PHPINSERTINTO不會將資料插入資料庫

所以基本上我有多步表單，我通過 jQuery 和 AJAX 請求處理。請求順利通過，但似乎我的 PHP 代碼不起作用（它沒有將資料插入資料庫表）。

這是我的代碼 jQuery AJAX 代碼：

$(".submit").click(function(){
var request;
    //moj kod 
    event.preventDefault();

// Abort any pending request
if (request) {
    request.abort();
}
// setup some local variables
var $form = $(this);

// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");

// Serialize the data in the form
var serializedData = $form.serialize();

// Let's disable the inputs for the duration of the Ajax request.
// Note: we disable elements AFTER the form data has been serialized.
// Disabled form elements will not be serialized.
$inputs.prop("disabled", true);

// Fire off the request to createticket.php
request = $.ajax({
    url: "createticket.php",
    type: "post",
    data: serializedData
});

// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
    // Log a message to the console
    console.log("Request uspje?no poslat!");
});

// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
    // Log the error to the console
    console.log(JSON.stringify(errorThrown));
    console.error(
        "Desio se sljedeci error: " 
        textStatus, errorThrown
    );
    
});

// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
    // Reenable the inputs
    $inputs.prop("disabled", false);
});

    //moj kod ->end
return false;
})

});

這是我的 createticket.php ：

<?php


$servername = "";
$username = "";
$password = "";
$dbname = "";
function clean_data($data)
{
    /* trim whitespace */
    $data = trim($data);
    $data = htmlspecialchars($data);
    return $data;
}
$make = isset($_POST['make']) ? $_POST['make'] : null;
$model = isset($_POST['model']) ? $_POST['model'] : null;
$godina = isset($_POST['godina']) ? $_POST['godina'] : null;
$engine = isset($_POST['engine']) ? $_POST['engine'] : null;
$termin = isset($_POST['termin']) ? $_POST['termin'] : null;
$usluga = isset($_POST['usluga']) ? $_POST['usluga'] : null;
$user_id = isset($_POST['user_id']) ? $_POST['user_id'] : null;
$request = "U obradi";

try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
    // set the PDO error mode to exception
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $sql = "INSERT INTO appointments (vl_id, make, model, godina, engine, opis, vrijeme, request) VALUES (?,?,?,?,?,?,?,?)";
    // use exec() because no results are returned
    $conn->exec($sql);
    echo "New record created successfully";
}
catch (PDOException $e) {
    echo $sql . "<br>" . $e->getMessage();
}

$conn = null;

?>

看起來一切都很好，當我重繪 頁面（phpMyAdmin）時，我的表中有 0 個資料，無論我收到訊息說一切都很好，從這一點開始我不知道我做錯了什么。我使用了這樣的 clean_data 函式，$godina = isset($_POST['godina']) ? clean_data($_POST['godina']) : null;因為我認為它有問題，但即使洗掉它，我的資料仍然沒有插入。

uj5u.com熱心網友回復：

我建議您使用準備好的陳述句并使用執行系結值。將您的查詢更改為這樣的內容

try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
    // set the PDO error mode to exception
    $sql = "INSERT INTO appointments (make, model, godina, engine, opis, vrijeme, request) VALUES (?,?,?,?,?,?,?)";
    $stmt = $conn->prepare($sql);
    $stmt->execute([$make, $model, $godina, $engine, $opis, $vrijeme, $request]);
    echo "New record created successfully";
}
catch (PDOException $e) {
    echo $sql . "<br>" . $e->getMessage();
}

標籤：php jQuery 阿贾克斯

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