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，但是这个api正在返回我作为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

每次循环都是在编写数组。

mysqli_

您应该使用mysqli_或PDO。以下是对<?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"); } ?>
的建议

pec

如何获取数组而不是json对象？

1 个答案: