Using MySQL results array more than once [duplicate] - php

Question

This question already has answers here:
Get all mysql selected rows into an array
(6 answers)
Closed 9 years ago.
I have the following code in my page
<?php
$dbc = mysql_connect();
$db = mysql_select_db();
$results= mysql_query("SELECT * FROM tbl_teams");
?>
<div class="datagrid"><table>
<thead><tr><th>header</th><th>header</th><th>header</th></tr></thead>
<tbody>
<tr>
<td>
<select name="game1_team1"><option value="0">Choose Team 1</OPTION><?php while($row = mysql_fetch_array($results)) {echo '<option value="'.$row['team_ID'].'">'. $row['team_name'].'</option>';}?></select>
</td>
<td>Vs.</td>
<td>
<select name="game1_team2"><option value="0">Choose Team 2</OPTION><?php while($row = mysql_fetch_array($results)) {echo '<option value="'.$row['team_ID'].'">'. $row['team_name'].'</option>';}?></select>
</td>
</tr>
The code shows the names of football teams from the MySQL table in dropdown game1_team1 but not in game1_team2; it's as though I can't use the same query twice. How do I remedy this? I'd like to use the same values for 60 identical drop downs on the page.
Can I store the values into an array and resuse them in each dropdown?

Answer 1

NOTE: The mysql_* function family is deprecated!
What does that mean? It means you should stop using it today. Old code should be updated as time and budget allow, but you should never, EVER write a new line of code that involves these functions
Another note: it is bad practice to use SELECT *... -- explicitly name the columns you are going to use in your query. That way, if the table structure changes, you can detect that because your query will fail, and react accordingly (in code). This is part of a broader concept known as defensive coding.
On to the problem at hand!
You can use PDO to fetch an array of results out of the database. You can then use that array as many times as you'd like!
// simple sample for connecting with PDO
$host="my.host.name"; // Host name
$username="my_user_name"; // username
$password="my password"; // password
$db_name="my_database_name"; // Database name
$pdo = new PDO('mysql:host='.$host.';dbname='.$db_name, $username, $password);
// actually do the query
$statement= $pdo->prepare('
SELECT
time_ID,
team_name
FROM
`tbl_teams`
');
$statement->execute();
// get an array with all the results
$results = $statement->fetchAll(PDO::FETCH_ASSOC);
After that, you can loop the array to make a set of options, or do anything else with the data. For example:
$options = array();
foreach ($results as $one_result){
$options[] = '<option value="'.$one_result['team_ID'].'">'.$one_result['team_name'].'</option>';
}
print '<select name="someSelectElement">'.implode('', $options).'</select>';
Documentation
Deprecation notice for mysql_* functions - http://www.php.net/manual/en/function.mysql-query.php
PHP PDO - http://php.net/manual/en/book.pdo.php
PHP mysqli - http://php.net/manual/en/book.mysqli.php

Answer 2

Using PDO you can easily fetch your results into an array, besides, the mysql_* functions are deprecated.
<?php
$db = new PDO( ... );
$stmt = $db->prepare("SELECT * FROM tbl_teams");
$stmt->execute();
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);
?>
<div class="datagrid">
<table>
<thead><tr><th>header</th><th>header</th><th>header</th></tr></thead>
<tbody>
<tr>
<td>
<select name="game1_team1">
<option value="0">Choose Team 1</OPTION>
<?php
foreach($results as $team) {
echo "<option value='{$team['team_ID']}'>{$team['team_name']}</option>";
}
?>
</select>
</td>
<td>Vs.</td>
<td>
<select name="game1_team2">
<option value="0">Choose Team 2</OPTION>
<?php
foreach($results as $team) {
echo "<option value='{$team['team_ID']}'>{$team['team_name']}</option>";
}
?>
</select>
</td>
</tr>
</tbody>
</table>
However, for efficiency I'd do this:
<?php
$db = new PDO( ... );
$stmt = $db->prepare("SELECT * FROM tbl_teams");
$stmt->execute();
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);
$teamList = '';
foreach($results as $team) {
$teamList .= "<option value='{$team['team_ID']}'>{$team['team_name']}</option>";
}
?>
...
<select name="game1_team1">
<option value="0">Choose Team 1</OPTION>
<?php echo $teamList; ?>
</select>
...
<select name="game1_team2">
<option value="0">Choose Team 2</OPTION>
<?php echo $teamList; ?>
</select>

Answer 3

you can also use MYSQLI_*
<?php
$dbc = mysqli_connect('host','user','password','db_name');
$results= mysqli_query("SELECT `team_ID`,`team_name` FROM tbl_teams");
?>
<div class="datagrid">
<table>
<thead><tr><th>header</th><th>header</th><th>header</th></tr></thead>
<tbody>
<tr>
<td>
<select name="game1_team1">
<option value="">Choose Team 1</OPTION>
<?php while($row1 = mysqli_fetch_array($dbc ,$results)) { echo '<option value="'.$row1['team_ID'].'">'. $row1['team_name'].'</option>';} ?>
</select>
</td>
<td>Vs.</td>
<td>
<select name="game1_team2">
<option value="">Choose Team 2</OPTION>
<?php while($row2 = mysqli_fetch_array($dbc ,$results)) {echo '<option value="'.$row2['team_ID'].'">'. $row2['team_name'].'</option>';}?>
</select>
</td>
</tr>
</tbody>
</table>
it will work now.. :)