output the result of a query of a database? [closed]

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Hello i got a problem with my mysql code it gives the following error upon trying to output:

Parse error: syntax error, unexpected $end in     C:\xampplite\htdocs\learncent\acksearch\search.php on line 25 

<?php
$db = new mysqli("localhost","root","","acksocial");

if(mysqli_connect_error())
{
printf("Connection failed:%s \n",mysqli_connect_error());
exit();
}

$name = mysqli_real_escape_string($db, $_POST['search']);
$table = 'acksearch';

if($result = $db->query("SELECT * FROM $table WHERE name = $name", MYSQLI_ASSOC))
{
    while($row = $result->fetch_object())
    {

    // $row is an associative array

    // Do something here

echo "Name: ".$row['name'];

echo " country: ".$row['country'];


}

$db->close();
?>

EDIT: added 1 more } and no error but it wont output the result?

Would be good if anyone could help me.

EDIT AGAIN : Now the code look like this

<?php
$db = new mysqli("localhost","root","","acksocial");

if(mysqli_connect_error())
{
printf("Connection failed:%s \n",mysqli_connect_error());
exit();
}

$name = mysqli_real_escape_string($db, $_POST['search']);
$table = 'acksearch';

if($result = $db->query("SELECT * FROM $table WHERE name = $name", MYSQLI_ASSOC))
{
while($row = $result->fetch_object())
{

// $row is an associative array

// Do something here

echo "Name: ".$row['name'];

echo " country: ".$row['country'];

}

The error its giving its:

Parse error: syntax error, unexpected $end in C:\xampplite\htdocs\searcher.php on line 25

Thanks Fredrik

---

you did not close the while loop:

while($row = $result->fetch_object())
    {

    // $row is an associative array

    // Do something here

echo "Name: ".$row['name'];

echo " country: ".$row['country'];

}///// I added this } here

you need to add } after the last row.

---

This error occurs if somewhere a } is missing.

In your case there should be a second } prior to $db->close();

---

You need to close one more brace "}" before you end "?>"