Looping through an mysql array and printing each row into a <select> - php

Question

I want to retrieve each row from one column in one table and put all of them in a element.
I'm pretty much flailing in the dark right now, what am I doing wrong/missing?
$db_connect = mysql_connect("$host", "$db_username", "$db_password");
$db_select = mysql_select_db("$db_name");
mysql_set_charset('utf8', $db_connect);
$db_query_project="SELECT * FROM $tbl_name";
print('<form name="report" method="post" action="">');
print('<label for="project">Projekt </label>');
print('<select name="project_select">');
print('<option>Test</option>');
foreach ($db_query_project as $key => $value){
print('<option>'.$key['project_name'].'</option>');
}
print('</select>');
print('</form>');
Is there a better way to go about doing this?

Answer 1

You need to connect to the database and perform the query in order to retrieve the data and fetch it. To do so, you can use either mysqli or PDO. Here's an example on how to fetch data from a table $tbl_name with a column project_name using mysqli.
$mysqli = new mysqli("localhost", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ")";
exit();
}
$query = "SELECT * FROM `$tbl_name`";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
printf ("%s \n", $row["project_name"]);
}
/* free result set */
$result->free();
}
/* close connection */
$mysqli->close();

Answer 2

i prefer this one:
function getFieldSelect($name, $selected, $values, $attributes) {
$r = "<select name=\"$name\" $attributes>";
foreach ($values as $k => $v) {
if ($k == $selected) {
$sel = "selected";
} else {
$sel = "";
}
$r .= "<option value=\"$k\" $sel>$v</option>\n";
}
$r .= "</select>";
return $r;
}
i got the values from DB and give it to my function in array format. I think this is the most flexible way to print a select field. You can use a database to get the values or any array.

Answer 3

<?php
$db_query_project= mysql_query("SELECT * FROM $tbl_name");
echo '<form name="report" method="post" action="">';
echo '<label for="project">Projekt </label>';
echo '<select name="project_select">';
echo '<option>Test</option>';
while( $row = mysql_fetch_assoc( $db_query_project ) )
{
echo '<option>'.$row['project_name'].'</option>';
}
echo '</select>';
echo '</form>';