在PHP中显示查询结果

Question

I have a query:

$string = "SELECT COUNT(id) as sponsered FROM `$database`.`$mem` where parent_id = 2 group by plan";

Which result in:

 sponsored plan
         2 gold
         1 silver
         1 mitra

This result is shown when I run this query in MySQL.

Now I want this result in PHP in an array data[] where

  data[0] contains 2
  data[1] contains 1
  data[2] contains 1

I have tried this

$string = "SELECT COUNT(id) as sponsered FROM $database.$mem where parent_id = 2 group by plan";
$res = mysqli_query($con, $string);
while($data = mysqli_fetch_assoc($res)) {
    echo $data['sponsered'];
}

But it results in data['sponsered'] containing 211

It would be better if this can be done without using any loop.

Answer 1

Inside the while loop assign the values to array.

$string = "SELECT COUNT(id) as sponsered FROM $database.$mem where parent_id = 2 group by plan"; 
$res = mysqli_query($con, $string); 
$data = array();
while($result = mysqli_fetch_assoc($res)) {
   $data[] = $result['sponsered']; 
} 
print_r($data);

Output:

data[0] contains 2
data[1] contains 1
data[2] contains 1

Answer 2

Your display is correct you just need a separator so the data isn't all clumped. In a browser:

echo $data['sponsered'] . "<br>";

in your loop should do it, or command line,

echo $data['sponsered'] . PHP_EOL;

should do it.

As is it throws 2 , then 1 , then 1 which views as 211 .

Answer 3

Try this:

Please refrain from passing data directly into query in a prepare statement, use ? or : ? or : as deemed necessary.

Below is a sample code:

<?php
    $table_name = $table; //not recommended to pass table_name as variable, unless necessary.
    $parent_id = 2;
    $p = $plan;

    $string = "SELECT COUNT(id) as sponsered FROM $table_name where parent_id = ? group by ?";
    if ($stmt = $db->prepare($string))
    {
        $stmt->bind_param('is',  $parent_id, $p);
        $stmt->execute();

        $stmt->bind_result($sponsored, $plan);
        $stnt->store_result();

        $html ='';

        while ($stmt->fetch() !== FALSE)
        {
            // Do what you like with the variables here, i.e. $sponsored, $plan
            $html .= "<table>
                        <th> Sponsored </th>
                        <th> Plan</th>
                        <tr>
                        <td>$sponsored</td>
                        <td>$plan</td>
                        </tr>
                    </table>";
        }
        else {
            die("Query failed");
        }
        var_dump($html);
    }
?>

NOTE: If your result from your question is just an HTML (table/column) , you don't need to do bind, you can just output the HTML formatted.

Answer 4

使用mysqli_fetch_array($result, MYSQLI_NUM)将数据作为数字数组检索