成功通過 Ajax 插入條目后，我想查看同一條目的 ID 和 url 是什么，并在不重繪 頁面的情況下將其顯示在模式視窗中
有沒有辦法從成功中獲取這些資料：函式（回應）{}？這是我必須使用 ajax 創建一個新條目的代碼，它完美地作業：
<script>
$("#enquiry_email_form").on("submit", function (event) {
event.preventDefault();
var form= $(this);
var ajaxurl = form.data("url");
var detail_info = {
post_title: form.find("#post_title").val(),
post_description: form.find("#post_description").val()
}
if(detail_info.post_title === "" || detail_info.post_description === "") {
alert("Fields cannot be blank");
return;
}
$.ajax({
url: ajaxurl,
type: 'POST',
data: {
post_details : detail_info,
action: 'save_post_details_form' // this is going to be used inside wordpress functions.php// *esto se utilizará dentro de las functions.php*
},
error: function(error) {
alert("Insert Failed" error);
},
success: function(response) {
modal.style.display = "block"; * abre la ventana modal*
body.style.position = "static";
body.style.height = "100%";
body.style.overflow = "hidden";
}
});
})
</script>
<button id="btnModal">Abrir modal</button>
<div id="tvesModal" class="modalContainer">
<div class="modal-content">
<span class="close">×</span> <h2>Modal</h2> * Ventana modal mostrar le url y ID generado *
<p><?php ***echo $title_post, $url, $ID*** ?></p>
</div>
</div>
存檔功能.php
function save_enquiry_form_action() {
$post_title = $_POST['post_details']['post_title'];
$post_description = $_POST['post_details']['post_description'];
$args = [
'post_title'=> $post_title,
'post_content'=>$post_description,
'post_status'=> 'publish',
'post_type'=> 'post',
'show_in_rest' => true,
'post_date'=> get_the_date()
];
$is_post_inserted = wp_insert_post($args);
if($is_post_inserted) {
return "success";
} else {
return "failed";
}
}
uj5u.com熱心網友回復：
當您使用
轉載請註明出處，本文鏈接：https://www.uj5u.com/houduan/370786.html