[SOLVED] php code for querying mysql database?

[intenziti]

Hi there,

 

I came upon this thread while looking for code for a project I am completing for university. I have made a form with a drop down menu so that the user can query the database about which course is available in each month. However I keep getting this error message

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /Applications/xampp/xamppfiles/htdocs/paul/month_drop_down.php on line 41

 

The php code I actually got from this page: http://www.webmasterworld.com/forum88/4831.htm I have tried to adapt it to suit my purposes and it seems that all my database information is correct but for the likes of me I cannot work out why the php code is not processing the form data from the drop down menu.

 

Would anyone here please be able to have a quick look at my code and see if there is anything that may be stopping it from functioning?

 

There are two fields in my table which are called 'month' and 'course', the table itself is called months.

 

Any help would be greatly appreciated!

 

Form code:

 

<form name="month_search" method="post" action="month_drop_down.php"><table width="600" align="center" border="0" cellspacing="1" cellpadding="2">

<tr><td width=300><select name="month" id="selCourse" style="width: 200px;"><option value="Select month to attend..." selected>Select month to attend...</option><option value="jan">January</option><option value="feb">February</option><option value="mar">March</option><option value="apr">April</option><option value="may">May</option><option value="jun">June</option><option value="jul">July</option><option value="aug">August</option><option value="sep">September</option><option value="oct">October</option><option value="nov">November</option><option value="dec">December</option></select></td><td><input type="submit" name="search" value="Search"/></td>

</tr>

 

</table></form>

 

 

PHP code:

 

<?php

 

// get variable after selecting something from the dropdown with name 'month'

$select = $_POST['search'];

 

// if something has been chosen

if (!empty($select)) {

 

// get the chosen value

$month = $_POST['month'];

 

// select the type from the database

// database connection details (change to whatever you need)

$HOST = 'localhost';

$DATABASE = 'sworphe_paul';

$USER = 'sworphe_paul';

$PASSWORD = 'sworphe_paul';

 

// connect to database

if(!$conn=mysql_connect('localhost','sworphe_paul' ,'sworphe_paul')) {

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

$sql_query = "SELECT * FROM months WHERE type='$month'";

 

// get the data from the database

$result = mysql_query($sql_query,$conn);

 

// output data

while (($deails = mysql_fetch_assoc($result))) {

// print out the name

echo('Courses available in '.$details['month'].' - ');

// print out the type

echo('Type: '.$details['course'].'<br>');

}

 

// close mysql connection

mysql_close($conn);

 

}

?>

 

 

Thank you so much in advance!  ???

[xyn]

could yo show me line 41?

[intenziti]

Sure line 41 sits just under // output data as follows:

 

while (($deails = mysql_fetch_assoc($result))) {

 

[xyn]

try.

while($details = mysql_fetch_array($result)) {

[intenziti]

Unfortunately it gives the exact same error

[xyn]

what is the error line... like

 

PARSE ERROR or something

[Fjerpje]

Hi there,

 

It means you didnt correctly close a string.

Look again at your code and look if all "" and ; are correctly placed

[intenziti]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /Applications/xampp/xamppfiles/htdocs/paul/month_drop_down.php on line 41

[Fjerpje]

hi there,

 

And that means that your SQL syntax isnt correct post it here

and i and other might help you

[intenziti]

SQL syntax? Sorry I'm new to this so where exactly would I find that? Is it something I would define myself?

[xyn]

the bit should look like ...

$result = mysql_query("SELECT * FROM

[Fjerpje]

Or just the line 41

[xyn]

ok try replace

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

 

with this

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

[intenziti]

No luck unfortunately. It keeps coming up with the exact same error

[xyn]

what error?

[intenziti]

 

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /Applications/xampp/xamppfiles/htdocs/paul/month_drop_down.php on line 41

[xyn]

ok thats this bit

// 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 
$sql_query = "SELECT * FROM months WHERE type='$month'";

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

// output data 
while (($deails = mysql_fetch_assoc($result))) { 

 

I would suggest to uyse the below then tell me what the error says:

// 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 
$result = mysql_query("SELECT * FROM months WHERE type='$month'") or die("My Error Line: ".mysql_error());

// output data 
while (($deails = mysql_fetch_assoc($result))) { 

[intenziti]

OK now getting the following error message:

 

My Error Line: Unknown column 'type' in 'where clause'

 

[xyn]

means type doesnt exist

[intenziti]

so 'type' is referring to the table data?

[xyn]

type does not exist in the months table

[intenziti]

Great! Thank you so much for your help!