在ajax請求後從php獲取響應文本

Question

我想將數據添加到我的數據庫，並取回訪問數據庫的php響應。

javascript代碼：

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"); 
} 
}); 

所以當一個名稱是無效或已經存在，我可以在我的html添加此消息span元素。

這是我的PHP代碼：

<?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); 
} 
?> 

那麼，這可能是從我的PHP代碼回聲文本將被傳遞到來自更迭功能的「迴應」？

編輯：

這是我的html代碼

<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

我假設你的PHP代碼是正確的。那麼你需要改變的是你的jQuery代碼。

只是改變了下面的代碼：

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

到以下幾點：

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

我不知道你會用

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

那麼該怎麼做，我現在把它排除在外

希望它有幫助

Answer 2

是它應該工作。試着改變你的JS到

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"); 
    } 
}); 

而且雖然它取決於用戶是誰，我不會迴應了，你回來給客戶

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

任何錯誤，但你不希望任何人未經授權看到這些錯誤