Question

我想从一个[]调用所有名称，数组不能使用foreach php吗？或者我不能做我想做的事情，我使用了foreach（“id”，“NOT LIKE”，$ id），但是当使用foreach的echo形式时，只显示一个名字，这是我的PHP代码：

if ($getusers !== FALSE) {

    $json["error_two"] = FALSE;

    foreach($db->get("users",array('id', 'NOT LIKE', $id))->results() as $u){

        $user = new user( $u->id );

        $json = array(
            'error_two' => FALSE,
            'getusers' => $getusers,
            'call_two' => [
                array(
                    'user_id' => $user->data()->id,
                    'fname' => $user->data()->fname,
                    'lname' => $user->data()->lname,
                    'username' => $user->data()->username,
                    'gender' => $user->data()->gender,
                ),
            ]
        );   
        echo json_encode($json);
    }
}

出：

{"error_two":false,"getusers":true,"call_two":[{"user_id":"1","fname":"Ned","lname":"Stark","username":"","gender":"Male"}]}{"error_two":false,"getusers":true,"call_two":[{"user_id":"3","fname":"Danyal","lname":"Join","username":"","gender":"Male"}]}{"error_two":false,"getusers":true,"call_two":[{"user_id":"4","fname":"dnianas","lname":".co","username":"","gender":"Male"}]}{"error_two":false,"getusers":true,"call_two":[{"user_id":"5","fname":"Mr.","lname":"Robot","username":"","gender":"Male"}]}

我想要的是：

{
"call_two" : [
        {
            "user_id" = "1",
            "fname" = "Ned",
            "lname" = "Stark",
            "username" = "",
            "gender" = "Male"
        },
        {
            "user_id" = "2",
            "fname" = "Jon",
            "lname" = "Snow",
            "username" = "",
            "gender" = "Male"
        },
        {
            "user_id" = "3",
            "fname" = "Danyal",
            "lname" = "Join",
            "username" = "",
            "gender" = "Male"
        }
    ]
}

Answer 1

...我使用了foreach（“id”，“NOT LIKE”，$ id），但是当使用foreach中的echo形式时，只显示一个名字......

那是因为你在$json循环的每次迭代中都覆盖了foreach。此外，这不是您提到的那里的无效JSON字符串。要实现有效（并且有点预期的JSON字符串），您的if块应该是这样的：

if ($getusers !== false) {
    $json = array('error_two' => false, 'getusers' => $getusers);
    foreach($db->get("users",array('id', 'NOT LIKE', $id))->results() as $u){
        $user = new user( $u->id );
        $json['call_two'][] = array(
                                'user_id' => $user->data()->id,
                                'fname' => $user->data()->fname,
                                'lname' => $user->data()->lname,
                                'username' => $user->data()->username,
                                'gender' => $user->data()->gender,
                            );
    }
    echo json_encode($json);
}

Answer 2

您正在覆盖'call_two'而不是向该数组添加出现

if ($getusers !== FALSE) {

    $json = array();
    $json["error_two"] = FALSE;
    $json['getusers'] = $getusers;

    foreach($db->get("users",array('id', 'NOT LIKE', $id))->results() as $u){
        $user = new user( $u->id );

        $call_two[] = array(
                            'user_id' => $user->data()->id,
                            'fname' => $user->data()->fname,
                            'lname' => $user->data()->lname,
                            'username' => $user->data()->username,
                            'gender' => $user->data()->gender,
                            );
        );   
    }
    $json['call_two'] = $call_two;

}
echo json_encode($json);

JsonArray得到所有的名字 - php

2 个答案: