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

Question

dean.ong.14 -11 Junior Poster in Training

12 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
292 Views
2 Days Discussion Span
Latest Post 12 Years Ago Latest Post by phorce

All 35 Replies

neha_jaltare 0 Newbie Poster

12 Years Ago

write ?> not ?php>

phorce 131 Posting Whiz in Training

12 Years Ago

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!

phorce 131 Posting Whiz in Training

12 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 :)

Reply to this topic

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.

diafol · Answer 1 · 2012-08-31T09:42:37+00:00

diafol

12 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

12 Years Ago

diafol:still the same

dean.ong.14 -11 Junior Poster in Training · Answer 3 · 2012-08-31T11:04:00+00:00

dean.ong.14 -11 Junior Poster in Training

12 Years Ago

neha jaltare:still the same

phorce 131 Posting Whiz in Training Featured Poster · Answer 4 · 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 5 · 2012-08-31T11:09:40+00:00

dean.ong.14 -11 Junior Poster in Training

12 Years Ago

yes it work but the database not showing

diafol · Answer 6 · 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 7 · 2012-08-31T11:11:02+00:00

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

dean.ong.14 -11 Junior Poster in Training · Answer 8 · 2012-08-31T11:11:58+00:00

dean.ong.14 -11 Junior Poster in Training

12 Years Ago

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

Edited 12 Years Ago by dean.ong.14

phorce 131 Posting Whiz in Training Featured Poster · Answer 9 · 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 10 · 2012-08-31T11:19:58+00:00

phorce 131 Posting Whiz in Training

12 Years Ago

Is there a database called "ong_assesment"

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

dean.ong.14 -11 Junior Poster in Training

12 Years Ago

i got no databse selected

dean.ong.14 -11 Junior Poster in Training · Answer 12 · 2012-08-31T11:22:47+00:00

dean.ong.14 -11 Junior Poster in Training

12 Years Ago

yes. that is my database name

phorce 131 Posting Whiz in Training Featured Poster · Answer 13 · 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 14 · 2012-08-31T11:24:41+00:00

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

dean.ong.14 -11 Junior Poster in Training · Answer 15 · 2012-08-31T11:29:07+00:00

dean.ong.14 -11 Junior Poster in Training

12 Years Ago

but i already insert data in

phorce 131 Posting Whiz in Training Featured Poster · Answer 16 · 2012-08-31T11:29:45+00:00

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

dean.ong.14 -11 Junior Poster in Training · Answer 17 · 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 18 · 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 19 · 2012-08-31T11:40:05+00:00

dean.ong.14 -11 Junior Poster in Training

12 Years Ago

ok thank you. i will try it now.

Edited 12 Years Ago by dean.ong.14

phorce 131 Posting Whiz in Training Featured Poster · Answer 20 · 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 21 · 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