为什么一个人不应该使用mysql_*功能的技术原因是什么？(例如mysql_query(),mysql_connect()或mysql_real_escape_string())？
即使他们在我的网站上工作,为什么还要使用其他东西？
如果他们不在我的网站上工作,为什么我会收到错误
警告:mysql_connect():没有这样的文件或目录
我试图从MySQL表中选择数据,但我收到以下错误消息之一:
mysql_fetch_array()期望参数1是资源,给定布尔值
要么
mysqli_fetch_array()期望参数1为mysqli_result,给定布尔值
要么
在布尔/非对象上调用成员函数fetch_array()
这是我的代码:
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');
while($row = mysql_fetch_array($result)) {
echo $row['FirstName'];
}
这同样适用于代码
$result = mysqli_query($mysqli, 'SELECT ...');
// mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
while( $row=mysqli_fetch_array($result) ) {
...
和
$result = $mysqli->query($mysqli, 'SELECT ...');
// Call to a member function fetch_assoc() on a non-object
while( $row=$result->fetch_assoc($result) ) {
...
和
$result = $pdo->query('SELECT ...', PDO::FETCH_ASSOC);
// Invalid …嗨,我收到错误" Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/**/**/locate.php on line 16".
我已经仔细检查了一切,谷歌/ Stackoverflow搜索,无法找出它为什么这样做.不胜感激任何想法!
getdate.php
function getDeals($the_type) {
$result = mysql_query("
SELECT *
FROM deals
WHERE deal_type = '" . $the_type . "'
");
}
locate.php？类型=乐趣
$type = $_GET['type'];
include("getdata.php");
getDeals($type);
if (mysql_num_rows($result)) {
echo '<ul>';
while($row = mysql_fetch_array($result))
{
echo '<a href="deal.php?i=' . $row["i"] . '">';
echo '<li class="deal ' . $row["deal_type"] . 'deal">';
echo '<h3>' . $row["deal_title"] . '</h3>';
echo '</li>';
echo …我在下面的代码(在查询的末尾)收到以下错误消息:
警告:mysql_num_rows():提供的参数不是第28行的../view-ind-order.php中的有效MySQL结果资源
该脚本应该检索订单(从列出订单表中所有order_id行的页面),订单内容,订购用户和产品信息.我想我在哪里得到的错误是订单中有多个产品,但我不能确定我哪里出错了.(标题有会话启动命令)
<?php
$page_title = 'Order';
include ('./includes/header.html');
if ( (isset($_GET['id'])) && (is_numeric($_GET['id'])) )
{
$id = $_GET['id'];
} elseif ( (isset($_POST['id'])) && (is_numeric($_POST['id'])) )
{
$id = $_POST['id'];
} else {
echo 'This page has been accessed in error';
include ('./includes/header.html');
exit();
}
require_once ('mydatabase.php');
$query = "SELECT us.users_id, us.users_first_name, us.users_surname, us.users_business,
ord.order_id, ord.users_id, ord.total, ord.order_date,
oc.oc_id, oc.order_id, oc.products_id, oc.quantity, oc.price
prd.products_id, prd.products_name, prd.price
FROM users AS us, orders AS ord, order_contents AS oc, products AS …我对 PHP 很陌生，试图编写一个连接到 MySQL 数据库的脚本，并在每个标题下以列表格式简单地显示内容；
我的表包含一个 ID（自动增量）、FName、SName 和 EAddress 字段。
该数据库名为 iphonehe_MGFSales，用户名是 iphonehe_MGFSale - 我已将该用户添加到具有完全权限的数据库中。
我正在尝试使用带有此代码的 mysql 函数建立与数据库的连接；
mysql_connect ("localhost", "iphonehe_MGFSale", "xxxxxxx") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ("iphonehe_MGFSales");
我创建的表称为 MGFSales DB。我正在使用此代码尝试构建查询；
$query = mysql_query("SELECT * FROM MGFSales_DB");
最后，我尝试使用以下代码显示结果；
while ($row = mysql_fetch_array ($query)) {
echo "<br /> ID: " .$row['ID']. "<br /> First Name: ".$row['FName']. "<br /> Last Name: ".$row['LName']. "<br /> Email: ".$row['EAddress']. "<br />";
}
我已将文件 index.php 命名并上传到我的服务器，运行时出现以下错误“警告：mysql_fetch_array()：提供的参数不是 /home/iphonehe/public_html/pauldmorris.co.uk 中的有效 …
当我运行我的php页面时,我收到此错误并且不知道什么是错的,有人可以帮忙吗？如果有人需要更多信息,我会发布整个代码.
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in H:\Program Files\EasyPHP 2.0b1\www\test\info.php on line 16
<?PHP
$user_name = "root";
$password = "";
$database = "addressbook";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM tb_address_book";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
print $db_field['ID'] . "<BR>";
print $db_field['First_Name'] . "<BR>";
print $db_field['Surname'] . "<BR>";
print $db_field['Address'] . "<BR>";
}
mysql_close($db_handle);
}
else …if(mysql_num_rows($result))
{
echo "no match found!";
}
这是一个错误 -
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\Hosting\6448289\html\includes\getQuestion.php on line 72
我有下一堂课:
<?php
class DbHelper{
private $databaseURL;
private $databaseUName;
private $databasePWord;
private $databaseName;
private $nameOfDbWithWorkers;
private $connection;
function __construct($dbURL, $dbUserName, $dbPword, $dbName, $nameOfDbWithWorkers){
$this->databaseURL = $dbURL;
$this->databaseUName = $dbUserName;
$this->databasePWord = $dbPword;
$this->databaseName = $dbName;
$this->nameOfDbWithWorkers = $nameOfDbWithWorkers;
}
function setConnectionToDb(){
$this->connection = mysql_connect($this->databaseURL,$this->databaseUName,$this->databasePWord) OR DIE("can't connect to DB");
mysql_select_db($this->databaseName, $this->connection)or die ("Error while connecting to database");
}
function getUser($login, $pass){
echo "$login, $pass";
$query = "SELECT type FROM $this->nameOfDbWithWorkers WHERE login = '$login' and password = '$pass';";
$queryResult …我做了一次搜索,找不到任何可以帮助我的东西.
我希望你能提供帮助.
我想执行一个MySQL查询,在表中搜索符合两个条件的条目(type = green AND on = yes)
我看到:警告:mysql_fetch_array():提供的参数不是36行/link/to/my/file.php中有效的MySQL结果资源
以下是代码摘录(第36行):
`$green = "SELECT * FROM homepage_vars WHERE type = 'green' AND on = 'yes'";
$green = mysql_query($green);
$green = mysql_fetch_array($green);`
我的网络应用程序中有一个PHP警告.我有点像菜鸟.我想根据每个用户配置文件中的'adjusttimezone'整数在'更新'时间戳中添加时区偏移量？
这是代码:
$timezonequery = mysql_query('SELECT adjusttimezone FROM members WHERE member_id=' . $_SESSION["myid"]);
$timezone = (int)mysql_result('$timezonequery', 0);
$updatedquery = mysql_query('SELECT DATE_ADD(updated, INTERVAL ' . $timezone . ' HOUR) FROM projects WHERE project_id=' . $i);
它有效,但我担心这个警告:
警告:mysql_result():提供的参数不是第9行xxx.php中的有效MySQL结果资源
有人可以帮忙吗？谢谢
在db.php我有:
<?php
class connect {
private $host = "localhost";
private $user = "root";
private $pass = "";
private $database = "databasename";
private $connect = null;
function connect() {
$this->connect = mysql_connect($this->host, $this->user, $this->pass) or die("Can't connect database");
mysql_select_db($this->database, $this->connect);
}
function getData() {
$data = array();
$sql = 'Select * From test';
$query = mysql_query($sql);
while($row = mysql_fetch_assoc($query)) {
$data[] = array($row['id'], $row['name']);
}
return $data;
}
}
?>
在index.php我有:
<?php
include 'db.php';
$connect = new connect();
$connect->connect(); …我从PHP中获取MySQL数据库中的数据,但它给出了如下错误:
警告:mysql_fetch_assoc():提供的参数不是97行上的/home/content/i/h/u/ihus235/html/cs/emrapp/surveyList.php中的有效MySQL结果资源[]
以下是我用来选择数据的查询:
$query = mysql_query("SELECT * form survey_Profile where user_Id='".$user_id."' ");