我的 php 文件不以 JSON 形式返回所有值

Question

所以我有一个数据库，我编写了一个php 文件，该文件将值插入table，然后返回此table 中的所有值。

但是，我的代码只返回表中的一个随机值，而不是全部。我不知道为什么，但这是我的代码：

<?php
include_once "init.php";

if(!empty($_POST['names'])){
    $contactname = $_POST['names'];

    $query = "INSERT INTO contacts (contactID, names) 
                VALUES('NULL', ?)";
    $results = $conn -> prepare($query);
    $results->bind_param('s', $contactname);
    $results->execute();
    $results->close();
    echo json_encode("Success"); 

    $query_two = "SELECT names FROM contacts";
    $result = mysqli_query($conn, $query_two);
    $response = array();

    if(mysqli_num_rows($result)){
        while($row = mysqli_fetch_assoc($result)){
            $response["names"][] = $row["names"];
        }
    }
    echo json_encode($response);
}
else{
    echo json_encode("Something went wrong");
}

?>

编辑：谢谢你们这么快为我提供解决方案！我修复了它，它可以工作，但第一个 echo json_encode("Success"); 没有被执行。

Answer 1

在你这里：

$response["names"] = $row["names"];

您正在替换 $response["names"] 的值。相反，请尝试将其添加到数组中：

$response = array("names" => array());

if(mysqli_num_rows($result)){
    while($row = mysqli_fetch_assoc($result)){
        $response["names"][] = $row["names"];
    }
}

---

要使用一个 JSON 对象，您需要在顶部初始化 $response 并根据需要更改值。

<?php
include_once "init.php";

$response = ['success' => false, 'message' => null, 'names' => []];

if(empty($_POST['names'])) {
    $response['message'] = 'No names were provided in the request';

} else {
    $contactname = $_POST['names'];

    $query = "INSERT INTO contacts (contactID, names) VALUES('NULL', ?)";
    $results = $conn->prepare($query);
    $results->bind_param('s', $contactname);
    $results->execute();
    $results->close();

    $response['success'] = true;

    $query_two = "SELECT names FROM contacts";
    $result = mysqli_query($conn, $query_two);

    if(mysqli_num_rows($result)){
        while($row = mysqli_fetch_assoc($result)){
            $response['names'][] = $row['names'];
        }
    }

}

echo json_encode($response);