Query data after an Ajax insert

Question

After a successful insertion with Ajax of an entry I would like to see what the ID and url of that same entry is and show it in a modal window

Any way to get that data?

<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";
    }
}

Answer 1

The function wp_insert_post() returns the post ID on success or the value 0 on error.

Your PHP function should look like this :

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();
    }
}

Finally, in the Ajax return we get the information to send to the popup.

<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>