable-fatal-error-object-of-class-mysqli_result-could-not-be-conver/</a>


I suspect you're trying to concatenate an object-variable. This is not possible with objects of the class "mysqli_result".

Do a var_dump() on each variable you concatenate to make $q and you'll find out which one it is.


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