我創建了一些 ajax 上傳圖像并將其加載到頁面中。它會在頂角創建一個帶有 X 按鈕的圖像，我正在嘗試獲取它，因此當單擊此按鈕時，我會運行另一段 php 代碼，將刪除正確的圖像并重新加載圖像。我無法讓我的 ajax 代碼來獲取 php 代碼，我不知道為什么。任何指出都會非常有幫助。我發現 dymanically 創建的 elemets 不會被拾取所以不得不將我的 ajax 代碼更改為$("body").on("click", "#deleteform button", function(e){所以我達到了這一點，但它仍然沒有獲取我的 php 代碼，我不知道為什么。任何指針都會非常有幫助AJAX JS：$(document).ready(function(){ $("#uploadForm").on('submit',function(e) { e.preventDefault(); $.ajax({ url: "include_advert/advert_new_gun_add_image.inc.php", type: "POST", data: new FormData(this), contentType: false, cache: false, processData:false, success: function(data) { $("#targetLayer").html(data); }, error: function() { } }); }); }); $(document).ready(function(){ $('button.deleteimage').on('click', function(){ var image_id = parseInt($(this).parent().attr('id').replace('deleteform', '')); console.log(image_id); // You can comment out this. Used for debugging. e.preventDefault(); $.ajax({ url: "include_advert/advert_new_gun_delete_image.php", type: "POST", data: {image_id: image_id}, contentType: false, cache: false, processData:false, success: function(data) { $("#targetLayer"+image_id).html(data); // targetLayer is dynamic and is different for each record }, error: function(xhr, status, error) { alert(xhr.responseText); } }); }); });
3 回答
POPMUISE
TA貢獻1765條經驗 獲得超5個贊
$(document).ready(function(){
$('button.delete-button').on('click', function(){
var image_id = parseInt($(this).parent().attr('id').replace('deleteform', ''));
console.log(image_id); // You can comment out this. Used for debugging.
e.preventDefault();
$.ajax({
url: "include_advert/advert_new_gun_delete_image.php",
type: "POST",
data: {image_id: image_id},
contentType: false,
cache: false,
processData:false,
success: function(data)
{
$("#targetLayer"+image_id).html(data); // targetLayer is dynamic and is different for each record
},
error: function(xhr, status, error) {
alert(xhr.responseText);
}
});
});
});
include 'advert_new_gun_save_image_script.inc.php';
include 'advert_new_dropdown_populate/advert_new_gun_image_populate.php';
$imagecount = 0;
echo ('<div class=row sm>');
foreach ($getadvertimages as $getadvertimages_row) {
echo ( '<div class="image-area" >
<form id="deleteform'.$getadvertimages_row['image_id'].'" method = "POST" action ="include_advert/advert_new_gun_delete_image.php" >
<img src="'. $getadvertimages_row['image_src'] . '" alt="Preview">
<button type="button" name="deleteimage" value="" class="remove-image delete-button" style="display: inline;" >X</button>
</form>
</div>');
}
echo ('</div>');
同樣，您可以使用 image_id 值使“targetLayer”動態化，就像我使用表單的屬性 id deleteform 所做的那樣。
慕碼人2483693
TA貢獻1860條經驗 獲得超9個贊
上面的答案幾乎就在那里，但我不得不將其更改 $('button.delete-button').on('click', function(){為$("body").on("click", "#deleteimage", function(e){
我還刪除了：contentType: false, cache: false,processData:false,
感謝 Ghulam 朝著正確的方向努力
$(document).ready(function(){
$("body").on("click", "#deleteimage", function(e){
var image_id = parseInt($(this).parent().attr('id').replace('deleteform', ''));
console.log(image_id); // You can comment out this. Used for debugging.
e.preventDefault();
$.ajax({
url: "include_advert/advert_new_gun_delete_image.php",
type: "POST",
data: {image_id: image_id },
success: function(data)
{
$('#imageareadiv').hide();
$("#targetLayer").html(data); // targetLayer is dynamic and is different for each record
},
error: function(xhr, status, error) {
alert(xhr.responseText);
}
});
});
});
翻翻過去那場雪
TA貢獻2065條經驗 獲得超14個贊
$(document).delegate('#deleteform button', 'click', function (e) {代替
$("body").on("click", "#deleteform button", function(e){- 3 回答
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