[SOLVED] Drop Down

peranha · January 17, 2007

I am trying to get a site up and running. I have 4 tables. A camera table, a facility table, a misc table,and a table that has foreign keys to each of these 3 tables. I have a dropdown list and need to input the id from each of the tables into the corresponding columns of the last table. I can put the page online if needed. Like I said the drop downs work, but the insert doesnt insert the id of what the user selects. If anyone can help that would be great. I can give as much info as needed, and even export the sql statements if needed, and the PHP script I have so far.
Thanks in advance for the help.

linuxdream · January 17, 2007

If you need specific answers you really need to post at least the main portions of your code. Or else how can we help?

peranha · January 17, 2007

Here is the code.

<HTML>
<HEAD><TITLE>Test Page</TITLE>
</HEAD>
<BODY>

<form action="dropdownupload.php" method="post">
<input type="submit" name="submit">

<?PHP
// set server access variables
$host = "localhost";
$user = "user";
$pass = "password";
$db = "equipment";

// open connection
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");

// select database
mysql_select_db($db) or die ("Unable to select database!");

//this code is bringing in the values for the dropdown.
$query="SELECT * FROM facilities";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo "<select name=\"facilities\" value=''>Facilities Name</option>";
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo "<option value=\"$nt[facilityid]\">$nt[facilityname]</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box

//this code is bringing in the values for the dropdown.
$query="SELECT * FROM cameras";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo "<select name=\"cameras\" value=''>Cameras model</option>";
// printing the list box select command

while($nt1=mysql_fetch_array($result)){//Array or records stored in $nt1
echo "<option value=$nt1[cameraid]>$nt1[model]</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box

//this code is bringing in the values for the dropdown.
$query="SELECT * FROM misc";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo "<select name=\"misc\" value=''>misc building</option>";
// printing the list box select command

while($nt2=mysql_fetch_array($result)){//Array or records stored in $nt2
echo "<option value=$nt2[miscid]>$nt2[building]</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box

// create query
$query = "INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ('$nt', '$nt1', '$nt2')";

// execute query
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());

// close connection
mysql_close($connection);
?>
</form>

</BODY>
</HTML>

linuxdream · January 17, 2007

Your Insert statement is incorrect. The statement should look something like:

$query = "INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ("$_POST['nt']", "$_POST['$nt1']", "$_POST['$nt2']")";

PHP cannot process form data unless it was submitted from a previous request since all its processing is server side. Also, You should think about separating the form and processing pages or at least throw a conditional in there or else you will be making an empty INSERT request every time the page loads when you only want that happening when someone submits a page.

B

peranha · January 17, 2007

I tried this, but had no luck working. I added a $ in the first NT field, and no luck, so I tried it without the $ signs in it at all, and no luck. I will try to split the pages up when I get this working on one page though. I never thought that it would make an empty insert request. When I try to load it, it just comes up with a blank white page, no errors. If I comment out the $query line, it loads the page fine. Here are the last lines of the code. The rest stayed the same.

echo "</select>";// Closing of list box

// create query
$query = "INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ("$_POST['$nt']", "$_POST['$nt1']", "$_POST['$nt2']")";

// execute query
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());

// close connection
mysql_close($connection);

Thanks again.

linuxdream · January 17, 2007

Sorry I forgot to remove the "$" from within the $_POST[] variable. Remove those $'s and it should be good. So:

$query = 'INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ("$_POST["nt"]", "$_POST["nt1"]", "$_POST["nt2"]")';

If it still doesn't work, then check you error log file to see the exact error. If you are *nix: tail /var/log/messages should do it.

Sorry, just woke up, still not thinking clearly yet!!! Notice I changed the " and ' in the statement as well.

B

peranha · January 17, 2007

Did this, and still no luck. This is the error I get when loading the web page.

Error in query: INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ("$_POST["nt"]", "$_POST["nt1"]", "$_POST["nt2"]"). You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'nt"]", "$_POST["nt1"]", "$_POST["nt2"]")' at line 1

linuxdream · January 17, 2007

I must be really off today, this should be a very easy thing. Try this:

$query = "INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ('{$_POST['nt']}', '{$_POST['nt1']}', '{$_POST['nt2']}')";

I usually filter my POST and GET variables thus changing the name so I don't have to deal with so many "s and 's..I've forgotten how to deal with them. The curly braces tell PHP that the enclosed data is a variable and it should replace everything with the proper values before it gets to MySQL.

Give it a shot, if not, then I must be completely off the deep end today. If MySQL still complains, then ditch the {} and the 's inside the $_POST variable. Technically you are supposed to have them (or "s) but it will work without them.

peranha · January 17, 2007

I tried this, at no luck. Here is the error I am getting if it helps any. This is all new to me, but it looks as though it is not getting the ID for some reason.

Error in query: INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ('', '', ''). Out of range value adjusted for column 'facilityid' at row 1

Jessica · January 17, 2007

You could also try:
$query = "INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ('".$_POST['nt']."', '".$_POST['nt1']."', '".$_POST['nt2']."')";

run print_r($_POST) at the top to make sure the post vars are being set as you think they should be.

peranha · January 17, 2007

This is all new to me, I just started with PHP, MYSQL about 2 weeks ago, and am still getting the hang of it. When I put in the print_r($_POST) at the beginning of the PHP code, I just get a blank white screen. Is there any thing else I need to put in front of it or behind.

Jessica · January 17, 2007

You need to end it with the same syntax you end any statement with, a ;

peranha · January 17, 2007

OK, I did the print, and here is what I get when I select different dropdown items in the web page. It changes to correspond with the item selected. Array ( [submit] => Submit Query [facilities] => 8 [cameras] => 4 [misc] => 9 ) This is with the following.

// create query
//$query = "INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ("$_POST['$nt']", "$_POST['$nt1']", "$_POST['$nt2']")";

// execute query
//$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());

Thanks for the help and the patients.

Jessica · January 17, 2007

so you can't use $_POST['nt'] because it doesn't exist. You need to check your variable names and the form.

peranha · January 18, 2007

Changed the code to look like this, and tried all that I can think of at no luck of it working.

<?PHP
// set server access variables
$host = "localhost";
$user = "user";
$pass = "passwork";
$db = "equipment";

// open connection
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");

// select database
mysql_select_db($db) or die ("Unable to select database!");

//this code is bringing in the values for the dropdown.
$query="SELECT * FROM facilities";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);
echo "<select name=\"facilities\" value=''>Facilities Name</option>";
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo "<option value=\"$nt[facilityid]\">$nt[facilityname]</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box

//this code is bringing in the values for the dropdown.
$query1="SELECT * FROM cameras";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result1 = mysql_query ($query1);
echo "<select name=\"cameras\" value=''>Cameras model</option>";
// printing the list box select command

while($nt1=mysql_fetch_array($result1)){//Array or records stored in $nt1
echo "<option value=$nt1[cameraid]>$nt1[model]</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box

//this code is bringing in the values for the dropdown.
$query2="SELECT * FROM misc";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result2 = mysql_query ($query2);
echo "<select name=\"misc\" value=''>misc building</option>";
// printing the list box select command

while($nt2=mysql_fetch_array($result2)){//Array or records stored in $nt2
echo "<option value=$nt2[miscid]>$nt2[building]</option>";
/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box

// create query
//$query = "INSERT INTO facility_cameras (facilityid, cameraid, miscid) VALUES ('".$_POST['nt']."', '".$_POST['nt1']."', '".$_POST['nt2']."')";

// execute query
//$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());

$query4 = print_r($_POST);

// close connection
mysql_close($connection);
?>

Every time I uncomment the $query Insert line, the page doesnt load, and this is what shows up in my Apache error log. Undefined index: nt, nt1, nt2.

peranha · January 20, 2007

I am trying to figure this out, but cannot. If you could give an example, that would be great. Thanks for the help so far.

Jessica · January 20, 2007

the variables in POST relate to the names of inputs, selects, and textareas. None of your form elements have those names.

peranha · January 20, 2007

If I cant use the POST statement in the insert statement, why does it work in the Print_r statement. ???

Jessica · January 20, 2007

$_POST is an array. The KEYs in it are the NAMES of your INPUTS. None of your inputs have name="n1" so you can't see $_POST['n1'] because it doesn't EXIST.

When you did print_r($_POST) it showed you the KEYS and VALUES. The KEYS are the names of your inputs.
I don't know how to make it clearer.

<input type="text" name="foo" value="bar" />
Will result in $_POST['foo'] being = to "bar";
You can't access a variable that doesn't exist.

Jessica · January 20, 2007

[quote author=peranha link=topic=122745.msg507465#msg507465 date=1169083955]
echo "<select name=\"facilities\" value=''>Facilities Name</option>";
echo "<select name=\"cameras\" value=''>Cameras model</option>";
echo "<select name=\"misc\" value=''>misc building</option>";
[/quote]

1 .The name of a select is what's inside of name="".
2. Select doesn't have a value="". The value goes on the <option>s only.
3. You should use single quotes on strings which do not have variables. These lines and many of your others should be changed:
[code]echo '<select name="facilities">Facilities Name</option>';[/code]

When you do print_r($_POST) you should see keys of facilities, cameras, and misc. THOSE are the posted variables.

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[SOLVED] Drop Down

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