Populating Dropdown List

Anon

I am trying to populate a dropdown box from a MySQL table as follows:

<?php
... connection stuff ...

$sql_select = "select id, code, name, city, state from stores order by id";
$result = @($sql_select, $connection) or die("couldn't execute query.");
?>

Then in the form:

All I get is the id in my dropdown. How do I get the rest of the columns to show, concat()? How do I add concat to this statement if this is the case?

Thanks!

Jeff

Anon

Hey Jeff,

Here is an example of some code I use:

After connecting to the database -

// 4.1 Query database for golfers (drop down)

$result = mysql_query("SELECT * FROM golfer order by lastName",$db);

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

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

echo ' <option value="'.$myrow["golferID"].'">'.$myrow["firstName"]." ".$myrow["lastName"]."</option>\n";
}
echo " </select>\n";

So I use the syntax $myrow["field name"] wherever I want to insert one or more of the table fields.

Hope this helps. happy form filling

Anon

<?
$sql_select = "select id, code, name, city, state from stores order by id";
?>

for the above query USE the code block BELOW
<?
while ( $row = mysql_fetch_row($result) )
echo "<option value=".$row[0].">".$row[1].$row[2].$row[3].$row[4]."</option>\n" ;
?>

Tell me...

Anon

You could always make it easier and use the for() command..

Anon

I use this codingstyle because it is easy to read and it gives you error if not correct.

<?php
// connect... blabla

$sql_select = "select id, code, name, city, state from stores order by id";
// Blablabla
//...blablabla
?>

<?php
while( list( $id, $code, $name, $city, $state ) = mysql_fetch_array( $result) )
{
echo "<option value='$id'>$id $code $name $city $state</option>";
}
?>

Anon

one addition to the for statement:

for($x=0; $x < $row; $x++) {
echo $row[$x];
}

should be

for($x=0; $x <count($row); $x++) {
echo $row[$x];
}

$countrow=count($row);

for($x=0; $x <$countrow; $x++) {
echo $row[$x];
}

either way...
stuart

Anon

<select name=\"store\">
<?
while( list( $id, $code, $name, $city, $state ) = mysql_fetch_array( $result) )
{
echo \"<option value=\'$id\'>$code $city $state</option>\";
}
?>
</select>

I tried this and it populates the drop down list except the variable $id doesn\'t carry to the form action script but everything else does... at least it ($id) doesn\'t show up in the table using the following insert query:

$sqlquery = \"INSERT INTO $table VALUES(NULL,\'$id\',\'$date_sent\',\'$fax_count\',\'$coupon_sent\',\'$fax_returned\',\'$double_checked\')\";

=====================================

Anon

Ok... I give up... I admit, I'm a rookie!

I am trying to make a simple input form that records data about some fax broadcasting stats from different stores. I already have the store info located in the stores table and I am recording the input in the fax_info table. I want to save the store.id as fax_info.s_id. For some reason the $id from the <select> form field is not working with the form action script. It won't insert the store id as s_id in the fax_info table. Everything else works as it is supposed to. Any criticism is welcome as well... 🙂

Here is what I am trying to do in its entirety: http://www.webtinker.com/test/stuff.txt

Thanks!

Jeff

Anon

Hi Jeffy,

You are trying to insert into field [fax_info.id] a NULL value in the
Form-action script.

According to the Table structure of fax_info
that you have given, the field [fax_info.id] contains a NOT_NULL Auto_increment constraint with Default value as '0'
ie:
CREATE TABLE fax_info (
id int(10) unsigned DEFAULT '0' NOT NULL auto_increment,.....

If this is the case
Modify your insert statement to :

INSERT INTO $table (s_id, date, fax_count, coupon_sent, fax_returned, double_checked)
VALUES('$id','$date_sent','$fax_count','$coupon_sent','$fax_returned','$double_checked')";

What you were doing was trying to insert a NULL value in a NOT NULL field. If a NOT NULL field contains a DEFAULT constraint then don't have to include it in your insert statement if you want it to take a default value.

Did this help ?...... tell me.

Anon

The variable $id has been put into a new variable now, this new variable is the SELECT form element

Replace the $id in the INSERT statement to $store

This is because you have assigned the value of $id to the SELECT form element which is the new variable that will be passed on to the next page on a form submit

Neville.

marcdry

Hi!
Don´t give up - I hope I can help you with this snippet:

<?
$count_select=0;
while ($count_select<$max_count1):?>
<tr>
<form name=<?form.$count_select?> method="post" action="">
<select name="select">
<option value=<? echo $markus[t1]?>>
<?echo "none selected";?>
</option>
<?
while (list($temp1,$temp2)=each($Pers_Array)):
?>
<option value=<? echo $markus[t1]?>>
<?echo $Pers_Array ["$temp1"]["surname"]." "
.$Pers_Array ["$temp1"]["name"]." "
.$Pers_Array ["$temp1"]["status"]." ";
?>
</option>
<?
endwhile;?>
</select>
<?
$count_select++;
reset($Pers_Array);?>
</form>
</tr>
<?
endwhile;?>
The $Pers_Array contains the array. The loop while (list($temp1,$temp2) provides the id of the array (former database) and the id can be displayed with $temp1 in the loop.

I hope this is helpfull. If you have need for more help to this topic, don´t matter to ask again.

Best regards M.