我决定在以前的网站的基础上开发一个新网站。 一切在我当前的网站上都可以正常运行,但是我无法理解为什么在新网站上看不到用户个人资料图片。
我有以下代码:
functions.php
function get_inst() {
$username = $_POST["login"];
$response = file_get_contents("https://www.instagram.com/$username/?__a=1");
$acccount = json_decode($response, true);
echo $acccount['graphql']['user']['profile_pic_url'];
die();
}
add_action( 'wp_ajax_nopriv_get_inst', 'get_inst' );
add_action( 'wp_ajax_get_inst', 'get_inst' );
custom.js
$('.buy .select_package .next').click(function(e){
e.preventDefault()
var current_price = $('.buy .select_package .nice-select .current .price').data("price");
let package = $('.buy .select_package .nice-select .current').html()
$('.buy .change_package .price').html(package)
//let packagePrice = $('.buy .select_package .nice-select .current .price').text()
let userName = $('.buy .select_package .user_name').val()
$('.buy .change_package .user_name').text(userName)
var price_we = $('.buy .payment .total_price .price').text('$'+cal_sell_price)
//alert(price_we);
$('.buy .select_package').hide()
$('.buy .change_package').fadeIn(300)
$('.buy .payment').removeClass('inactive')
$("input[name='price']").val(cal_sell_price);
$("input[name='count']").val($('.buy .select_package .nice-select .current .price').data("count"));
$("input[name='login']").val(userName);
var dataForAjax = "action=get_inst&login="+userName;
var addressForAjax = myajax.url;
$.ajax({
type:'POST',
data:dataForAjax,
url:addressForAjax,
success: function(response){
$(".buy .change_package .icon img").attr("src", response);
}
});
,这就是我显示化身的方式:
<div class="icon">
<img data-src="<?php bloginfo('template_url'); ?>/assets/img/placeholder.png" alt="ig profile image" class="lozad">
</div>
UPD。
控制台中没有错误。但是在代码中,我看到以下内容:
<img data-src="https://mywebsite.com/wp-content/themesmywebsite/assets/img/logo_mini.png" alt="ig profile image" class="lozad loaded" src="" data-loaded="true">