使用 json 和 cURL 访问禁止的错误 403 Web 服务答案

【问题标题】：Access forbidden Error 403 web service using json and cURL使用 json 和 cURL 访问禁止的错误 403 Web 服务
【发布时间】：2024-01-24 12:05:01
【问题描述】：

我必须做一个网络服务。因此我参考了网上的一些教程并想出了以下代码

index.php

<html>
<head>
    <title>Form page</title>
</head>

<body>

    <form action="http://localhost:81/my%20web%20service/webservice" method="get">
        Table name:<br>
        <input type="text" name="s" value=""><br>
        <input type="submit" value="Submit">
    </form>
</body>
</html>

webservice.php

<?php

include('connectdb.php');

$something = $_POST['s'];

$sql = "SELECT * FROM $something";
$result = $conn->query($sql);
$myArray = array();
if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $myArray[] = $row;
        echo "name: " . $row["name"]. " - town: " . $row["town"]. " - telephone: " . $row["telephone"]. "<br>";
    }
    //$data = json_encode($myArray);
    //var_dump($data);

    $url = 'http://localhost:81/my%20web%20service/show%details.php';
    $ch=curl_init($url);
    $ch = curl_init();

    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_POST, 1);

    $string = http_build_query($myArray);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $string);
    $applist = curl_exec($ch);

    curl_close($ch);

} else {
    echo "0 results";
}
//$result->close();
$conn->close();
?>

connectdb.php

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "webservice_trial";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else{
    echo "connection successful.<br>";
}

?>

现在我想做的是让数据显示在另一个页面上。即我想将数据从 webservice.php 从 json 格式传输到另一个页面（显示 details.php）。

显示详细信息.php

<html>
<head>
    <title>Show details page</title>
</head>

<body>
    <?php
    print_r($_POST['data']);
    ?>
</body>
</html>

如何将 json 字符串从 web 服务重定向到 show details.php

我得到的是这个

connection successful.
name: hilton - town: colombo - telephone: 774933705
name: galadari - town: colombo - telephone: 112894143
name: mt. lavinia - town: mt. lavinia - telephone: 773580324
string(257) "[{"name":"hilton","town":"colombo","telephone":"774933705","description":"excellent"},{"name":"galadari","town":"colombo","telephone":"112894143","description":"best"},{"name":"mt. lavinia","town":"mt. lavinia","telephone":"773580324","description":"good"}]"
Access forbidden!

You don't have permission to access the requested object. It is either read-  protected or not readable by the server.

If you think this is a server error, please contact the webmaster.

Error 403

【问题讨论】：

你是在本地主机还是虚拟主机上工作
我正在使用 localhost (xampp)
检查您的主机文件、Apache2 配置文件和虚拟主机文件
@NimmiRashinika，您的查询有问题：` $sql = "SELECT * FROM $something";` ，$something 我猜不是您的表名
@Nehal 女士，我已经检查过了

标签： php mysql json web-services curl

【解决方案1】：

试试下面这段代码，把它添加到你的虚拟主机配置中

<Directory "E:/MyProjects/projectname/">
        Options Indexes FollowSymLinks MultiViews
        AllowOverride all
        Order Deny,Allow
        Allow from all
        Require all granted
</Directory>

还有

Options FollowSymLinks

到：

Options +FollowSymLinks

我通过这种方式解决了同样的问题。希望对您有所帮助。

【讨论】：

【解决方案2】：

你好尼米拉希尼卡，

--> 首先，您在表单中使用 GET 方法并使用 $_POST 获取数据。我想在这里你无法得到任何回应。 --> 其次认为你有 init curl twise time ，所以请删除该代码。

在我的代码下面，工作正常..

Index.php

<html>
<head>
    <title>Form page</title>
</head>

<body>

    <form action="http://localhost/csvToExcle/sof35545870/webservice/webservice.php" method="get">
    Table name:<br>
        <input type="text" name="s" value=""><br>
        <input type="submit" value="Submit">
    </form>
</body>
</html>

connection.php

<?php

$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "telecom";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else{
    echo "connection successful.<br>";
}
?>

webservice.php

<?php error_reporting(E_ALL); ini_set('display_errors', 1); include('connection.php');
$something = $_GET['s'];
$sql = "SELECT * FROM $something";
$result = $conn->query($sql);
$myArray = array();
if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $myArray[] = $row;
        echo "auth_key: " . $row["auth_key"]. " - token: " . $row["token"]. " - telephone: " . $row["caller_id_caller"]. "<br>";
    }

    $url = 'http://localhost/csvToExcle/sof35545870/showDetails.php';
    $string = http_build_query($myArray);

    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_POST, 1);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $string);
    $applist = curl_exec($ch);

    if($applist === false)
    {
        echo 'Curl error: ' . curl_error($ch);
    }
    curl_close($ch);

} else {
    echo "0 results";
}
$conn->close();
?>

showDetails.php

<html>
<head>
    <title>Show details page</title>
</head>

<body>
    <?php
    print_r($_POST);
    ?>
</body>
</html>

【讨论】：