Php 修复了预期的数组，但得到了对象？

我试图使用API显示数据库中的数据，但出现以下错误：

"Expected array but received: {"data":{"useri":[{"ID":"1","username":"asdas","city":"asdasd","age":"20"},"

我如何才能使其只获取ID为1的数组

PHP:

JS:

我如何才能使其只获取ID为1的数组

此外，html：

<tr ng-repeat="user in users| filter:search_query">
            <td>
            <span>{{user.username}}</span></td>
            <td>{{user.city}}</td>
            <td>{{user.age}}</td>
            <td>
</tr>

{{user.username}
{{user.city}
{{user.age}

当您将“useri”作为索引时，json\u编码输出总是一个json对象

要输出json数组，要编码的数组应该只是索引数组

$indexed_array = array($user1, $user2);

$associative_array = array("user1" => $user1, "user2" => $user2);

所以你可以这样做

if($num > 0){
    $users_array = array(); //stocam userii intr-un array

    while($num = $rdej->fetch(PDO::FETCH_ASSOC)){
        extract($num);

        $users_item = array(
            "ID" => $ID,
            "username" => $username,
            "city" => $city,
            "age" => $age
        );

        array_push($users_array, $users_item);


    }

    // 200 OK - request reusit
    http_response_code(200);

    echo json_encode($users_array);
}

如果您需要对useri的引用，那么您可以在客户端的javasript中这样做

当您将“useri”作为索引时，json_编码输出总是一个json对象

要输出json数组，要编码的数组应该只是索引数组

$indexed_array = array($user1, $user2);

$associative_array = array("user1" => $user1, "user2" => $user2);

所以你可以这样做

if($num > 0){
    $users_array = array(); //stocam userii intr-un array

    while($num = $rdej->fetch(PDO::FETCH_ASSOC)){
        extract($num);

        $users_item = array(
            "ID" => $ID,
            "username" => $username,
            "city" => $city,
            "age" => $age
        );

        array_push($users_array, $users_item);


    }

    // 200 OK - request reusit
    http_response_code(200);

    echo json_encode($users_array);
}

---

如果您需要对useri的引用，那么您可以在客户端的javasript中执行该操作

创建额外信息的是您的代码，因此只需停止添加您不想要的子数组

if($num > 0){
    $users = [];

    while($user = $rdej->fetch(PDO::FETCH_ASSOC)){

        $users[] = array(
                        "ID" => $user['ID'],
                        "username" => $user['username'],
                        "city" => $user['city'],
                        "age" => $user['age']
                    );
    }

    // 200 OK - request reusit
    http_response_code(200);

    echo json_encode($users);

不要在全局范围内特别尝试

---

是您的代码创建了额外的信息，所以不要再添加您不想要的子数组了

if($num > 0){
    $users = [];

    while($user = $rdej->fetch(PDO::FETCH_ASSOC)){

        $users[] = array(
                        "ID" => $user['ID'],
                        "username" => $user['username'],
                        "city" => $user['city'],
                        "age" => $user['age']
                    );
    }

    // 200 OK - request reusit
    http_response_code(200);

    echo json_encode($users);

不要在全局范围内特别尝试