PHP: Warning: mysql_fetch_array() expects parameter 1 to be resource

Question

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];
}

Possible duplicate of mysql_fetch_array() expects parameter 1 to be resource (or mysqli_result), boolean given — Mr. Engineer
– Mr. Engineer, Commented Feb 3, 2016 at 7:49
while($row = mysql_fetch_assoc($result)), not while($row = mysql_fetch_assoc($query)) — n-dru
– n-dru, Commented Feb 3, 2016 at 7:53
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
– Kickstart, Commented Feb 3, 2016 at 11:12

Louis B · Accepted Answer · 2016-02-03 07:55:47Z

1

Some tips:

Use mysqli better than mysql
Split the vars in the query, like "SELECT ".$rowname." FROM ".$tablename;

answered Feb 3, 2016 at 7:55

Louis B

3061 gold badge6 silver badges20 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

CAMELYON Over a year ago

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];

Louis B · Accepted Answer · 2016-02-03 11:38:20Z

0

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();

answered Feb 3, 2016 at 11:38

Louis B

3061 gold badge6 silver badges20 bronze badges

Collectives™ on Stack Overflow

PHP: Warning: mysql_fetch_array() expects parameter 1 to be resource

2 Answers 2

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related