PLEASE HELP!! I AM GOING MAD!! Warning: mysql_fetch_array(): supplied argument is not

[gpittingale]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/sites/miraclehosting.co.uk/public_html/scdc.php on line 169
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Can anyone help here this is doin ma ed in!! gggrrrhhhhh
i get the above result when Load the script into the browser.
here is the form:

<form action="scdc.php" method="post" name="chooser">
  <label>Select Parish </label>     
  <select name="chooser" id="chooser">
    <option value="1">Abington Pigotts</option>
    <option value="2">Arrington</option>

    <option value="3">Barbraham</option>
    <option value="4">Balsam</option>
    <option value="5">Bar Hill</option>
    <option value="6">Barrington</option>
    <option value="7">Barton</option>
    <option value="8">Bassingbourn-Cum-Kneesworth</option>
</select>
  <input type="submit" name="select" value="select"> 
</form>

although there is 97 options i just havent given all of them for obvious reasons.
Here is the php:

<?php

// get variable after selecting something from the dropdown with name 'chooser'
$select = $_POST['select'];

// if something has been chosen
if (!empty($select)) {

// get the chosen value
$chooser = $_POST['chooser'];

// select the type from the database
// database connection details (change to whatever you need)
$HOST = 'localhost';
$DATABASE = '******';
$USER = '********';
$PASSWORD = '********';

// connect to database
if(!$conn=mysql_connect('localhost','web244-scdc','hartley07')) {
echo("<li>Can't connect to $HOST as $USER");
echo("<li>mysql Error: ".mysql_error());
die;
}

// select database
if (!mysql_select_db($DATABASE,$conn)) {
echo("<li>We were unable to select database $DATABASE");
die;
}

// if everything successful create query
// this selects all rows where the type is the one you chose in the dropdown
// * means that it will select all columns, ie name and type as i said above
$sql_query = "SELECT * FROM base_data WHERE type='$chooser'";

// get the data from the database
$result = mysql_query($sql_query,$conn);

// output data
while ($details = mysql_fetch_array($result, MYSQL_ASSOC)) {  
// print out the name
echo('Parish: '.$details['ParishName'].' - ');
// print out the type
echo('Owner Occupier Household: '.$details['OwnerOccupierHousehold'].'<br>');
}

// close mysql connection
mysql_close($conn);

}

?>

The line highlighted in red is causing the problems. please help?

[jrdark13]

try your query as:

$sql_query = "SELECT * FROM base_data WHERE type='".$chooser."'";

notice $chooser has been properly removed from the leteral sentence.

[ShawnCplus]

Your query might be failing. Use

$result = mysql_query($sql_query,$conn) or die('Invalid query: ' . mysql_error());

[gpittingale]

ah sorry just tried any other ideas?

[gpittingale]

Your query might be failing. Use

$result = mysql_query($sql_query,$conn) or die('Invalid query: ' . mysql_error());

OK this correct but.........Now iget this error Invalid query: Unknown column 'type' in 'where clause'

[jrdark13]

what are the columns called in the database?

That error just told you it couldn't find the asked column? and what is you query right now? (SELECT....)

[gpittingale]

But i would like to bring to view a full row thatt i plan to display in a table? do you think i have used the right script or have a messedup?

[ShawnCplus]

Just look at the error

Invalid query: Unknown column 'type' in 'where clause'

There is no type column to perform the where clause against.
Show us the layout of the table( what are the columns in base_data?)

[gpittingale]

ok the table was too big to copy and paste so i decided to export it to word and make it downloadble hope this is enough for you
download here thanks again guys

[ShawnCplus]

All you had to post was the structure, not the dump of the database. I would explain to you how much is wrong with the design of the table but instead I'll give you the quick fix.

<form action="scdc.php" method="post" name="chooser">
  <label>Select Parish </label>
  <select name="chooser" id="chooser">
    <option value="Abington Pigotts">Abington Pigotts</option>
    <option value="Arrington">Arrington</option>
    <option value="Barbraham">Barbraham</option>
    <option value="Balsam">Balsam</option>
    <option value="Bar Hill">Bar Hill</option>
    <option value="Barrington">Barrington</option>
    <option value="Barton">Barton</option>
    <option value="Bassingbourn-Cum-Kneesworth">Bassingbourn-Cum-Kneesworth</option>
  </select>
  <input type="submit" name="select" value="select">
</form>

$sql_query = "SELECT * FROM base_data WHERE ParishName LIKE '$chooser'";

[gpittingale]

You my freind are an absolute GENIUS!! thanks for the help can i add you as a contact incase i ever get stuck? Also i saw what i did wrong in the script i get it now also what was wrong with the table breifly