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如何从mysqli输出json数组,而[重复]

mysqli api json arrays php

-1

Mr FoX · 技术社区 · 1 年前

我需要这个代码的帮助

$token = $_POST['token'];
$get_data = mysqli_query($db_con,"SELECT * FROM `app_login` WHERE `token`='$token'");
while ($d_token_data = mysqli_fetch_array($get_data)){
    $code = $d_token_data['code'];
    $get_acc_data = mysqli_query($db_con,"SELECT * FROM `students` WHERE `login_code`='$code'");
    $d_get_data = mysqli_fetch_all($get_acc_data,MYSQLI_ASSOC);
    echo json_encode(array('status' => '1' , 'data' => $d_get_data));
}

我有多排的

multiple rows

单行

single row

我也试过 fatch_all

$token = $_POST['token'];
$get_data = mysqli_query($db_con,"SELECT * FROM `app_login` WHERE `token`='$token'");
$d_token_data = mysqli_fetch_all($get_data,MYSQLI_ASSOC);
$code = $d_token_data['code'];
$get_acc_data = mysqli_query($db_con,"SELECT * FROM `students` WHERE `login_code`='$code'");
$d_get_data = mysqli_fetch_all($get_acc_data,MYSQLI_ASSOC);
echo json_encode(array('status' => '1' , 'data' => $d_get_data));

1 回复 | 直到 1 年前

1

-1

Dhruv Pandya 1 年前

$token = $_POST['token'];
$get_data = mysqli_query($db_con,"SELECT * FROM `app_login` WHERE `token`='$token'");
$dataToReturn = [];
while ($d_token_data = mysqli_fetch_array($get_data)){
    $code = $d_token_data['code'];
    $get_acc_data = mysqli_query($db_con,"SELECT * FROM `students` WHERE `login_code`='$code'");
    $d_get_data = mysqli_fetch_all($get_acc_data,MYSQLI_ASSOC);
    $dataToReturn[] = $d_get_data;
}
echo json_encode(array('status' => '1' , 'data' => $dataToReturn));

您应该先准备数据,然后只返回一次数据。