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我对JavaScript并不熟悉，我确定我遇到的问题有一个简单的解决方案，我只需要一些方向。调整预制URL缩短脚本
下面的脚本不是我写的，它是一个URL缩短脚本。当用户输入长URL并按提交时，脚本会创建一个随机的短URL，并在输入字段中显示短URL。我想知道的是javascript的哪一部分控制输入中短URL的显示？我想更改URL缩短后显示的内容。
page.php文件
<script src="script.js" type="text/javascript"></script>
<form action="#" id="form-add-url" class="profile" method="post" onsubmit="return add_url()">
<input type="text" id="urls-url" name="url" class="widefat-main" placeholder="Paste a link" tabindex="1" title="URL">
<input type="hidden" name="action" value="add_url">
<button type="submit" class="button-main" tabindex="3">Submit</button>
</form>
的script.js
function add_url() {
jQuery("#front-url .loading-dark").fadeIn(200);
jQuery.post(url_base+"ajax.php", jQuery("#form-add-url").serialize(),
function(return_data) {
jQuery("#front-url .loading-dark").fadeOut(200);
data = jQuery.parseJSON(return_data);
var status = data.status;
if (status == "OK") {
jQuery("#urls-url").val(data.url);
} else if (status == "OK2") {
jQuery("#search_query").val("");
jQuery("#page_number").val("");
reload_urls("", 1);
jQuery("#urls-url").val(data.url);
} else if (status == "ERROR") {
show_notification("error", data.message, 3000);
} else {
show_notification("error", "Internal error. Please contact administrator.", 3000);
}
}
);
return false;
}
我的PHP：
case 'add_url':
if ($options['only_registered'] == 'yes' && !$active_user)
{
$return_object = new stdClass();
$return_object->message = 'URL shortening is available for registerd users only.';
$return_object->status = 'ERROR';
echo json_encode($return_object);
exit;
}
$url = trim(stripslashes($_POST['url']));
if (substr(strtolower($url) , 0, 7) != "http://" && substr(strtolower($url) , 0, 8) != "https://") $url = 'http://' . $url;
$error = '';
if ($url == '')
{
$error = 'Hey, seems you forgot to paste a link.';
}
else
if (!preg_match('|^http(s)?://[a-z0-9-]+(.[a-z0-9-]+)*(:[0-9]+)?(/.*)?$|i', $url))
{
$error = 'Are you sure you submitted the correct URL?';
}
else
if (sizeof($url) > 255)
{
$error = 'Hey, seems URL is too long.';
}
if (!empty($error))
{
$return_object = new stdClass();
$return_object->message = $error;
$return_object->status = 'ERROR';
echo json_encode($return_object);
exit;
}
if (substr(strtolower($url) , 0, strlen($url_base)) == strtolower($url_base))
{
$return_object = new stdClass();
$return_object->message = 'Hey. Seems this URL is short enough. ;-)';
$return_object->status = 'ERROR';
echo json_encode($return_object);
exit;
}
if ($active_user) $user_id = $active_user['id'];
else $user_id = 0;
$url_details = $icdb->get_row("SELECT * FROM " . $icdb->prefix . "urls WHERE url = '" . mysql_real_escape_string($url) . "' AND deleted = '0' AND user_id = '" . $user_id . "'");
if ($url_details)
{
$icdb->query("UPDATE " . $icdb->prefix . "urls SET created = '" . time() . "' WHERE id = '" . $url_details['id'] . "'");
$url_code = $url_details['url_code'];
}
else
{
$icdb->query("INSERT INTO " . $icdb->prefix . "urls (user_id, url, url_code, redirects, created, blocked, deleted, short) VALUES ('" . $user_id . "', '" . mysql_real_escape_string($url) . "', '', '0', '" . time() . "', '0', '0', '" . $_POST[short] . "')");
$url_code = url_code($icdb->insert_id);
$icdb->query("UPDATE " . $icdb->prefix . "urls SET url_code = '" . $url_code . "' WHERE id = '" . $icdb->insert_id . "'");
}
$htaccess = url_rewrite();
$return_object = new stdClass();
if ($active_user)
{
$return_object->status = 'OK2';
}
else $return_object->status = 'OK';
$return_object->url = $url_base . ($htaccess ? '' : '?u=') . $url_code;
echo json_encode($return_object);
exit;
break;
你知道吗，你只需要以其长格式编写'jQuery'一次？通过将代码封装在'（function（$）{....}）（jQuery）;'中，无论是否使用了noConflict，都可以使用'$'。此外，**请更新您的问题标题是有意义的** – ThiefMaster
@ThiefMaster - 请注意，它不是一个DOM就绪处理程序，这将是'jQuery（function（$）{....}）;' – adeneo
我认为你正在寻找的代码的一部分是在服务器端，由Ajax返回...你有PHP代码？ – Sergio