检索和显示数据可能是简单的PHP / MYSQL问题

Question

I have been racking my brains over this for a while now. Here is the data I have in the SQL data base as an example:

ID | TYPE | DATA  
1 | TXT | TEST  
2 | PHP | php  
3 | JS | JAVASCRIPT

That is just an example, there are multiple listing for TXT, PHP and JS throughout the table. What I want to do is retrive all the data and display it all into separate drop down/select boxes. Meaning, select box one would list all data with type TXT, select box two would list all data with type PHP and select box 3 would list all data with type JS. The only way I have came about doing this is doing individual sql queries for each different type. I know there is a way to do it all in 1 query and then display it the way I want to but I just can't seem to figure out how and I know its going to drive me nuts when someone helps and I see just how they did it.

Answer 1

The only way that I know of to get all of the data in one query is just to do a generic SELECT * FROM tbl, and then you can group them in the code:

$res = mysqli_query('SELECT * FROM tbl');

$data = array();
while($row = mysql_fetch_assoc($res)) {
    $type = $row['type'];
    $data[$type][] = $row;
}

// $data contains all of the record, grouped by the TYPE column

foreach($data as $type => $records) {
    echo "records for $type: <select>";
    foreach($records as $record) {
        echo "<option value='$id'>$id</option>";
    }
    echo "</select>";
}