CODEWITHSUNDEEP
phpmysql
Chathurika
Chathurika

Reputation: 419

MYSQL Error when making Auto complete text box

I want to make a auto complete text box for select employee name from DB. But it makes query error which is

Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in

Following is my code.

<?php
  include 'func/db_connect.php';
  if(!empty($_POST["keyword"])) {
    $query ="SELECT * FROM employee WHERE name like '" . $_POST["keyword"] . "%' ORDER BY name LIMIT 0,6";

    $result=mysql_fetch_array($query);

    if(!empty($result)) {
        ?>
        <ul id="name-list">
            <?php
            foreach($result as $name) {
                ?>
                <li onClick="selectName('<?php echo $name["name"]; ?>');"><?php echo $name["name"]; ?></li>
            <?php } ?>
        </ul>
<?php } } ?>

What is the wrong with this code, can anyone help me !

Upvotes: 2

Views: 88

Answers (5)

nishant02
nishant02

Reputation: 23

You can try this:

$cn=mysql_connect("localhost","root","");
if(!$cn)
{
    echo "Unable to connect";
    die();
}

    $query = "Select * from table";
    $result= mysql_query($query,$cn);
    $n = mysql_num_rows($result);

    if($n>0)
{
    while($rw=mysql_fetch_array($result))
    {

    $owenername=$rw["owenername"];?>
    <p>Owner Name:<?php echo $owenername;?> </p>
    <?php}
}
else
{
    ?>data not found
    <?PHP 
}
    ?>

Upvotes: 0

JYoThI
JYoThI

Reputation: 12085

you have execute the query first

Deprecated features in PHP 5.5.x

The original MySQL extension is now deprecated, and will generate E_DEPRECATED errors when connecting to a database. Instead, use the MYSQLi or PDO_MySQL extensions.

 <?php
 // include 'func/db_connect.php';


 global $conn;

$servername = "localhost";  //host name

$username = "username"; //username

$password = "password"; //password

$mysql_database = "dbname"; //database name

//mysqli prepared statement 

$conn = mysqli_connect($servername, $username, $password) or die("Connection failed: " . mysqli_connect_error());

 mysqli_select_db($conn,$mysql_database) or die("Opps some thing went wrong");


if(!empty($_POST["keyword"])) {

$name_val = '%'.$_POST["keyword"].'%';

$stmt = $conn->prepare("SELECT * FROM employee WHERE name like ? ORDER BY name LIMIT 0,6");

            $stmt->bind_param('s',$name_val);

            $qry_res=$stmt->execute();


if($row_count>0) {
?>
<ul id="name-list">
    <?php
  while($row = $qry_res->fetch_assoc())
    {
        ?>
        <li onClick="selectName('<?php echo $row["name"]; ?>');"><?php echo $row["name"]; ?></li>
    <?php } ?>
</ul>

     $stmt->close();

Upvotes: 1

kiran mulmi
kiran mulmi

Reputation: 111

Try this

$result=mysql_query($query);
while($data = mysql_fetch_assoc($result))
{
    $row[] = $data;
}

And change !empty($result) to count($row) > 1

Upvotes: 3

Maninderpreet Singh
Maninderpreet Singh

Reputation: 2587

Try with this , you need to use mysql_query() function and you pass string directly to mysql_fetch_array()

$query  = mysql_query("SELECT * FROM employee WHERE name like '" . $_POST["keyword"] . "%' ORDER BY name LIMIT 0,6");
$result = mysql_fetch_array($query);

Note : mysql_* functions deprecated and removed in PHP 7.x. Use MySQLi or PDO_MySQL extension

Upvotes: 2

Karl Wolf
Karl Wolf

Reputation: 198

You need to actually perform the query before being able to fetch the results:

$result = mysql_query("SELECT id, name FROM mytable");
$rows = mysql_fetch_array($result);

Check out the PHP docs for more in-depth examples: http://php.net/manual/de/function.mysql-fetch-array.php

On a sidenote: using mysql_* function has been deprecated for a while, have a look into mysqli!

Upvotes: 2

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