如何在不使用 ajax 的情况下将数据从 php 文件发布到另一个 php 文件以将数据添加到数据库

Question

我有一个 php 文件 (todo.php & todo.js)，它从用户那里收集一些数据并使用 ajax 将收集到的数据发布到另一个要添加到数据库的 php 文件 (add.php) 中。我的问题是我不想使用 ajax 将数据发布到第二个 php 文件 (add.php)。我该如何改变这个。我已提取代码并显示如下

//todo.js 从 todo.php 调用来验证新的 todoEntry

function addToDo(){
    var descField = document.getElementById('nDesc');
    var percField = document.getElementById('nPerc');

    if( !validateField(descField) ){
        alert("Description " + errorMessage);
        return;
    }
    if( !validatePercentField(percField) ){
        alert("Percentage " + errorMessage);
        return;
    }

    //use ajax to post new entries to add.php which will add them
    ajaxCall(encodeURI('add.php?description=' + descField.value  
           + '&percentage=' + percField.value), function(response){
        if( response.status == 200 ){
            if( response.responseText.toString().indexOf
        ("success") != -1 ){
                //clear values
                descField.value = "";
                percField.value = "";

                //refresh table
                fetchToDo();
            }
            else
                alert(response.responseText);
        }
    });
    }

     From add.php:
    <?php
    session_start();

    if( !isset($_SESSION['username']) ){
        die("Your Session has expired. Please re-login to continue");
    }

    //get required data
    $description = trim(isset($_GET['description']) ? urldecode($_GET   
        ['description']) : "");
    $percentage = trim(isset($_GET['percentage']) ? urldecode($_GET
        ['percentage']) : "");

    //validate data
    if( empty($description) || empty($percentage) ){
        die("All fields are required!");
    }

    //connect to database
    include('connect.php');

    //insert data
    $stmt = $mysqli->prepare("INSERT INTO todo_entry VALUES (NULL, ?, NOW 
        (), NOW(), ?, ?)");
    $stmt->bind_param("ssi", $_SESSION['username'], $description,
         $percentage);
    $stmt->execute();

    if( $stmt->affected_rows > 0 ){
        $stmt->close();
        echo "success";
    }
    else{
        echo "An Error Occured: " . $stmt->error;
        $stmt->close();
    }

    //close database
    $mysqli->close();
    ?>

Answer 1

cURL 之于 PHP 就像 AJAX 之于 JS。如果你想直接从 PHP 调用一个脚本而不需要浏览器向服务器发出额外的请求，你可以使用 cURL。这是一个简单的函数。

/*
 * Makes an HTTP request via GET or POST, and can download a file
 * @returns - Returns the response of the request
 * @param $url - The URL to request, including any GET parameters
 * @param $params - An array of POST values to send
 * @param $filename - If provided, the response will be saved to the 
 *    specified filename
 */
function request($url, $params = array(), $filename = "") {
    $ch = curl_init();
    $curlOpts = array(
        CURLOPT_URL => $url,
        // Set Useragent
        CURLOPT_USERAGENT => 
            'Mozilla/5.0 (Windows NT 6.3; WOW64; rv:29.0) 
                    Gecko/20100101 Firefox/29.0',
        // Don't validate SSL 
        // This is to prevent possible errors with self-signed certs
        CURLOPT_SSL_VERIFYPEER => false,
        CURLOPT_SSL_VERIFYHOST => false,
        CURLOPT_RETURNTRANSFER => true,
        CURLOPT_FOLLOWLOCATION => true
    );
    if(!empty($filename)){
        // If $filename exists, save content to file
        $file2 = fopen($filename,'w+') 
            or die("Error[".__FILE__.":".__LINE__."] 
                    Could not open file: $filename");
        $curlOpts[CURLOPT_FILE] = $file2;
    }
    if (!empty($params)) {
        // If POST values are given, send that shit too
        $curlOpts[CURLOPT_POST] = true;
        $curlOpts[CURLOPT_POSTFIELDS] = $params;
    }
    curl_setopt_array($ch, $curlOpts);
    $answer = curl_exec($ch);
    // If there was an error, show it
    if (curl_error($ch)) die(curl_error($ch));
    if(!empty($filename)) fclose($file2);
    curl_close($ch);
    return $answer;
}