在 JSON 字典中返回 Mysqli

Question

我正在从 mysql 数据库中检索数据，并希望将其作为 JSON 字典返回。

<?php
if($db->connect_errno > 0){
   die('Unable to connect to database [' . $db->connect_error . ']');
}

$sql = <<<SQL
SELECT tbldomainpricing.extension, tblpricing.msetupfee
FROM `tblpricing`
INNER JOIN tbldomainpricing ON tbldomainpricing.id = tblpricing.relid 
WHERE tblpricing.type = 'domainregister'
SQL;

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
}

$all_rows = array();

while($row = $result->fetch_assoc()) {
 $all_rows []= $row['extension'] . ":" . $row['msetupfee'];
}
header("Content-Type: application/json");
echo json_encode($all_rows);
?>

回报：

{"extension":".com","msetupfee":"6.99"},{"extension":".net","msetupfee":"6.99"},

我如何获得回报

{"com":"6.99"},{"net":"6.99"}

谢谢

Answer 1

听起来像你想要的：

while($row = $result->fetch_assoc()) {
 $all_rows[$row['extension']] = $row['msetupfee'];
}

参考 u_mulder 的评论，你可能还想要这个：

while($row = $result->fetch_assoc()) {
 $all_rows[][$row['extension']] = $row['msetupfee'];
}

Answer 2

我没有测试它，但可能你需要这样的东西：

while($row = $result->fetch_assoc()) {
 $all_rows[] = array($row['extension'], $row['msetupfee']);
}