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Display all databases in dropdown menu.


prathameshkakade

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Hello everyone.I was trying to display all the databases available on a server in a dropdown list.But I could not get it.Here is my code.

<html>
<head>
<title>
List Databases
</title>
</head>
<body>
<?php
$connection = mysqli_connect('localhost','root','123456789');
if(!$connection)die(mysqli_connect($connection));else echo "Step 1:<br/>Successfully connected to MySQLi server<br/><hr/>";
$query = mysqli_query($connection,"SHOW DATABASES");
echo "Step 2:<br/>The databases available on this server are :<br/>";
while(($result = mysqli_fetch_assoc($query)) != NULL)
{
	echo $result['Database']."<br/>";
}
echo "<hr/>";
echo "<form method = 'POST' action = 'list-db.php'>";
echo "<select name = 'db[]'>";
$a = 0;
while($result[$a]!= NULL)
{
	echo "<option>$result[$a]</option>";
	$a++;
}
 
?>

Please help.

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The result set (records returned), generated by your query "SHOW DATABASES", has already been looped through to display the databases.

while($record = mysqli_fetch_assoc($result_set))
{
	echo $record['Database']."<br />";
}

You now need to explicitly set the pointer to the first record (row) of the result set to loop through it again to create the select box.  

mysqli_data_seek($result_set,0); // set pointer to first record because already at end of records
                                 // now can loop through records (result set) again because we set to first record
while($record = mysqli_fetch_assoc($result_set))
{
	echo "<option>{$record['Database']}</option>";
}

As Josh pointed out, after fetching all of the records to display the databases, the function mysqli_data_seek() will be used to set the internal pointer of the result set to the first record. This will allow you to loop through the result set again starting from the first record, or any record for that matter specified by mysqli_data_seek() s offset parameter, to create the select box.

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