车把:当#个前面的字符不一致时,抓取特定的子字符串

时间:2019-07-08 15:45:07

标签: handlebars.js

我正在使用把手,需要获取值的特定部分。

通常,我知道我可以只使用<?php ob_start(); include_once __DIR__.'/header2.php'; if (!$_SESSION['u_uid']) { echo "<meta http-equiv='refresh' content='0;url=index.php?donation=notlogin'>"; exit(); } else { include_once __DIR__.'/includes/dbh.php'; if ($_SERVER['REQUEST_METHOD'] != 'POST') { header("Location: index.php"); exit(); } $ch = curl_init(); curl_setopt($ch, CURLOPT_SSLVERSION, CURL_SSLVERSION_TLSv1_2); curl_setopt($ch, CURLOPT_URL, 'https://ipnpb.sandbox.paypal.com/cgi-bin/webscr'); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0); curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_POSTFIELDS, "cmd=_notify-validate&" . http_build_query($_POST)); $response = curl_exec($ch); curl_close($ch); if ($response == "VERIFIED" && $_POST['receiver_email'] === "admin@pianocourse101.com") { $cEmail = strip_tags($_POST['payer_email']); $firstname = strip_tags($_POST['first_name']); $lastname = strip_tags($_POST['last_name']); $price = strip_tags($_POST['mc_gross']); $currency = strip_tags($_POST['mc_currency']); $item = strip_tags($_POST['item_number']); $paymentStatus = strip_tags($_POST['payment_status']); if ($item == "Donation" && $currency == "USD" && $paymentStatus == "Completed" && $price == 100) { $sql = "INSERT INTO donation (user_email, firstname, lastname, amount) VALUES (?,?,?,?);"; $stmt = mysqli_stmt_init($conn); if(!mysqli_stmt_prepare($stmt, $sql)) { echo "SQL error"; } else { mysqli_stmt_bind_param($stmt, "sssi", $cEmail, $firstname, $lastname, $price); mysqli_stmt_execute($stmt); } } } } ?> ;但是,该值位于长字符串的末尾,并且其前的字符数将不会是一致的长度。

在下面的示例中,我试图在两种情况下都获取982。注意,参数“ referral_code”将始终相同。

substring

http://www.website.com/?referral_code=982

是否有任何Handlebars函数(或使用http://www.website.com/subpage/?referral_code=982的方法)可以使我从以上两个实例中获得substring

谢谢!

0 个答案:

没有答案