Question

尝试从API将一些json数据存储到MySQL。

我正在使用这个简单的PHP代码

NSArray

但是我收到了这个错误

<?php 

    require 'config.php';

    $url = "https://www.widgety.co.uk/api/operators.json?app_id<key>&token<token>";
    $string = file_get_contents('https://www.widgety.co.uk/api/operators.json?app_id<key>&token=<token>');
    $arr = json_decode($string, true);

    foreach($arr as $item){
        $title          = $item['operator']['title'];
        $id             = $item['operator']['id'];
        $profile_image  = $item['operator']['profile_image_href'];
        $cover_image    = $item['operator']['cover_image_href'];
        $href           = $item['operator']['href'];
        $html_href      = $item['operator']['html_href'];
        $unique_text    = $item['operator']['unique_text'];
        $reasons_to_book = $item['operator']['reasons_to book'];
        $members        = $item['operator']['members_club'];
        $brochures      = $item['operator']['brochures'];
        $ships          = $item['operator']['ships'];
        $video_url      = $item['operator']['video_url'];

        $query("INSERT INTO operator (title, id, profile_image) VALUES('$title', '$id', '$profile_image')");

        $dataatodb = mysqli_query($con, $query);

        if(!$dataatodb){
            die("Error" . mysqli_error($con));
        }

    }


    ?>

我做错了什么？

PS在网址上我有正确的app_id和令牌

Answer 1

当你的foreach ($jsonObject as $item) { $title = $item->title; }数组是一个对象时，你正在对它进行处理。

->

注意['']而不是var_dump()

要进一步调试，请使用$apiURL = 'url'; $response = file_get_contents($apiUrl); $jsonResponse = json_decode($response); var_dump($jsonResponse);检查您的变量是否包含实际对象，或者您的api请求是否失败。

var maxValue = 50, minValue = 1;
var yourArray = []; // You can have your numbers array in this one.
var sortedArray = yourArray.sort(); // this will sort the array 

var resultArray =  myarray.filter(function(value){
  return (value > minValue && value < maxValue);
});

Answer 2

/* create a connection */
$mysqli = new mysqli("localhost", "root", null, "yourDatabase");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

/* let's say we're grabbing this from an HTTP GET or HTTP POST variable called jsonGiven... */
$jsonString = $_REQUEST['jsonGiven'];
/* but for the sake of an example let's just set the string here */
$jsonString = '{"name":"jack","school":"colorado state","city":"NJ","id":null}
';

/* use json_decode to create an array from json */
$jsonArray = json_decode($jsonString, true);

/* create a prepared statement */
if ($stmt = $mysqli->prepare('INSERT INTO testdemo (name, school, city, id) VALUES (?,?,?,?)')) {

    /* bind parameters for markers */
    $stmt->bind_param("ssss", $jsonArray['name'], $jsonArray['school'], $jsonArray['city'], $jsonArray['id']);

    /* execute query */
    $stmt->execute();

    /* close statement */
    $stmt->close();
}

/* close connection */
$mysqli->close();

将json内容从API存储到mysql数据库

2 个答案: