“如何解决此php文件中的此null错误?”

时间:2019-05-28 10:34:57

标签: php mysql api

我正在按主要类别打印子类别ID和名称。我在此表中将mscategory_id作为foreign键。所以我与前键匹配并以json格式返回。在我的数据库中,我有针对1个主要类别的图标。执行后,我得到空值。 enter image description here

This is Database

<?php
require 'conn.php';
require 'headers.php';
$method=$_SERVER['REQUEST_METHOD'];
if($method == "GET"){

    $id=$_GET['mscategory_id'];
    $sql = "SELECT  sscategory_id, sscategory_name FROM t_srv_sscategory WHERE mscategory_id='$id' ";
    $result = mysqli_query($conn,$sql);
    $response=array(); 
    if (!empty($result)) {
            // check for empty result
            if (mysqli_num_rows($result) > 0) {

                $result = mysqli_fetch_array($result);
                $user = array();
                $user["sscategory_id"] = $result["sscategory_id"];
                $user["sscategory_name"] = $result["sscategory_name"];
                // success
                $response["success"] = 1;
                // user node
                $response["user"] = array();
                 array_push($response["Sub Category Services"], $user);
                // echoing JSON response
                echo json_encode($response);
            } 
            else {
                // no user found
                $response["success"] = 0;
                $response["message"] = "No user found";

                // echo no users JSON
                echo json_encode($response);
            }} 
    else {
        // no user found
        $response["success"] = 0;
        $response["message"] = "Credentials Error";


        echo json_encode($response);
    }
}
else {
    $json["error_msg"] = "Required parameters (phone or password) is missing!";
    echo json_encode($json);
    $conn->close();
}
?>

0 个答案:

没有答案