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please someone help me . im in urgent situation. i cant solved this.


this is my result form

<?php
session_start();
?>



<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">












Laporan Bayaran









Nama Koperasi :



--Nama Koperasi--
<?php
//First cari fail untuk connection ke database
include_once ("./includes/dbconnection.php");

$sql = "SELECT * FROM coop01";
$result = mysqli_query($conn, $sql);

while ($coop = mysqli_fetch_array($result)) {
$c_id = $coop['coopId'];
$c_name = $coop['coopName'];

echo "
".$c_name."

";
}

?>





Jenis Carian :



--Sila Pilih--
Resit
Baucer
Invois Jualan
Invois Belian






Tarikh:
































/* Style the search field */
form.example input[type=text] {
padding: 8px;
font-size: 15px;
border: 1px solid grey;
float: left;
width: 80%;
background: #f1f1f1;
}

/* Style the submit button */
form.example button {
float: left;
width: 20%;
padding: 10px;
background: #2196F3;
color: white;
font-size: 17px;
border: 1px solid grey;
border-left: none; /* Prevent double borders */
cursor: pointer;
}

form.example button:hover {
background: #0b7dda;
}

/* Clear floats */
form.example::after {
content: "";
clear: both;
display: table;
}

What I have tried:

this i my outcome after submit button search





<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">





Posted
Updated 4-May-18 6:34am
v2

1 solution

This is a common error when a mysqli command does not succeed. Please see mysqli boolean given - Google Search[^], or read the documentation for the failing command.
 
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