Question

我有一个我想访问的SQL表名为＆＃39; name_scraper＆＃39;并在一个设定范围之间检索数据，然后将生成的数组返回到我的node.js文件以供使用。基本上，Node js请求，php $ _GET [＆＃39;＆＃39;]，mysqli_query，然后？

这是我的节点js代码联系php文件

var request = require('request');
var http = require('http');

    var post_options = {
        host: 'localhost',
        path: '/name_request.php?pos1=0&pos2=5',
        method: 'POST',
        headers: {
            'User-Agent': 'Super Agent/0.0.1',
            'Content-Type': 'application/x-www-form-urlencoded',
        }
    }; 
var post_req = http.request(post_options, function (res) {
    res.on('data', function (chunk) {
        console.log('Response: ' + chunk);
        chnk = chunk; 
            console.log("chunk: " +chnk[0]); //random test I did that doesn't work
        });
    });

    post_req.end();

现在上面的代码可以联系我的php文件，然后将该数组输出到控制台。

这是我的php：

$servername = "localhost"; //localhost 
$username = "root"; 
$password = ""; 
$dbname = "test"; 

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

$start = $_GET['pos1'];
$end =  $_GET['pos2'];

$sql = mysqli_query($conn, "SELECT * FROM name_scraper WHERE ID BETWEEN '$start' AND '$end'");
$array = mysqli_fetch_all($sql);
print_r($array); //can change to specific index and it works fine

?>

这总体上设法检索数据但不以可用的形式返回数据，该数组不是返回的数组，只是响应的一部分。我需要它以可用的形式返回数组，这样我就可以在每个索引中操作和提取数据。

Answer 1

您应该更改输出方法，如下所示;

$sql = mysqli_query($conn, "SELECT * FROM name_scraper WHERE ID BETWEEN '$start' AND '$end'");
$array = mysqli_fetch_all($sql,MYSQLI_ASSOC); //If you add MYSQLI_ASSOC your output will be better
echo json_encode($array); //That will output an json array

希望这有帮助

将node.js的请求发送到PHP，返回一个数组

1 个答案: