[RESOLVED] creating a drop-down list

[snoopgreen]

Hello everyone

im hoping someone may be able to help me with an problem i have.

well in my database i have a field called type of acred
this field is set with nine different values
eg. PART 1, PART 2, Project and so on

SO on my system i have a form which is used to add programmes
and what i wanted was a drop -down box, which lists all the 9 values from the database
and a user can chose more than one of these values

I dont understand how i will do this because i know i will have to connect to the database 1st to get all the info from that table
but can u do that if im also inserting into different tables?
coz at present my form is inserting values into 2 tables, but it will end up being more

any help on either of these issues would be greatly appreciated
thanks in advance
snoopgreen

[jmrouleau]

Sure you can...just transfer the values to an array and print from the array wherever you want.

[snoopgreen]

thanks for the reply
im sorry i understand what your saying but dont know how to code it

an example of how i wud go about coding this would be great or even some more advise on how i wud go about completing this

thanks again
snoopgreen

[bradgrafelman]

but can u do that if im also inserting into different tables?

Of course, you can have as many different types of queries on a page as you want. They're individual queries.

I'm assuming you already know how to connect to a database, so here's an example setup of how to query the MySQL table for the values of the of the drop-down and form the HTML:

$query = 'SELECT theValues FROM thetable';
$exec = mysql_query($query) or die('MySQL Error: ' . mysql_error() );

echo "<select name=\"dropdown\">\n";

while( $result = mysql_fetch_row($exec) ) {

 echo '<option>' . $result[0] . "</option>\n";

}

echo "</select>\n";

[snoopgreen]

hello brad
thanks for your help

im getting another problem now
Im stuck on trying to get the values position in my form correctly

i have put the code in below, but the values appear above my form not in the field i want them too!

Ive cut some of the form cut off coz its quite a long form, but to get the idea

$db = mysql_connect("database", "username", "password");
mysql_select_db("username",$db);
$sql = 'SELECT type_of_acred FROM acreditation';
$acred = mysql_query($sql) or die ('MySQL Error: ' . mysql_error());

echo("<form action= \"" . $PHP_SELF . "\" method=\"POST\">");
echo("<table width=\"626\" border=\"1\" align=\"center\">");
echo("<tr>");
echo("<td width=\"160\">Programme UCAS code: </td>");
echo("<td colspan=\"4\"><input type=\"text\" name=\"ucas\" size=\"10\"></td>");
echo("</tr>");
echo("<tr>");
echo("<td>Validation Notes: </td>");
echo("<td colspan=\"4\"><textarea name=\"Validnotes\"></textarea></td>");
echo("</tr>");
echo("<tr>");
echo("<td>Acreditation Body:</td>");
echo("<td colspan=\"4\"><input type=\"text\" name=\"AcredBody\"></td>");
echo("</tr>");
echo("<tr>");
echo("<td>Acreditation Type:</td>");
echo("<td><select name=\"TypeAcred\" multiple></td>");
While ($result = mysql_fetch_array($acred) ) {
echo '<option>' . $result[0] . "</option>\n";
}
echo("</select>");
echo("</tr>");
echo("<tr>");
echo("<td>Acreditation Start Date:</td>");
echo("<td colspan=\"4\"><input type=\"text\" name=\"AcredStart\"></td>");
echo("</tr>");

this code is making the acred type values appear above the form not in the place i want, which is Acreditation Type part.
Could you please advise me some more on correcting this please

thanks again for the help
snoopgreen

[bradgrafelman]

You don't need separate echo statements for each line. In fact, it will not create a new line when you simply enter down and start a new echo statement.

It's early in the morning for me right now, so maybe that's why, but I really don't understand what you're saying it's doing by looking at the code - it looks fine to me. Can you post a link to this page so we can see exactly what it's doing?

[Houdini]

echo("<td>Acreditation Type:</td>");
echo("<td><select name=\"TypeAcred\"multiple>");
while ($result = mysql_fetch_array($acred) ) {
 extract($result); 
echo "<option ='$varname'>$varname\n";

Change this part and it will list them all, or remove the multiple for a drop down.

[snoopgreen]

hello thanks for both your replies

but i still cant get it working
the code im using is a bit different to what ive posted, but im trying to do exactly the same thing.
code below:

<?php
$db = mysql_connect("", "", "");
mysql_select_db("",$db);
$result3 = mysql_query("SELECT prog_id FROM programme", $db);

echo ("<SELECT name=\"dropdown\" multiple>");
While ($rec = mysql_fetch_row($result3)) {
extract ($result3);
echo '<option>' . $result3[0] . "</option>\n";
//echo ("option = '$varname'>$varname\n");
echo("</option>");
}

i am getting a box in the top corner but im not getting any data in it
it should just have a list of numbers, but it is blank
i dont know where im going wrong?

thanks for the help / advice apprecaited
snoopgreen

[snoopgreen]

thanks guys

ive gotten it working it lists all the programme ids now

thanks for all your help

the code ive used is

echo ("<SELECT name=\"dropdown\">");
While ($rec = mysql_fetch_array($result3)) {
echo "<option>" . $rec["prog_id"] . "</option>\n";

and this is bringing the id's for me
thanks again
snoopgreen

[Houdini]

Change this

echo '<option>' . $result3[0] . "</option>\n";
//echo ("option = '$varname'>$varname\n");
echo("</option>");
}

To this

echo "<option= '$prog_id'>$prog_id</option>\n";
}
echo("</select></td>")

To enter code select your code and click the ~ in the little dog eared page next to the quote bubble, just before the seperator in the format rows.

[snoopgreen]

thnaks

ive always wondered how to input code like u have been doing

thanks for help i will have a look at your code
thanks again
snoopgreen