Question

当用户单击我的提交按钮时，我想向基于表单信息创建的特殊URL发送POST请求。我还希望它在WordPress MYSQL数据库中插入一行。

我附加到admin-ajax.php的末尾：

function add_query_db_callback(){

global $wpdb;

    $id_instagram = $_POST['id_instagram'];
    $table_name = 'wp_insta_email';

        $data_array = array(
            'id_instagram' => $id_instagram
        );

        $rowResult = $wpdb->insert($table_name, $data_array, $format=NULL);

echo $_POST['data'];

if (!$rowResult) {
    echo "FAILED TO UPDATE";
} else {
    $rowResult;
    echo "WILL UPDATE SUCCESSFULLY - CALL RESULT FUNCTION";
};

wp_die();
die();

}`

和文件header.php中：

    <!-- Search_box --> 
<div class="s006">
    <form  method="post" id="form_id" onsubmit="return myFunction()">
        <fieldset>
            <div class="inner-form">
                <div class="input-field">
                    <button class="btn-search" type="button submit" value="Submit" name="BtnSubmit">
                        <svg xmlns="http://www.w3.org/2000/svg" width="24" height="24">
                            <path d="M15.5 14h-.79l-.28-.27C15.41 12.59 16 11.11 16 9.5 16 5.91 13.09 3 9.5 3S3 5.91 3 9.5 5.91 16 9.5 16c1.61 0 3.09-.59 4.23-1.57l.27.28v.79l5 4.99L20.49 19l-4.99-5zm-6 0C7.01 14 5 11.99 5 9.5S7.01 5 9.5 5 14 7.01 14 9.5 11.99 14 9.5 14z"></path>
                        </svg>
                    </button>
                    <input id="search" type="text" placeholder=" INPUT YOUR ID INSTAGRAM " value="" name="id_instagram"/>
                </div>
            </div>
        </fieldset>
    </form>
</div>

和

<script type="text/javascript">
function myFunction() {
    var act= "//app.neolyze.com/public/"+document.getElementById("search").value;
    var name = document.getElementById("search").value;
    document.getElementById("form_id").action = act;
    document.getElementById("form_id").submit();
    if (name == '') {
        alert("Please Fill All Fields");
    } else {
        // AJAX code to submit form.
        var ajaxData = {
            'action': 'add_query_db',
            'id_instagram': name
        }
        jQuery.ajax({
            type: "POST",
            url: '<?php echo admin_url('admin-ajax.php'); ?>',
            data: ajaxData,
            success: function( response ) {
                console.log("Data returned: " + response );
                $statusSelectCell.parent().css({"background-color": "#b3e6b3"});
                $statusSelectCell.parent().animate({backgroundColor: currentBackgroundColor}, 1200);
            },
            error: function() {
                alert("FAILED TO POST DATA!!");
            }

    });
    }
    return act;
}
</script>

但它不会保存到我的数据库中（仅传递到我的预期URL）。

Answer 1

为什么您要在JavaScript代码中调用document.getElementById（“ form_id”）。submit（）函数？实际上.submit（）函数会将您重定向到所需的url，并且不执行其下面的行。或者，如果您想重定向到特定页面，则需要在ajax成功部分的最底部添加代码

https://www.googleapis.com/blogger/v3/blogs/(blogId)/posts/search?key=(myKey)&q=(someSearchTerm)&pageToken=(stringOfNextPageToken)

希望这会对您有所帮助

Answer 2

我对此解决方案做出了回应 `

        global $wpdb;

        $table_name = 'wp_insta_email';
        $_POST['id_instagram']= str_replace(array('@'), '',$_POST['id_instagram']);
        $data_array = array(
            'id_instagram' => $_POST['id_instagram']
        );

        $rowResult = $wpdb->insert($table_name, $data_array, $format=NULL);

        $url ='Location: https://exampleUrl.com' . $_POST['id_instagram'].'/compatitor/';

        header($url); //for redirect
        die();
    }
            ?>`

如何在wordpress中用php写ajax？

2 个答案: