Question 379307 in Mysql Question posted in
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AJAX – insert data in mysql

The code below displays a message after typing a value into the field

Poprawnie zapisano: - but without this value

Also does not write anything to the database. Did I make a mistake somewhere?

HTML file

<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
</head>
<body>

        <input type="text" id="wartosc" placeholder="Wartość do zapisania">
        <button id="zapiszBtn">Zapisz do bazy danych</button>


        <script>
            $(document).ready(function(){
                $("#zapiszBtn").click(function(){
                    var wartosc = $("#wartosc").val();
                    $.ajax({
                        url: "zapisz.php",
                        type: "POST",
                        data: wartosc,
                        success: function(wartosc){
                            alert("Poprawnie zapisano: " + wartosc);
                        }
                    });
                });
            });
        </script>
</body>

ZAPISZ.php

<?php

$sname= "localhost";
$unmae= "root";
$password = "root";

$db_name = "vote2";

$conn = mysqli_connect($sname, $unmae, $password, $db_name);

if (!$conn) {
    echo "Połączenie się nie powiodło!";
}

// Uzyskanie wartości z AJAX
if(isset($_POST['wartosc'])){

    $wartosc = $_POST['wartosc'];

    // Wstawienie wartości do bazy danych
    $stmt = $pdo->prepare("INSERT INTO 'test' ('r1') VALUES ('$wartosc')");
    $result = mysql_query($stmt);


    if($result) {
        echo "Wartość została pomyślnie zapisana do bazy danych.";
        } else {
            echo "Wystąpił błąd podczas zapisywania wartości do bazy danych.";
        }
}

?>

I would like the value I enter in the field to be saved in the database

Tags:

2

Answers


  1. 0

    You’re using mysqli_connect() to establish a connection, but then you’re trying to prepare a statement using $pdo->prepare(). You should use either mysqli functions throughout or PDO functions throughout, but not mix them.

    <input type="text" id="wartosc" placeholder="Wartość do zapisania"/>
<button id="zapiszBtn">Zapisz do bazy danych</button>

<script src="https://code.jquery.com/jquery-3.6.0.min.js"></script>
<script>
$(document).ready(function(){
    $("#zapiszBtn").click(function(){
        var wartosc = $("#wartosc").val();
        $.ajax({
            url: "zapisz.php",
            type: "POST",
            data: { wartosc: wartosc }, // Send data as an object
            success: function(response){
                alert(response);
            }
        });
    });
});
</script>

    PHP (zapisz.php):

    <?php
$sname= "localhost";
$unmae= "root";
$password = "root";
$db_name = "vote2";

$conn = mysqli_connect($sname, $unmae, $password, $db_name);

if (!$conn) {
echo "Połączenie się nie powiodło!";
}

if(isset($_POST['wartosc'])){
$wartosc = $_POST['wartosc'];

// Prepare and execute statement
$stmt = $conn->prepare("INSERT INTO test (r1) VALUES (?)");
$stmt->bind_param("s", $wartosc);
if ($stmt->execute()) {
    echo "Wartość została pomyślnie zapisana do bazy danych.";
} else {
    echo "Wystąpił błąd podczas zapisywania wartości do bazy danych.";
}
$stmt->close();
}
$conn->close();
?>

    This code should properly insert the value into your database and display the appropriate success or error message.

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  2. 0

    You are not sending the data correctly. You need data: { wartosc: wartosc } otherwise $_POST["wartosc"] won’t find anything – the parameter you send must have a name, not just a value.

    Also you’re using mysqli_connect() to establish a connection database with a variable called $conn, but then you’re trying to prepare a statement using a variable $pdo which doesn’t exist, and then finally executing the query using mysql_query which ia function that (since PHP 7) also doesn’t exist (and never accepted a statement object as input when it did excist). I suggest you go back and look at tutorials and examples of this process a bit more carefully.

    All of this should have produced errors and warnings in your PHP code, which ought to be visible in the response you are viewing with alert – if they’re not, you should enable PHP error reporting while doing development. And you don’t need to check for mysqli errors manually – mysqli report mode should be enabled, to allow it to throw them automatically.

    Here’s an improved version of the code:

    <input type="text" id="wartosc" placeholder="Wartość do zapisania"/>
<button id="zapiszBtn">Zapisz do bazy danych</button>

<script src="https://code.jquery.com/jquery-3.6.0.min.js"></script>
<script>
$(document).ready(function(){
    $("#zapiszBtn").click(function(){
        var wartosc = $("#wartosc").val();
        $.ajax({
            url: "zapisz.php",
            type: "POST",
            data: { wartosc: wartosc }, // Send data as an object
            success: function(response){
                alert(response);
            }
        });
    });
});
</script>

    PHP (zapisz.php):

    <?php
$sname= "localhost";
$unmae= "root";
$password = "root";
$db_name = "vote2";

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = mysqli_connect($sname, $unmae, $password, $db_name);

if(isset($_POST['wartosc']))
{
  $wartosc = $_POST['wartosc'];

  // Prepare and execute statement
  $stmt = $conn->prepare("INSERT INTO test (r1) VALUES (?)");
  $stmt->execute([$wartosc]);

  echo "Wartość została pomyślnie zapisana do bazy danych.";
  $stmt->close();
}
?>

    This code should properly insert the value into your database and display the appropriate success or error message.

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