如何获取数组而不是json对象？

Question

<?php 

$link = mysql_connect('localhost', 'root', 'admin')
or die('There was a problem connecting to the database.');
mysql_select_db('hospitalmaster');

$hnum = (int)$_POST["hnum"];
$sql = "SELECT d.doctorid, d.doctorname 
        from hospitalmaster.doctor_master d 
            inner join pharmacymaster.pharbill e 
        where e.hnum = '$hnum' 
        and e.presid = d.d_empno 
        group by e.presid";

$result = mysql_query($sql);
$response = array();

if (mysql_num_rows($result) > 0) {
    while ($row = mysql_fetch_assoc($result)) {
        $response = $row;
    }

    echo json_encode(array("Doctor"=>$response));
} else {
    echo ("no DATA");
}
?>

我有上面显示的api，但这个api返回我作为json对象而不是json数组？ 我想知道如何将dotorid作为一个阵列的医生名称，因为我有很多医生的名字和身份证，我希望每个医生和他们相应的身份证作为一个单独的数组，现在他们作为单独的对象返回。 由于这是我第一次编写api，我不知道如何修改代码。

Answer 1

你每次循环都在编写数组。

while ($row = mysql_fetch_assoc($result)) {
    $response[] = $row;
    //change ^^
}

你应该使用mysqli_或PDO。 这是对mysqli_的建议

<?php 

$link = mysqli_connect('localhost', 'root', 'admin','hospitalmaster')
or die('There was a problem connecting to the database.');

$sql = "SELECT d.doctorid, d.doctorname 
        from hospitalmaster.doctor_master d 
            inner join pharmacymaster.pharbill e 
        where e.hnum = ?
        and e.presid = d.d_empno 
        group by e.presid";

$stmt = $link->prepare($sql);
$stmt->bind_param('i', (int)$_POST["hnum"]);
$stmt->execute();

if ($stmt->num_rows() > 0) {
    $result = $stmt->get_result();
    $response = array();

    while ($row = $result->fetch_assoc()) {
        $response[] = $row;
    }

    echo json_encode(array("Doctor"=>$response));
} else {
    echo ("no DATA");
}
?>