[SOLVED] Populating a drop-down

[rawky1976]

Hi all; hope you're well!!!! I found a script that displays the field names from a table in a drop-down but I want to show the values in that field.

 

I've tried to amend that script but think I've got the logic of the functions incorrect.

 

Thanks for any help you might be able to offer

 

<?php
$connection = mysql_connect("localhost","xxx","xxx");
$select_db = mysql_select_db("dbname", $connection);
//$fields = mysql_list_fields("dbname", "tablename", $connection);
//$columns = mysql_num_fields($fields);
$query = "select field from dbname.table ORDER BY field";
$result = mysql_query($query) or die(mysql_error());
$num_rows = mysql_num_rows($result);
$columns = mysql_fetch_array($result);
echo "System Name: <select name=Field>";
for ($i = 0; $i < $columns[$num_rows]; $i++) {
echo "<option value=$i>";
//echo mysql_field_name($fields, $i);
}
echo "</select>";
?>

[revraz]

This should get you on track.  This assumes the value is the first field and the name is the 2nd.

 

echo "System Name: <select name=Field>";

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

 

echo "<option value={$row[0]}>";

echo $row[1];

echo "</option>";

}

echo "</select>";

 

[rawky1976]

Thanks, I'm now getting undefined output on

 

echo $row[1];

 

Is it because I'm only selecting one field and not *?

[rawky1976]

OK, I figured that part out, but I'm only getting one row from the table, I need all of them please?

 

Thank you, Mark

[revraz]

Yes, select what you want displayed.  If just field, and you only want the field as the value and text, change $row[1] to $row[0]

[revraz]

Repost the new code

 

OK, I figured that part out, but I'm only getting one row from the table, I need all of them please?

 

Thank you, Mark

[rawky1976]

Thanks, here it is: -

 

<?php

$connection = mysql_connect("localhost","user","pass");

$select_db = mysql_select_db("dbname", $connection);

//$fields = mysql_list_fields("dbname", "tablename", $connection);

//$columns = mysql_num_fields($fields);

$query = "select * from dbname.tablename ORDER BY fieldname";

$result = mysql_query($query) or die(mysql_error());

$num_rows = mysql_num_rows($result);

$columns = mysql_fetch_array($result);

echo "System Name: <select name=Field>";

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

 

echo "<option value={$row[4]}>";

echo $row[0];

echo "</option>";

}

echo "</select>";

?>

[rawky1976]

Fixed it, there were two variables assigned the the fetch_array. Commented the top one out!!!

 

Thank you, Mark