没有实际错误,但是当非买方或卖方的用户访问该页面时,它仍然显示错误消息.是否可以抑制错误消息？
直接链接到图像
alt text http://i37.tinypic.com/2m5ijvq.jpg
替代文字http://i36.tinypic.com/148kads.jpg
//transaction id
$transactionid = $_GET['id'];
//Retrieve info about transaction
$query = "SELECT ads.*, feedback.*, transactions.* FROM (ads INNER JOIN transactions ON ads.id=transactions.ad_id) INNER JOIN feedback ON transactions.id=feedback.transaction_id WHERE transaction_id = '$transactionid'";
$data = mysqli_query($dbc, $query);
$row = mysqli_fetch_array($data);
$seller = $row['seller'];
$buyer = $row['buyer'];
//check if user is buyer or seller
if ($_SESSION['user_id'] == $seller) {
$query = "SELECT * FROM feedback WHERE transaction_id = '$transactionid' AND seller_comment IS NULL";
$data1 = …嗨,
我上个月开始学习PHP,现在我知道如何使用基本的php创建网站.我想进一步了解,但我不知道转向哪种方式.
我听说过codeigniter,zend,cakePHP等框架,但不知道它们的用途.它们应该是我的下一步吗？我目前正在使用NetBeans.
我看看其他php脚本以及它们的直接结构如何整洁并分类到文件夹中,而我的所有文件都在根目录下.我如何学习正确创建脚本结构？
我还想学习如何使用模板,框架有助于此吗？
另外,我在大学学习Java.OO PHP应该是我的下一步吗？
基本上,从初学者到掌握者的路线图是什么？
<?php
$userid = $_SESSION['user_id'];
$price = $ad['price'];
$owner = $ad['owner']; //owner of advertisement
//make a bid for this advertisement
$query = "INSERT INTO bids (id, ad, bidder, bid, bidwhen, owner, quantity)
VALUES (NULL, '$adid', '$userid','$price', now(), '$owner', 1)";
$bidData = mysqli_query($dbc, $query);
$message = $_POST['message']; //message to owner
if ($message != "") { //if message box is not empty insert comment
$title = $ad['title'];
$bidid = mysql_insert_id($bidData); //Line 123 get last id of bid insert and put it into message …$teams = array(1, 2, 3, 4, 5, 6, 7, 8);
$game1 = array(2, 4, 6, 8);
$game2 = array();
如果teams[x]不在game1那么插入game2
for($i = 0; $i < count($teams); $i++){
for($j = 0; $j < count($game1); $j++){
if($teams[$i] == $game1[$j]){
break;
} else {
array_push($game2, $teams[$i]);
}
}
}
for ($i = 0; $i < count($game2); $i++) {
echo $game2[$i];
echo ", ";
}
我期待结果是:
1, 3, 5, 7,
但是,我得到:
1, 1, 1, 1, 3, 3, 3, 3, …这是我的尝试.我能想到的最好的方法就是在chrome扩展程序中打开链接.我希望它能在新标签页中打开.
<!DOCTYPE html>
<html>
<head>
<title></title>
<link href="css/style.css" rel="stylesheet"/>
<script type="text/javascript">
var items = [];
var background;
function init() {
background = chrome.extension.getBackgroundPage();
items = background.items;
createItemTable();
}
function createItemTable() {
var content = document.getElementById("content");
var list = document.createElement("div");
list.setAttribute("class", "list");
content.appendChild(list);
for (x in items) {
var item = items[x];
var link = document.createElement("a"); // create the link
link.setAttribute('href', "'" + item["link"] + "'"); // set link path
link.setAttribute("onclick", "openTab('" + item["link"] + "');");
var titleNode = document.createElement("div");
titleNode.setAttribute("class", …$goutte = new GoutteClient();
$crawler = $goutte->request('GET', 'https://www.website.com');
$reviewContent = $crawler->filter('.review-content');
$rows = $reviewContent->filter('.row');
foreach ($rows as $row) {
$col1 = $row->filter('.col-md-3');
$col2 = $row->filter('.col-md-9');
}
给出错误$col1
我可以使用这个让它工作，但你不能使用，break因为它不是真正的for loop
$crawler->filter('.row')->each(function (Crawler $row, $i) {
$col1 = $row->filter('.col-md-3');
$col2 = $row->filter('.col-md-9');
...
...
}
php ×5
arrays ×1
dom ×1
domcrawler ×1
for-loop ×1
goutte ×1
javascript ×1
loops ×1
nested-loops ×1
sql ×1