php - 如何在php中从mysql编码多个结果…我得到5个结果，但是当我运行API时，它没有以JSON格式显示任何数据

我在php中创建了一个API以从mysql数据库中搜索，数据库给了我5个结果，但是运行时在我的php API中...它没有显示任何数据，因为以下响应是我的API代码。

这是我的Php的API，我将包名作为标头中的输入，然后在mysql数据库中找到该输入，这应返回5个结果给我，而且我必须将这5个结果编码为json并作为响应发送。

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "package";

//Create Connection
$conn = new mysqli($servername, $username, $password, $dbname);
//Check Connection
if($conn->connect_error) {
    die("Connection Failed: " . $conn->connect_error);
}

$packagename = (print_r($_SERVER['HTTP_PACKAGENAME'],true));
//$pass = (print_r($_SERVER['HTTP_PASSWORD'],true));
//print_r($_SERVER);

//print_r(apache_request_headers());
$sql = "select * from  `package_details` where Package_Name='$packagename' ";
$result = $conn->query($sql);

if($result->num_rows>0) {
    //$response["package_details"] = array();
    //Output data of each row
    while($row = $result->fetch_assoc()) {
        //temp user array
        $user = array();

        $user["id"] = $row["id"];
        $user["Package_Name"] = $row["Package_Name"];
        $user["Package_Day"] = $row["Package_Day"];
        $user["Package_Description"] = $row["Package_Description"];

        //push single product into final response array
        //array_push($response["package_details"], $user);
    }
    //success
    //$response["success"] = "valid";
    //echoing JSON response
    echo json_encode($user);
} else {
    //no products found
    $response["package_details"] = array();
    while($row = null) {

    $user = array();
        $user["id"] = $row[null];
        $user["Name"] = $row[null];
        $user["Contact_Number"] = $row[null];
        $user["Email_Id"] = $row[null];


        //push single product into final response array
        array_push($response["package_details"], $user);
    }
        $response["success"] = "invalid";
    //$response["success"] = "invalid";
    //$response["message"] = "No Products Found";

    //echo no users JSON
    echo json_encode($response);
}
?>

最佳答案

您只是对最后一个$user进行json编码，而不是应放入$response['package_details']的用户数组

请参阅下面的修改后的代码

$response = array();

if($result->num_rows>0) {

    while($row = $result->fetch_assoc()) {
        $user = array();

        $user["id"]                  = $row["id"];
        $user["Package_Name"]        = $row["Package_Name"];
        $user["Package_Day"]         = $row["Package_Day"];
        $user["Package_Description"] = $row["Package_Description"];

        //push single product into final response array
        $response["package_details"][] = $user;
    }
    //success
    $response["success"] = "valid";

} else {

    //no products found
    $response["success"] = "invalid";
    $response["message"] = "No Products Found";
}

echo json_encode($response);

关于php - 如何在php中从mysql编码多个结果…我得到5个结果，但是当我运行API时，它没有以JSON格式显示任何数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/42788363/