如何從php發佈數據while循環與ajax

Question

我想創建一個朋友按鈕像fb.I有一個while循環回聲的名稱和一些其他數據我需要保存到我的數據庫中的每個用戶的表請求seperatly.My的問題是，我不知道如何傳遞他們的Ajax，因爲我有相同的文本框buttons.I我試圖通過他們與相同的ID，但這傳遞了3次相同的數據或只有他們之一。我試圖使用id =「something $ row [id]」來定義每個文本框和按鈕，但我不知道如何觸發jquery函數並將它們與ajax一起發佈。 a picture of what i am trying

我的PHP文件

echo "<div class='col-xs-12'> 
         <div class='row fixed-table'> 
          <div class='table-content'> 
           <table class='table table-striped text-muted' id='mytable'> 
            <tbody>"; 
             while($row = mysqli_fetch_array($requests)){ 
              $username=("SELECT * FROM users WHERE user_id=".$row['patient']); 
              $found=mysqli_fetch_array(mysqli_query($sql_link,$username)); 
              echo "<tr> 
               <td class='text-center'> 
                <p> 
                 $found[username] 
                <p>                
               </td> 
               <td class='text-center'>"; 
                $patient_username=$found['username']; 
                $patient_id=$found['user_id']; 
               echo"<input type='text' id='patient_id$row[id]' class='hidden' value='$patient_id'> 
                <input type='text' id='patient_username$row[id]' class='hidden' value='$patient_username'> 
                <input type='text' id='doctor_id$row[id]' class='hidden' value='$doctor_id'> 
                <input type='text' id='doctor_fname$row[id]' class='hidden' value='$doctor_fname'> 
                <input type='button' id='accept_btn$row[id]' class='btn btn-primary' value='Accept'>                                        
               </td> 
              </tr>"; 
             } 
            echo "</tbody> 
           </table> 
          </div> 
         </div> 
        </div>"; 

我的JS文件

function decision(){ 
     var patient_id = $('#patient_id').val(); 
     var patient_username = $('#patient_username').val(); 
     var doctor_id = $('#doctor_id').val(); 
     var doctor_fname = $('#doctor_fname').val(); 

      $.ajax({ 
       type:"post", 
       url:"php/add.php", 
       data: {"patient_id": patient_id,"patient_username": patient_username,"doctor_id": doctor_id,"doctor_fname": doctor_fname}, 
       success:function(data){ 
        $("#decision").html(data); 
       } 
      }); 

$('#accept_btn').click(function(){ 
     decision(); 
    }); 

Answer 1

我終於做到了。它花了我幾小時，但它的工作原理，我wanto謝謝大家的幫助，特別是伊戈爾伊利奇。 該解決方案

PHP

<tbody>"; 
             while($row = mysqli_fetch_array($requests)){ 
              $username=("SELECT * FROM users WHERE user_id=".$row['patient']); 
              $found=mysqli_fetch_array(mysqli_query($sql_link,$username)); 
              echo "<tr> 
               <td class='text-center'> 
                <p> 
                 $found[username] 
                <p>                
               </td> 
               <td class='text-center'>"; 
                $patient_username=$found['username']; 
                $patient_id=$found['user_id']; 
               echo"<buuton id='accept_btn$row[id]' class='btn btn-primary' data-patient_id='$patient_id' data-patient_username='$patient_username' data-doctor_id='$doctor_id' data-doctor_fname='$doctor_fname' 
               href='#' onclick='decision(this);'> 
               Accept</button> 
               </td> 
              </tr>"; 
             } 
            echo "</tbody> 

和js

function decision(p){ 

    var patient_id = $(p).data('patient_id'); 
    var patient_username = $(p).data('patient_username'); 
    var doctor_id = $(p).data('doctor_id') 
    var doctor_fname = $(p).data('doctor_fname'); 

     $.ajax({ 
      type:"post", 
      url:"php/add.php", 
      data: {"patient_id": patient_id,"patient_username": patient_username,"doctor_id": doctor_id,"doctor_fname": doctor_fname}, 
      success:function(data){ 
       $("#decision").html(data); 
      } 
     }); 

} 

Answer 2

這很容易與jQuery爲你甚至需要知道表單中的ID 。但是，首先您需要將所有輸入包裝到表單中，併爲表單分配一個ID併爲每個輸入分配一個名稱。

 var data = $("#myForm").serialize(); 
 
     $.post('myPHPFILEURL.php', data,function(retData){ 
 
      //CODE THAT HANDLES SERVER RESPONSE 
 
      
 
      
 
      });

  echo "<div class='col-xs-12'> 
 
           <div class='row fixed-table'> 
 
            <div class='table-content'> 
 
             <form id='myForm'> 
 
             <table class='table table-striped text-muted' id='mytable'> 
 
              <tbody>"; 
 
               while($row = mysqli_fetch_array($requests)){ 
 
                $username=("SELECT * FROM users WHERE user_id=".$row['patient']); 
 
                $found=mysqli_fetch_array(mysqli_query($sql_link,$username)); 
 
                echo "<tr> 
 
                 <td class='text-center'> 
 
                  <p> 
 
                   $found[username] 
 
                  <p>                
 
                 </td> 
 
                 <td class='text-center'>"; 
 
                  $patient_username=$found['username']; 
 
                  $patient_id=$found['user_id']; 
 
                 echo"<input type='text' name='patient_id$row[id]' id='patient_id$row[id]' class='hidden' value='$patient_id'> 
 
                  <input type='text' id='patient_username$row[id]' name='patient_username$row[id]' class='hidden' value='$patient_username'> 
 
                  <input type='text' id='doctor_id$row[id]' name='doctor_id$row[id]' class='hidden' value='$doctor_id'> 
 
                  <input type='text' id='doctor_fname$row[id]' name='doctor_fname$row[id]' class='hidden' value='$doctor_fname'> 
 
                  <input type='button' id='accept_btn$row[id]' name='accept_btn$row[id]' class='btn btn-primary' value='Accept'>                                        
 
                 </td> 
 
                </tr>"; 
 
               } 
 
              echo "</tbody> 
 
             </table> 
 
             </form> 
 
            </div> 
 
           </div> 
 
          </div>";

只需選擇您的形式和序列化（）的數據，然後通過新的序列化的字符串到您的AJAX功能的數據參數。

Answer 3

您可以添加data-id="userID" 屬性的輸入，而不是在ID有它的

<input type='button' id='accept_btn$row[id]' class='btn btn-primary' value='Accept'>

所以，你的投入會是這個樣子：

<input type='button' id='accept_btn' data-id='$row[id]' class='btn btn-primary' value='Accept'> 

然後在你的jQuery/javascript代碼使用像這樣：

$("#accept_btn").on("click", function(){ 
    var userID = $(this).data("id"); //this will get the value from data-id from the clicked button 
    //rest of your code here... 
}); 

您也可以爲所有輸入字段添加此項，並從中檢索數據。

$("#patient_username").data("id") 

希望這會有所幫助。你可以檢查這個樣本jsfiddle https://jsfiddle.net/Lz3k1he1/