Question

我很努力地將一個叫做$numbers的數組傳遞給$.ajax，就像是1,2,4那樣。它沒有被識別，我不知道如何在下一頁獲取這些信息。這是我的代碼：

<?php 
    // $numbers = 1,2,4; 
     print 
     "<div class='col-lg-6 col-md-6 col-sm-6 col-xs-12 col-md-pull-6 col-sm-pull-6'>". 
         "<div class='form-group'>". 
          "<div class='col-md-12'><strong>Forename:</strong></div>". 
          "<div class='col-md-12'><input type='text' class='form-control' id='forename'></div>". 
         "</div>". 
         "<div class='form-group'>". 
          "<div class='col-md-12'><strong>Surname:</strong></div>". 
          "<div class='col-md-12'><input type='text' class='form-control' id='surname'></div>". 
         "</div>". 
          "<div class='form-group'>". 
           "<div class='col-xs-12 col-md-6'>". 
            "<button id='button' class='btn btn-success btn_add'>Add</button>". 
           "</div>". 
          "</div>". 
         "</div>". 
     "</div>"; 

     } 

?> 

<script type="text/javascript">  
    $(document).on("click", 'button.btn_add', function(){ 
     var forename = document.getElementById('forename').value; 
     var surname = document.getElementById('surname').value; 
     var number = <?php echo $numbers; ?>; 
     if((forename == null || forename == "") && (surname == null || surname == "")){ 
      alert("Please fill in all fields"); 
     } 
else { 
    $.ajax({ 
     url: adduserinfo.php, 
     method: 'post', 
     dataType: 'json', 
     data: 'forename=' + forename + '&surname=' + surname + '&numbers=' + number 
    }); 

</script> 

Answer 1

您需要傳遞對象而不是查詢字符串。

使用

如果$號碼被陣列然後使用

var number = <?php echo json_encode($numbers); ?>; // cant echo array 

如果$號碼被串接着使用

var number = "<?php echo $numbers; ?>"; // cant echo array 

AJAX調用

$.ajax({ 
    url: adduserinfo.php, 
    method: 'POST', 
    dataType: 'json', 
    data: { forename : forename, surname : surname, numbers : number } 
});