我是新学习php,但我一开始就陷入困境.无论我在PHP中尝试什么方法,但我无法在输出中显示变量数据..下面是从互联网上复制粘贴的代码,即使在我看不到$ username的输出...请帮助我!
<html>
<body>
<h1>Hi Jacob</h1>
<h3>Demonstrates using a variable</h3>
<?php
$userName = "Jacob";
print Hi, $userName;
?>
</body>
</html>
在输出中它" Hi, $username "会显示出来.不是变量的数据
我有一个简单的 php 脚本，它将 POST 的内容存储到一个文本文件中。data.txt 看起来像这样：
Something - Line1
Something - Line2
Something - Line3
Something - Line4
我试图找出将每一行设置为变量的最佳方法，例如
$line1 = "Something - Line1";
$line2 = "Something - Line2";
$line3 = "Something - Line3";
$line4 = "Something - Line4;"
所以我可以用它做进一步的处理。请注意，行数可能会发生变化。有任何想法吗？:-)
谢谢！
嘿家伙我是整个数据库场景的新手,并尝试执行一个相对简单的任务,但显然我做错了什么.每次我尝试执行此语句时,都会收到1064错误,告诉我语法错误或服务器版本太旧.SQL Server版本是5.1.x,我正在运行PHP5.
这是我的代码:
$query = "INSERT INTO `cut_log` (`driver`, `date1`, `time`, `cut`, `flood`, `notes`) VALUES ($driver, $date, $time, $cut, $flood, $notes)";
$result = $mysqli->query($query);
if($result) {
echo "success";
} else {
echo "" . $mysqli->errno . $mysqli->error;
}
我试图通过添加已存在的原始数字值来更新我的数据库中的字段.我有一个系统,员工可以登录并更新普通用户的余额.目前我有一个测试用户和工作人员.用户余额设置为100.我有以下代码:
<?php
if(isset($_POST['search'])){
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$result = $mysqli->query( "SELECT * FROM Users WHERE Username ='$searchq'");
if ($result){
//fetch result set as object and output HTML
if($obj = $result->fetch_object())
{
echo '<div class="booksearched">';
echo '<form method="POST" id = "books" action="">';
echo '<div class="book-content"><h3>Student Username: '.$obj->Username.'</h3>';
echo '<br>';
echo '<div class="book-content"><i>First Name: <b>'.$obj->FirstName.'</b></i></div>';
echo '<div class="book-desc"><i>Last Name:<b> '.$obj->LastName.'</b></i></div>';
echo '<br>';
echo '<div class="book-qty"> Current Balance<b> '.$obj->Balance.'</b></div>';
echo 'New Balance: <input type="number" name="newBalance" value = "1" min = "1" …我昨天开始学习PDO,我以为我把它弄下来但是在尝试提交一个简单的表格时我又遇到了一个错误.我在w3schools上使用这个例子,但是有一个表单输入.
的index.php
<form action="submit.php" method="post">
<input type="text" id="name" placeholder"Enter Your Name">
<button type="submit">Submit</button>
</form>
这是我处理表单的页面:
submit.php
<?php
$servername = "localhost";
$username = "testuser";
$password = "testpassword";
$dbname = "testdb";
$nickname = $_POST['name'];
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
//Set PDO Error Mode to Exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "INSERT INTO test (name)
VALUES (:nickname)";
// use exec() because no results are returned
$conn->exec($sql);
echo "New record created successfully";
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage(); …我正在为一个招聘机构制作一个原型网站,我在将任何数据上传到为这个原型系统创建的数据库时遇到了问题.这是注册新用户页面,它不会将html文本字段中的任何信息发送到数据库.
这是我的代码 -
<?php
require (__DIR__.'/connections/connections.php');
if(isset($_POST['Register'])){
session_start();
$Title = $_POST['Consultant_Title'];
$Fname = $_POST['Consultant_First_Name'];
$Lname = $_POST['Consultant_Last_Name'];
$Cbranch = $_POST['Consultant_Branch'];
$Email = $_POST['Email'];
$Username = $_POST['Username'];
$Password = $_POST['Password'];
$StorePassword = password_hash($Password, PASSWORD_BCRYPT, array('cost' => 10));
$spl = $con->query("INSERT INTO consultant_details (Title, Fname, Lname, Cbranch )Values('{$Title}', '{$Fname}', '{$Lname}', '{$Cbranch}')");
$spl = $con->query("INSERT INTO users ( Email, Username, Password )Values('{$Title}', '{$Email}', '{$Username}', '{$StorePassword}')");
header('location: Login.php');
}
?>
Html -
<h2>Staff Register </h2>
<div class="clr"></div>
<form action="" method="post" name="Registerform" id="Registerform">
<div class="FormElement"><select …目前我想在用户登录系统时从两个表中获取两个不同的数据.权利表是"user","user_staff"和"user_group".但是当用户输入其用户名和密码并提交它时,提示"致命错误:在第26行的C:\ xampp\htdocs\auditsystem\index.php中的非对象上调用成员函数fetch_array()"
以下是代码:
if($username!= "" && $password != "")
{
//INNER JOIN user_group_module_role ON user.user_group_module_role_id = user_group_module_role.id
$result = $db->query("SELECT * FROM user
INNER JOIN user_staff ON user.user_staff_id = user_staff.id
WHERE username = '$username' AND password = '$password'");
if($result->num_rows == 1)
{
$validate = $result->fetch_assoc();
$query1 = "SELECT * FROM usergroup WHERE id = $validate[user_group_id]";
$result1 = $db->query($query1);
$row1 = $result1->fetch_array();
//change here for the authority
$_SESSION['user_staff'] = $validate['displayname'];
$_SESSION['usergroup'] = $row1['user_group_type'];
echo "<script language='javascript'>window.location='panel.php'</script>";
}
else
{
echo …我用PHP和MySQL数据库创建了票务系统.
在视图故障单页面上,我显示了数据库中的每个故障单,其中包括每个故障单的优先级.优先级设置为低,正常或高.
当数据传递到我的页面时,它会将div中的优先级设置为我在数据库表中所说的内容.
我想找到一种方法来改变背景颜色,具体取决于优先级字段中显示的值,所以红色表示高,橙色表示正常,绿色表示低.
HTML/PHP
<div class="priority">
<?php
$priority_query = "SELECT priority FROM tickets";
$priority = mysql_query($priority_query);
if ($priority = 'low') { ?>
<div class="priority_low"><?php echo 'Priority: ' . mysql_result ($result,$j,'priority') .''; ?></div>
<?php
} else if ($priority = 'normal') { ?>
<div class="priority_normal"><?php echo 'Priority: ' . mysql_result($result,$j,'priority') .''; ?></div>
<?php
} else if ($priority = 'high') { ?>
<div class="priority_high"><?php echo 'Priority: ' . mysql_result($result,$j,'priority') .''; ?></div>
<?php } ?>
</div>
CSS
.priority_low {
background-color: green;
}
.priority_normal …我有这个代码但我面临铬的错误,没有定义load_new_content_dr ...
<html>
<head>
<script lang="javascript" src="http://code.jquery.com/jquery-latest.js">
$(document).ready(function(){
$("#Dr_name").change(load_new_content_dr());
$("#day").change(load_new_content_day());
});
function load_new_content_dr(){
var selected_option_value=$("#Dr_name option:selected").val();
$.post("_add_reservasion.php", {option_value: selected_option_value},
function(data){
$("#day").html(data);
alert(data);
}
);
$.post("__add_reservasion.php",
function(data){
$("#time").html(data);
alert(data);
}
);
}
function load_new_content_day(){
var selected_day = $("#day option:selected").val();
$.post("__add_reservasion.php", {selected_day:selected_day},
function(data){
$("#time").html(data);
alert(data);
}
);
}
</script>
</head>
<body>
...
some codes for connecting database and other stiffs...
...
<!--======================doctore name=========================-->
<p>doctor name:</p>
<select name="Dr_name" id = "Dr_name" form="new" onchange="load_new_content_dr()">
<option value="null"></option>
<?php
while($row = mysqli_fetch_array($table))
{
$name = $row["name"]; …我正在尝试一个更安全的页面,我开始使用密码加密部分.我正在尝试实现password_hash +密码验证,但到目前为止,我还没有成功完成整个工作.所以,这是在我的登录区域:
$username = mysqli_real_escape_string($connection, $_POST['username']);
$password = mysqli_real_escape_string($connection, $_POST['password']);
$query = "SELECT username, password FROM `users` WHERE username='$username' and user_enabled='1'";
$result = mysqli_query($connection, $query) or die(mysqli_error($connection));
if($row = mysqli_fetch_assoc($result)) { $dbpassword = $row['password']; }
if(password_verify($password, $dbpassword)) {
echo "Successful login";
}else{
echo "Invalid Login Credentials.";
}
我总是获得无效的登录凭据.
当我修改用户的新密码时,我正在执行以下操作:
$pass = mysqli_real_escape_string($connection, $_POST['password']);
$options = [ 'cost' => 10,
'salt' => mcrypt_create_iv(22, MCRYPT_DEV_URANDOM),
];
$password = password_hash($pass, PASSWORD_BCRYPT, $options)."\n";
$query = "UPDATE users
SET `password` = '".$password."'
WHERE id = …