如何在不使用ajax的情况下将数据从一个php文件发布到另一个php文件以将数据添加到数据库

Question

I have a php file (todo.php & todo.js) which collects some data from user and uses ajax to post the collected data to another php file (add.php) to be added to database . My issue is that I don't want to use ajax to post the data to the second php file (add.php). How do I change this. I have extracted the codes and shown below

//todo.js is called from todo.php to validate new 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 is to PHP as AJAX is to JS. If you want to call one script from another directly from PHP without requiring the broswer to make additional requests to the server, you can use cURL. Here's a simple function.

/*
 * 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;
}