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
我需要帮助检查是否存在行.我收到"电子邮件不再存在publisher@example.com".
有没有更好的方法来检查mysqli是否存在行？
email no longer exists publisher@example.com
我尝试使用php登录,但是我得到了这个错误:Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given,我做错了什么？
register.php
<!doctype html>
<html lang"fi">
<head>
<link rel="icon" type='image/png' href='images/logo.png'>
<title>
asd
</title>
<link href="css/styles.css" type="text/css" rel="stylesheet">
</head>
<body>
<!--reg alkaa-->
<form action="register.php" method="post">
<p><input type="text" name="username" placeholder="Username">
<p><input type="email" name="email" placeholder="Email">
<p><input type="password" name="pass" placeholder="Password">
<p><input type="password" name="pass1" placeholder="Password">
<p><input type="submit" name="submit" value="Register">
</form>
<?php
if(isset($_POST['submit']))
{
$username = mysqli_real_escape_string($_POST['username']);
$pass = mysqli_real_escape_string($_POST['pass']);
$pass1 = mysqli_real_escape_string($_POST['pass1']);
$email = mysqli_real_escape_string($_POST['email']);
if($username && $pass && $pass1 && $email)
{
if($pass==$pass1)
{ …此代码出错:
致命错误:在第42行的C:\ Users\fel\VertrigoServ\www\login\validation.php中的非对象上调用成员函数prepare()
码:
function repetirDados($email) {
if(!empty($_POST['email'])) {
$query = "SELECT email FROM users WHERE email = ?";
$stmt = $pdo->prepare($query); // error line: line 42
$email = mysql_real_escape_string($_POST['email']);
$stmt->bindValue(1, $email);
$ok = $stmt->execute();
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);
if ($results == 0) {
return true;
} else {
echo '<h1>something</h1>';
return false;
}
}
}
可能的原因是什么？另一个问题,相当于mysql_num_rows什么？对不起,我是pdo的新手
码-
$q="SELECT * FROM tbl_quevote WHERE que_id = '".$qid."' and voteby='".$uid."'";
$result = $mysqli->query($q) or die(mysqli_error($mysqli));
$num_rows = mysql_num_rows($result);
echo $num_rows;
错误:
警告:mysql_num_rows()期望参数1是资源,第13行的C:\ xampp\htdocs\shizin\voting.php中给出的对象
how to check result is empty?
嘿伙计们,当我尝试运行此代码时,我收到了上述警告:
$mysqli=new mysqli("localhost", "***", "***","***") or die(mysql_error());
function checklogin($username, $password){
global $mysqli;
$result = $mysqli->prepare("SELECT * FROM users WHERE username = ?");
$result->bind_param("s", $username);
$result->execute();
if($result != false){
$dbArray=mysql_fetch_array($result);
我的代码中出现此错误,我不知道如何解决我的代码:
<?php
session_start();
include_once"connect_to_mysql.php";
$db_host = "localhost";
// Place the username for the MySQL database here
$db_username = "root";
// Place the password for the MySQL database here
$db_pass = "****";
// Place the name for the MySQL database here
$db_name = "mrmagicadam";
// Run the actual connection here
$myConnection= mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("mrmagicadam") or die ("no database");
$sqlCommand="SELECT id, linklabel FROM pages ORDER BY pageorder ASC";
$query=mysqli_query($myConnection, $sqlCommand) or die(mysql_error());
$menuDisplay="";
while($row=mysql_fetch_array($query)) {
$pid=$row["id"]; …致命错误:未捕获错误:在C:\ xampp\htdocs\phoenixproject\register.php中调用未定义函数mysql_escape_string():16堆栈跟踪:#0 {main}抛出C:\ xampp\htdocs\phoenixproject\register.php在第16行
如何解决这个问题？
<?php
require("config.php");
?>
<?php
if(isset($_POST['submit'])){
$email1 = $_POST['email1'];
$email2 = $_POST['email2'];
$pass1 = $_POST['pass1'];
$pass2 = $_POST['pass2'];
if($email1 == $email2) {
if($pass1 == $pass2) {
//All good. Nastavi broo.
$name = mysql_escape_string($_POST['name']);
$lname = mysql_escape_string($_POST['lname']);
$uname = mysql_escape_string($_POST['uname']);
$email1 = mysql_escape_string($email1);
$email2 = mysql_escape_string($email2);
$pass1 = mysql_escape_string($pass1);
$pass2 = mysql_escape_string($pass2);
mysql_query("INSERT INTO `users` (`id`, `name`, `lname`, `uname`, `email`, `pass`) VALUES (NULL, '$name', '$lname', '$uname', '$email1', '$pass1')") or die (mysql_error());
}else{
echo "Sorry, your password is …当我执行下面的PHP代码时,我得到致命错误,我不知道如何解决它.
谢谢您的帮助
错误
PHP致命错误:未捕获错误:在/Applications/MAMP/htdocs/lprapp/config.php:23中调用未定义函数mysql_query()堆栈跟踪:在/ Applications/MAMP/htdocs/lprapp/config中引发#0 {main}.第23行的PHP
码
<?php
$user = 'root';
$password = 'root';
$db = 'inventory';
$host = 'localhost';
$port = 8888;
$link = mysqli_init();
$success = mysqli_real_connect(
$link,
$host,
$user,
$password,
$db,
$port
);
?>
<?php
$username = $_POST['username'];
$password = $_POST['password'];
$sql = mysql_query("SELECT * FROM login WHERE username = '".$_POST['username']."' and password = '".md5($_POST['password'])."'");
$row = mysql_num_rows($sql);
if($rom > 0 )
{
session_start();
$_SESSION['username'] = $_POST['username'];
$_SESSION['password'] = $_POST['password'];
echo "login done"; …我试图从我的数据库中选择数据并在Word(DOCX)文件中显示数据.
我正在使用两种类型的数据.
第一种类型是我想要显示一次的数据.喜欢客户名称.
第二种类型的数据是使用动态行脚本导入的数据.我的数据库中的数据如下所示:行:
-----------------------------------------
| internal_id | quantity | product_name |
| 1 | 1 | One |
| 1 | 5 | Two |
| 1 | 4 | Three |
| 1 | 2 | Four |
| 1 | 6 | Five |
-----------------------------------------
在我的Word模板中,我将这些行定义如下:
{name}
{address}
{lines}
{quantity}
{product_name}
{/lines}
当我执行我的脚本时,我在我的Word文件中获得以下数据:
Name
Address 123
{!404_DOCXTemplate_quantity}
{!404_DOCXTemplate_product_name}
有人知道为什么我的代码的多行部分不起作用？
这是我的PHP脚本,我选择数据并生成Word文件:
$result1 = mysqli_fetch_assoc($res1) ;
$link2 = mysqli_connect("localhost", "root", "pass", "db");
if($link2 === false){
die("ERROR: Could not connect. …