在使用Ajax成功插入一个条目后，我希望看到该条目的ID和url是什么，并在模式窗口中显示它
有没有办法得到这些数据？
<script>
ajax_url = "<?php echo admin_url(\'admin-ajax.php\'); ?>";
</script>
<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
},
error: function(error) {
alert("Insert Failed" + error);
},
success: function(response) {
modal.style.display = "block";
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>
<p><?php ***echo $title_post, $url, $ID*** ?></p>
</div>
</div>
function.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";
}
}
SO网友:FS-GSW
功能wp_insert_post() 成功时返回post ID，错误时返回值0。
您的PHP函数应该如下所示：
function save_enquiry_form_action() {
$args = [
// Your postdata...
];
$is_post_inserted = wp_insert_post($args);
if( !empty($is_post_inserted) ) {
wp_send_json_success([
\'post_id\' => $is_post_inserted,
\'post_title\' => get_the_title( $is_post_inserted ),
]);
} else {
wp_send_json_error();
}
}
最后，在Ajax返回中，我们得到要发送到弹出窗口的信息。
<script>
$.ajax({
[...] // everything before success function was good.
success: function(response) {
if ( response.success == true ) {
// in case of `wp_send_json_success()`
var post_id = response.data.post_id;
var post_title = response.data.post_title;
$(\'#tvesModal .modal-content p\').html(post_id + "<span>" + post_title + "</span>");
modal.style.display = "block";
body.style.position = "static";
body.style.height = "100%";
body.style.overflow = "hidden";
} else {
// in case of `wp_send_json_error()`
alert("Insert Failed");
}
}
});
</script>
