help me pls Parse error: syntax error, unexpected $end in F:\xampp\htdocs\

Question

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

<html>
<head>
</head>
<body>

<table border='1'>
<tr>
<th>Agent_ID</th>
<th>Address</th>
<th>Bedrooms</th>
<th>Price</th>
<?php
$connect=mysql_connect("localhost","root","");

mysql_select_db("ong_assesment',$connect);
$query = mysql_query (SELECT Agent_ID,Address/Suburb,Bedrooms/Bathrooms,Price FROM suburbs);

{
echo '<tr>';
echo '<td>'. $row[Agent_ID].'</td>';
echo '<td>'. $row[Address].'</td>';
echo '<td>'. $row[Bedrooms].'</td>';
echo '<td>'. $row[Price].'</td>';

echo '</table>'
mysql_close();
}
?php>
</body>
</html>

php

4 Contributors
35 Replies
322 Views
2 Days Discussion Span
Latest Post 14 Years Ago Latest Post by phorce

Dani AI

Generated 9 Months Ago

Thread summary and practical fixes (concise)

This thread shows three common, related problems: a parse error (unexpected $end), a failing database selection / empty result, and "undefined constant" notices when printing row values. fixed most of the syntax mistakes and correctly pointed out the SELECT vs affected-rows issue. The items below are concentrated, actionable steps and a safer example using modern APIs.

Quick debugging checklist

"unexpected $end" usually means an unclosed string, missing semicolon, or an unmatched brace/parenthesis. Run a quick syntax check from the command line:
php -l path/to/file.php
Turn on full errors while developing:
ini_set('display_errors',1); error_reporting(E_ALL);
Use <?php to open PHP blocks (avoid short tags unless explicitly enabled). Check every if, while, { and } for balance.
"No database selected" means select_db failed (wrong name or connection). Confirm the DB name and connection. For SELECT queries, use mysql_num_rows (or the equivalent) to detect empty result sets, not mysql_affected_rows.
Column names with special characters (like Address/Suburb) are fragile — rename them or quote with backticks, or use aliases in the SQL.

Safer, modern approach (use PDO or mysqli, not deprecated mysql_*):

<?php
$pdo = new PDO('mysql:host=127.0.0.1;dbname=ong_assessment;charset=utf8mb4','root','',[
  PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
  PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC,
]);

$stmt = $pdo->query('SELECT agent_id, address, bedrooms, price FROM suburbs');
$rows = $stmt->fetchAll();

if (empty($rows)) {
  echo '<p>No entries in the database.</p>';
} else {
  echo '<table border="1"><tr><th>Agent</th><th>Address</th><th>Beds</th><th>Price</th></tr>';
  foreach ($rows as $r) {
    echo '<tr><td>'.htmlspecialchars($r['agent_id']).'</td><td>'.htmlspecialchars($r['address']).'</td><td>'.htmlspecialchars($r['bedrooms']).'</td><td>'.htmlspecialchars($r['price']).'</td></tr>';
  }
  echo '</table>';
}

Linking PHP and HTML / deployment notes
A PHP page is linked the same way as HTML, but it must be served by the web server. With XAMPP put files under F:\xampp\htdocs\ (or an htdocs subfolder) and visit http://localhost/YourFolder/Farcourts.php. Opening the file with a file:// URL will not execute PHP. Also, migrate away from mysql_* (removed in PHP 7+) and always escape output (htmlspecialchars) and use prepared statements when there is user input.

All 35 Replies

phorce 131 Posting Whiz in Training

14 Years Ago

Please post your script, how it looks now.. Removing any passwords etc..

phorce 131 Posting Whiz in Training

14 Years Ago

Look at the syntax, if you have to explain what the script does, then it's pretty self-explanatory.. I cannot do your work for you, that wouldn't be fair and I'm sure you actually want to learn something ;) I'll help you a bit though:

Explain what PHP is
Explain the code you've written (So it retrives data from a table inside a database)
Explain each line of code:

$connect=mysql_connect("localhost","root","");
$db_selected = mysql_select_db("ong_assessment", $connect);

This code establishes a connection with the database on the server. The mysql_connect accepts three parameters. The mysql_select_db attempts to select the database that data will be retrived from. It accepts two parameters, one is the name of the database and the other is the connection variable.

...

Something like this :)

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Member #120589 · Answer 1 · 2012-08-31T09:42:37+00:00

Member #120589

14 Years Ago

change to

echo '</table>';

dean.ong.14 -11 Junior Poster in Training · Answer 2 · 2012-08-31T10:23:32+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

diafol:still the same

neha_jaltare 0 Newbie Poster · Answer 3 · 2012-08-31T10:44:25+00:00

neha_jaltare 0 Newbie Poster

14 Years Ago

write ?> not ?php>

dean.ong.14 -11 Junior Poster in Training · Answer 4 · 2012-08-31T11:04:00+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

neha jaltare:still the same

phorce 131 Posting Whiz in Training Featured Poster · Answer 5 · 2012-08-31T11:04:39+00:00

Try this, bit different to your script.. Have a look at yours, you have many errors:

<html>
<head>
</head>
<body>

<table border='1'>
<tr>
<th>Agent_ID</th>
<th>Address</th>
<th>Bedrooms</th>
<th>Price</th>
<?php
$connect=mysql_connect("localhost","root","");

mysql_select_db("ong_assesment",$connect);
$query = "SELECT Agent_ID,Address/Suburb,Bedrooms/Bathrooms,Price FROM suburbs";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result))
{
    echo '<tr>';
    echo '<td>'. $row[Agent_ID].'</td>';
    echo '<td>'. $row[Address].'</td>';
    echo '<td>'. $row[Bedrooms].'</td>';
    echo '<td>'. $row[Price].'</td>';
}
    echo '</table>';

?>
</body>
</html>

e.g. in yours you have:

mysql_select_db("ong_assesment',$connect);

should be:

mysql_select_db("ong_assesment",$connect);

Hope this helps

dean.ong.14 -11 Junior Poster in Training · Answer 6 · 2012-08-31T11:09:40+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

yes it work but the database not showing

Member #120589 · Answer 7 · 2012-08-31T11:10:13+00:00

WHat editor / IDE are you using? Some editors can spot these errors for you, highlighting the problem lines.

phorce 131 Posting Whiz in Training Featured Poster · Answer 8 · 2012-08-31T11:11:02+00:00

phorce 131 Posting Whiz in Training

14 Years Ago

Is there an error showing? Is there anything in the table?

dean.ong.14 -11 Junior Poster in Training · Answer 9 · 2012-08-31T11:11:58+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

it came out table but no data.
message:No database selected

Edited 14 Years Ago by dean.ong.14

phorce 131 Posting Whiz in Training Featured Poster · Answer 10 · 2012-08-31T11:16:39+00:00

I don't think you're understanding me.. I asked is there an error message showing, NOT what text editor you're using... Ok, try this:

<html>
<head>
</head>
<body>

<table border='1'>
<tr>
<th>Agent_ID</th>
<th>Address</th>
<th>Bedrooms</th>
<th>Price</th>
<?php
$connect=mysql_connect("localhost","root","");

mysql_select_db("ong_assesment",$connect);
$query = "SELECT Agent_ID,Address/Suburb,Bedrooms/Bathrooms,Price FROM suburbs";
$result = mysql_query($query) or die(mysql_erro());

if(!mysql_affected_rows() >= 1)
{
    echo 'There are no entries inside the database';
}
while($row = mysql_fetch_array($result))
{
    echo '<tr>';
    echo '<td>'. $row[Agent_ID].'</td>';
    echo '<td>'. $row[Address].'</td>';
    echo '<td>'. $row[Bedrooms].'</td>';
    echo '<td>'. $row[Price].'</td>';
}
    echo '</table>';

?>
</body>
</html>

Do you get the output: There are no entries inside the database? OR do you get something else?

phorce 131 Posting Whiz in Training Featured Poster · Answer 11 · 2012-08-31T11:19:58+00:00

phorce 131 Posting Whiz in Training

14 Years Ago

Is there a database called "ong_assesment"

dean.ong.14 -11 Junior Poster in Training · Answer 12 · 2012-08-31T11:21:19+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

i got no databse selected

dean.ong.14 -11 Junior Poster in Training · Answer 13 · 2012-08-31T11:22:47+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

yes. that is my database name

phorce 131 Posting Whiz in Training Featured Poster · Answer 14 · 2012-08-31T11:24:35+00:00

Ok, try this:

where you have:

mysql_select_db("ong_assesment",$connect);

change it to:

$db_selected = mysql_select_db("ong_assesment", $connect);

Making sure, your database name is called "ong_assesment".

dean.ong.14 -11 Junior Poster in Training · Answer 15 · 2012-08-31T11:24:41+00:00

Parse error: syntax error, unexpected T_VARIABLE in F:\xampp\htdocs\movies4.php on line 14

phorce 131 Posting Whiz in Training Featured Poster · Answer 16 · 2012-08-31T11:26:40+00:00

Now you have to insert data, for the data to be shown :) Do you use mysqladmin? If so, you can just throw some example data in just to test!

dean.ong.14 -11 Junior Poster in Training · Answer 17 · 2012-08-31T11:29:07+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

but i already insert data in

dean.ong.14 -11 Junior Poster in Training · Answer 18 · 2012-08-31T11:31:58+00:00

<head>
</head>
<body>
<table border='1'>
<tr>
<th>Agent_ID</th>
<th>Address</th>
<th>Bedrooms</th>
<th>Price</th>
<?php
$connect=mysql_connect("localhost","root","");
$db_selected = mysql_select_db("ong_assessment", $connect)
$query = "SELECT Agent_ID,Address,Bedrooms,Price FROM suburbs";
$result = mysql_query($query) or die(mysql_error());
if(!mysql_affected_rows() >= 1)
{
echo 'There are no entries inside the database';
}
while($row = mysql_fetch_array($result))
{
 echo 'There are no entries inside the database';
}
while($row = mysql_fetch_array($result))
{
echo '<tr>';
echo '<td>'. $row[Agent_ID].'</td>';
echo '<td>'. $row[Address].'</td>';
echo '<td>'. $row[Bedrooms].'</td>';
echo '<td>'. $row[Price].'</td>';
}
echo '</table>';
?>
</body>
</html>

phorce 131 Posting Whiz in Training Featured Poster · Answer 19 · 2012-08-31T11:35:19+00:00

LOOK at the syntax..

<head>
</head>
<body>
<table border='1'>
<tr>
<th>Agent_ID</th>
<th>Address</th>
<th>Bedrooms</th>
<th>Price</th>
<?php
$connect=mysql_connect("localhost","root","");
$db_selected = mysql_select_db("ong_assessment", $connect);
$query = "SELECT Agent_ID,Address,Bedrooms,Price FROM suburbs";
$result = mysql_query($query) or die(mysql_error());
if(!mysql_affected_rows() >= 1)
{
echo 'There are no entries inside the database';
}

while($row = mysql_fetch_array($result))
{
echo '<tr>';
echo '<td>'. $row[Agent_ID].'</td>';
echo '<td>'. $row[Address].'</td>';
echo '<td>'. $row[Bedrooms].'</td>';
echo '<td>'. $row[Price].'</td>';
}
echo '</table>';
?>
</body>
</html>

Now, if the output is: 'There are no entries inside the database' then there are no records in the table 'suburbs'. There are NO errors now, don't create errors for yourself.

dean.ong.14 -11 Junior Poster in Training · Answer 20 · 2012-08-31T11:40:05+00:00

dean.ong.14 -11 Junior Poster in Training

14 Years Ago

ok thank you. i will try it now.

Edited 14 Years Ago by dean.ong.14

phorce 131 Posting Whiz in Training Featured Poster · Answer 21 · 2012-08-31T11:41:34+00:00

Yes, your script is wrong. I posted an altered version the script that fixed the errors you had.

dean.ong.14 -11 Junior Poster in Training · Answer 22 · 2012-08-31T11:42:47+00:00

Notice: Use of undefined constant Agent_ID - assumed 'Agent_ID' in F:\xampp\htdocs\movies4.php on line 23

Notice: Use of undefined constant Address - assumed 'Address' in F:\xampp\htdocs\movies4.php on line 24