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I got this error while adding this code. Would appreciate some help. It's for a CS jackpot site.

$sitename = "website.com"; // YOUR DOMAIN
$link = mysql_connect("localhost", "db_user", "db_pass"); // MYSQL , LOCALHOOST , USERNAME , PASSWORD 
$db_selected = mysql_select_db('db_name', $link); // MYSQL DATABASE
mysql_query("SET NAMES utf8");

function fetchinfo($rowname,$tablename,$finder,$findervalue) {
    if($finder == "1") $result = mysql_query("SELECT $rowname FROM $tablename");
    else $result = mysql_query("SELECT $rowname FROM $tablename WHERE `$finder`='$findervalue'");
    while($row = mysql_fetch_assoc($query))
        return $row[$rowname];
}
CAMELYON
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    Possible duplicate of [mysql\_fetch\_array() expects parameter 1 to be resource (or mysqli\_result), boolean given](http://stackoverflow.com/questions/2973202/mysql-fetch-array-expects-parameter-1-to-be-resource-or-mysqli-result-boole) – Mr. Engineer Feb 03 '16 at 07:49
  • shouldn't $query be $result? – Reinder Wit Feb 03 '16 at 07:53
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    `while($row = mysql_fetch_assoc($result))`, not `while($row = mysql_fetch_assoc($query))` – n-dru Feb 03 '16 at 07:53
  • still getting the error, now with "boolean give" – CAMELYON Feb 03 '16 at 08:06
  • Which suggests that the query failed. Echo out the query that it is trying to execute and hopefully we can see the error in it. – Kickstart Feb 03 '16 at 11:12

2 Answers2

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Some tips:

  • Use mysqli better than mysql

  • Split the vars in the query, like "SELECT ".$rowname." FROM ".$tablename;

Louis B
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  • I have no idea how to translate this into mysqli: `function fetchinfo($rowname,$tablename,$finder,$findervalue) { if($finder == "1") $result = mysql_query("SELECT $rowname FROM $tablename"); else $result = mysql_query("SELECT $rowname FROM $tablename WHERE `$finder`='$findervalue'"); while($row = mysql_fetch_assoc($query)) return $row[$rowname];` – CAMELYON Feb 03 '16 at 10:52
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Hope this help...

<?php
//Open a new connection to the MySQL server
$mysqli = new mysqli('host','username','password','database_name');

if ($mysqli->connect_error) {
    die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
}

//MySqli Select Query
$result = $mysqli->query("SELECT id, product_name FROM products");

while($row = $results->fetch_assoc()) {
    echo $row["id"].' - '.$row["product_name"].'<br>';
}  

$results->free();
$mysqli->close();
Louis B
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