Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given

Question

WTF? - this is becoming a nightmare, just seem to get error after error.. Im willing to pay someone for their time just to help me achieve what seems to be the unachievable, please.. alls I want is a dropdown menu that selects data from a table according to its category and displays it.. HELP! :(

<form action="portfolio.php" method="post">
 <select onload="displayProject(this.value);" onchange="displayProject(this.value);">
  <option value='none'>All</option>
  <option value='1'>Fencing</option>
  <option value='2'>Driveway</option>
 </select>
</form>

<?php 

$db = new mysqli('localhost', 'wlarter_user', 'pw', 'wlarter_portfolio');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$option= $_POST['option'];

$queries = "SELECT * FROM image"; 
if ($option != 'none'){
 $queries = "SELECT * FROM image where category=".$option; 
}

$result=@mysqli_query($db,"$queries");

while($row = mysqli_fetch_array($result))
{
?>

<div class="box-portfolio"> <?php echo $row['Img']; ?> </div>


<?php
}                      
mysqli_close($db);
?>

STOP writing big letters for HTML. It is html 3.0 standard I think. We are not using HTML 3.0 anymore! — Xatenev, Apr 14 '14 at 10:26
Im new to php, this is some code I have managed to scrape together, forgive me for any mistakes.. — user3505895, Apr 14 '14 at 10:27
$queries=$query; $result=@mysqli_query($db,"$query"); This actually won't work. $query is NOT Defined — Xatenev, Apr 14 '14 at 10:28

score 0 · Answer 1 · answered Apr 14 '14 at 10:29

0

can you try like this.......

$result=@mysqli_query($db,"$queries");

answered Apr 14 '14 at 10:29

V-T

756
2
9
22

score 0 · Accepted Answer · answered Apr 14 '14 at 10:30

0

Simply turn it around, is probably what you want:

$queries=$query;

into

$query = $queries;

FULL code:

<FORM action="portfolio.php" method="post">
    <SELECT onload="displayProject(this.value);" onchange="displayProject(this.value);">
        <OPTION VALUE='none'>All</OPTION>
        <OPTION VALUE='1'>Fencing</OPTION>
        <OPTION VALUE='2'>Driveway</OPTION>
    </SELECT>
</FORM>

<?php

$db = new mysqli('localhost', 'wlarter_user', 'pw', 'wlarter_portfolio');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$option= $_POST['option'];

$queries = "SELECT * FROM image";
if ($option != 'none'){
    $queries = "SELECT * FROM image where category=".$option;
}

$query=$queries;
$result=mysqli_query($db,"$query");

while($row = mysqli_fetch_array($result))
{
    ?>

    <div class="box-portfolio"> <?php echo $row['Img']; ?> </div>


<?php
}
mysqli_close($db);
?>

answered Apr 14 '14 at 10:30

Xatenev

6,383
3
18
42

Remove the @ before the mysqli_query so you see errors. – Xatenev Apr 14 '14 at 10:31
var_dump($option) please!. – Xatenev Apr 14 '14 at 10:33
Not sure what that is? is there anyway we could talk via skype? Im willing to pay for your services – user3505895 Apr 14 '14 at 10:34
My name is Xatenev in Skype. – Xatenev Apr 14 '14 at 10:35
You just should've checked the error, mate, like i answered. :3 – Maximilian Riegler Apr 14 '14 at 13:03

score 0 · Answer 3 · answered Apr 14 '14 at 10:37

0

replace this line

$result=@mysqli_query($db,"$queries");

with this:

$result=@mysqli_query($db,$queries);

it may help..

answered Apr 14 '14 at 10:37

Moax6629

378
4
14

score 0 · Answer 4 · answered Apr 14 '14 at 11:49

Because the result of a mysql query can be a resource or a boolean value (if it failed), you should check the result of your query like this after executing it:

if ($result !== false) {
    while($row = mysqli_fetch_array($result))
    {
        ?>
            <div class="box-portfolio"> <?php echo $row['Img']; ?> </div>
        <?php
    }
}
else {
    // do your error message stuff here
    echo mysqli_error($db);
}

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given

4 Answers4