在ajax请求之后从php获取响应文本

Question

我想将数据添加到我的数据库，并从php访问数据库获取响应。

javascript代码：

var request = $.ajax({
type: "POST",
url: "nieuwDuel.php",
data: dataString,
success: function(response)
{
    var responseText = "onbekend";
    responseText = jQuery(response);
    document.getElementById("duelToegevoegd").innerHTML=responseText;
    $( "#openstaandeDuels" ).load( "getOpenstaandeDuels.php" );
}
});

因此，当一个名称无效或已经存在时，我可以将此消息添加到我的html中的span元素。

这是我的PHP代码：

<?php
if($_POST)
{
    // Create connection
    $con=mysqli_connect("localhost","root","root","websitedb");

    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $date = getdate();
    $date_time = $date['year'] . "-" . $date['mon'] . "-" . $date['mday'] . " " . $date['hours'] . ":" . $date['minutes'] . ":" . $date['seconds'];

    $uitdagerArray = mysqli_query($con,"SELECT id FROM deelnemers WHERE naam='" . $_POST['uitdager'] . "'");
    $uitgedaagdeArray = mysqli_query($con,"SELECT id FROM deelnemers WHERE naam='" . $_POST['uitgedaagde'] . "'");

    $uitdager = mysqli_fetch_array($uitdagerArray);
    $uitgedaagde = mysqli_fetch_array($uitgedaagdeArray);

    $uitdagerId = $uitdager['id'];
    $uitgedaagdeId = $uitgedaagde['id'];

    $sql="INSERT INTO duels (uitdager, uitgedaagde, aanmaakdatum, gespeeld) VALUES ('" . $uitdagerId . "','" . $uitgedaagdeId . "','" . $date_time . "','" .  "0" . "')";

    if(!mysqli_query($con,$sql)) {
        echo mysql_errno() . ": " . mysql_error() . "\n";
    }
    else {
        echo "Duel Toegevoegd";
    }
    mysqli_close($con);
}
?>

那么我的php代码中echo的文本是否可能会从succes函数传递给'response'？

编辑：

这是我的HTML代码

<div id="nieuwDuel">
            <h2>Nieuw Duel</h2>
            <form name="form" method="post">
                Uitdager: <input type="text" id="naamUitdager" placeholder="naam uitdager">
                Uitgedaagde: <input type="text" id="naamUitgedaagde" placeholder="naam uitgedaagde">
                <input type="submit" class="nieuwDuelToevoegen">
            </form>
            <span id="duelToegevoegd" style="display:none"></span>
            <a href="index.php#home">Home</a>
        </div>

Answer 1

是的，它应该工作。 尝试将JS更改为

var request = $.ajax({
    type: "POST",
    url: "nieuwDuel.php",
    data: dataString,
    success: function(response)
    {
        console.log(response);//This will output the response you are getting to the console so you can check it
        $("#duelToegevoegd").html(response);
        $( "#openstaandeDuels" ).load( "getOpenstaandeDuels.php" );
    }
});

此外，我不会回应你回到客户端的任何错误

echo "Failed to connect to MySQL: " . mysqli_connect_error();

虽然这取决于用户是谁，但您不希望任何未经授权的人看到这些错误

Answer 2

我假设您的PHP代码是正确的。 那么你需要改变的是你的jQuery代码。

只需更改以下代码：

responseText = jQuery(response);
document.getElementById("duelToegevoegd").innerHTML=responseText;

如下：

responseText = response;
$('#duelToegevoegd').html(responseText);

我不知道你要做什么

$( "#openstaandeDuels" ).load( "getOpenstaandeDuels.php" );

所以，我现在就把它留下来

希望能帮助到你