相关疑难解决方法(0)

我刚开始使用PHP进行编码,并且在尝试创建待办事项列表时,我在创建一个名为"add.php"的文件后撤消了,该文件应该向表中添加新数据.主文件index.php调用表元素ok并在浏览器上显示它们,但我无法添加新项.

add.php中的代码在这里:

<?php

require_once 'app/init.php';

if(isset($_POST['name'])) {
    $name = trim($POST['name']);

    if(!empty($name)) {
        $addedQuery = $db->prepare("
            INSERT INTO items (name, user, done, created) 
            VALUES (:name, :user, 0, NOW() )
                ");

        $addedQuery->execute([
            'name' => $name,
            'user' => $_SESSION['user_id']
            ]);
    }
}

header('location: index.php');




?> 

如果它有帮助,我在WampServer 2.5上使用PHP 5.5.12和MySQL 5.6.17

我一直在尝试从我的数据库中选择两个默认日期之间的项目:现在和60天前.但是,我尝试的所有查询都不起作用.

这是我尝试过的:

$Now = date("Y-m-d");
$Before = date("Y-m-d", strtotime("-60 days");

// This is try1
$sql = "SELECT * FROM myTable WHERE myTimestamp BETWEEN " . $Before . " AND " . $Now;  

// This is try2
$sql = "SELECT * FROM myTable WHERE myTimestamp >= " . $Before . " AND myTimestamp <= " . $Now; 

我猜对了如何做到这一点.我已经查看了与此问题相同的其他问题,但没有提出任何解决方案.

请注意:这些查询不会出错.他们只是没有找回任何东西.我还习惯get_defined_vars()在页面上打印日期.这是他们展示的:

[Now] => 2016-01-07
[Before] => 2015-11-08

尝试使用以下代码连接到localhost sql db(此时不对查询执行任何操作只是希望查询执行):

<?php

$host     = "localhost";
$port     = 3306;
$socket   = "";
$user     = "root";
$password = "Password1";
$dbname   = "CIIP_WIKI";

$con = new mysqli($host, $user, $password, $dbname, $port);

if(!$con)
{
echo ("db connection failed!");
die ('Could not connect to the database server' . mysqli_connect_error());
}

else {
    echo ("db connection established.");
    echo ("<br/>");
    }

$query = sprintf("SELECT first_name FROM actor WHERE actor_id='1'");

$result = mysql_query($query);

$con->close();

?>

我一直得到以下......

Welcome

db connection established.

Warning: mysql_query(): Access denied for user …

当我尝试调用我的函数时,我正在调用非对象上的成员函数query().

我的代码看起来像这样:

function add_profile() {

    $hostname = "localhost";
    $dbusername = "username";
    $dbname = "dbname";
    $dbpassword = "password";
    $link = mysqli_connect($hostname, $dbusername, $dbpassword, $dbname); 
    if (!$link) { 
    die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error()); 
    } 

    $sql = "INSERT INTO payment_profiles(id, client_id) VALUES ( '','$profile_id')"; 

    $result = $mysqli->query($sql);
    if (!result) 
    { 
    echo 'Error: ', $mysqli->error;
    }
}

add_profile();

它说我的错误就行了: $result = $mysqli->query($sql);

我假设我没有正确地打电话.在此先感谢您的帮助

我是PHP的新手,我需要一点帮助.我很确定这可能是最常见的问题之一,并且很容易修复,但我一直在努力.我会这么愚蠢

好吧,所以这是我的问题:每次我去我的localhost尝试打开我的PHP站点它不会因为我每次都得到这个错误.

"解析错误:语法错误,第4行的C:\ xampp\htdocs\exam\_register.php中的意外'$ db'(T_VARIABLE)"

这是我的PHP代码.

    <?php
    session_start()

    $db = mysqli_connect("localhost", "root", "", "authentication");

    if (isset($_POST['register_btn'])) {
        session_start();
        $username = mysql_real_escape_string($_POST['username']);
        $emailaddress = mysql_real_escape_string($_POST['emailaddress']);
        $password = mysql_real_escape_string($_POST['password']);
        $password2 = mysql_real_escape_string($_POST['password2']);

        if ($password == $password2) {
            $password = md5($password);
            $sql = "INSERT INTO users(username, email, password) VALUES ('$username', '$emailaddress', '$password')";
            mysql_query($db, $sql);
            $_SESSION['message'] = "You are now registered!";
            $_SESSION['username'] = $username;
            header("location: home.php");
        }else {
            $_SESSION['message'] = "The two passwords do not match";
        }   
    }
?>

我一直收到这个错误,我已经查看过前一张海报提出的问题,并且向他们发出的建议对解决我的问题毫无帮助,如果已经完成,那么我当然不会提出这个问题所以请不要重定向!我正在开发一个项目,其中用户注册他们的详细信息名称年龄,电子邮件等但我希望他们具有编辑数据的能力将数据写入数据库很好,但试图检索它以供用户查看我得到错误mysql_fetch_array()期望参数1是资源,给定对象

这是我的代码

<?php
require_once'connect.php';

    $FirstName = $_POST['fName'];
    $UserName = $_POST['uName'];
    $Age = $_POST['age'];
    $Password = $_POST['password'];
    $Email = $_POST['email'];

    $sql = "INSERT INTO `user` (`First_Name`,`UserName`,`Age`,`Password`,`Email`) 
        VALUES ('$FirstName','$UserName','$Age','$Password','$Email')";

    if(! mysqli_query($con, $sql)){
        die("Echo ".mysqli_error($sql));
    }else{
        header('Location: BankDetails_Form.php');
    }

这样就可以将数据上传到数据库

现在我用来检索它的代码是:

<?php
    include 'connect.php';

    if(isset($_POST['view'])){

        $query = "SELECT `First_Name`,`Username`,`Age`,`Password`,`Email` FROM `user`";
        $result = mysqli_query($con, $query)
         or die('Error querying database');

         while ($row = mysql_fetch_array($result)){
             echo $row['First_Name'].''.$row['Username'].''.$row['Age'].$row['Password'].$row['Email'];
         }
    }
?>

任何人都可以指出我正确的方向,为什么我得到了这个错误,我再次意识到这个问题已经提出,但以前的解决方案并没有帮助我.

在登录时遇到这个错误,我安装了新版本的xampp/php仍然面临着很多mysqli的问题......

下面是login.php的代码,之后转到lock.php,代码如下......

<?php

include("config.php");
session_start();

if($_SERVER["REQUEST_METHOD"] == "POST")
{
// username and password sent from form 

$myusername=addslashes($_POST['username']); 
$mypassword=addslashes($_POST['password']); 


$sql="SELECT id FROM mc_admin WHERE username='$myusername' and passcode='$mypassword'";
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
$active=$row['active'];

$count=mysql_num_rows($result);


// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1)
{
session_register("myusername");
$_SESSION['login_user']=$myusername;

header("location: home.php");
}
else 
{
$error="Your Login Name or Password is invalid";
}
}
?>

以下是lock.php的代码:

<?php
include('config.php');
session_start();
$user_check=$_SESSION['login_user'];

$ses_sql=mysql_query("select username from mc_admin where username='$user_check' ");

$row=mysql_fetch_array($ses_sql);

$login_session=$row['username'];

if(!isset($login_session))
{
header("Location: login.php"); …

我正在创建一个登录表单,但我没有成功登录时遇到问题.我的表单看起来像这样

<form method="post" action="phpScripts/loginProcess.php">

    <input type="text" name="uname" value="" placeholder="Username">

    <input type="password" name="pass" value="" placeholder="Password">

    <input id="submit" type="submit" value="submit">

</form>

我已经检查过,我正在连接数据库,我的uname和pass都是正确的.我经过多次双重检查.问题在我的查询中.这是loginProcess.php文件

<?php

    require_once ('dbconn.php');



    if(isset($_POST['login'])){

        $uname  = mysqli_real_escape_string($_POST['uname']);
        $pass   = mysqli_real_escape_string($_POST['pass']);

        $query  = mysql_query ("SELECT * FROM staff WHERE uname='$uname'");
        $numrows = mysql_num_rows($query);

        if ($numrows !=0){
            die("Success!!");
        }
        else{
            die("That user doesnt exist");
        }

    }
    else{
        echo "Username or Password incorrect";
    }
?>

在Fred-ii的帮助下,我能够找出错误并真正快速解决登录问题!我想提供我用来登录的代码,以防其他人遇到这个问题.

这是我用来登录的代码.我觉得我可以使用execute和$ userSql/$ pwSql减少冗余.任何建议清理这一点将不胜感激!

<?php
ini_set('display_errors', 1); error_reporting(E_ALL);


require_once ('dbconn.php');


$uname = $_POST['uname'];
$pass …

main.php文件中的函数

public function get_email_details($email, $table_name)
{
    $query_verify_email = "SELECT * FROM $table_name where u_email='$email'";
    $result = mysqli_query($this->connection,$query_verify_email) or die(mysql_error());
    return $result;
}

我用它来称呼它

$result_get_email_details = $Object->get_email_details($email, $table_name);
$rows=mysql_fetch_array($result_get_email_details); // It is showing me error on this line

错误:mysql_fetch_assoc()期望参数1是资源,在我评论之后在上面的行给出的对象.

请帮我解决这个SQL查询中的错误

我是PHP的初学者.当我尝试做mysql_num_rows时,它给出了一个错误:

警告:mysql_num_rows()期望参数1是资源,第22行的D:\ xampp\htdocs\kcp\police\viewcase.php中给出的对象

这是代码:

<?php
session_start();
if(empty($_SESSION['pusername'])||empty($_SESSION['pid']))
{
    header('location://localhost/kcp/police_login.html');
    exit();

}
else
{
    $station=$_SESSION['station'];
    $con = mysqli_connect('localhost','root','','kcp');

    if(!$con)
    {
        echo 'NOT CONNECTED TO DB';
    }
    $sql="SELECT * FROM complaint WHERE station='$station'";
    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
    $res=mysqli_query($con,$sql) or die("llllllll");

    $numrow=mysql_num_rows($res);
    echo $numrow;
}

?>

所以我在显示查询输出时遇到了这个问题.我将展示我的代码并尝试解释我能做的最好的事情.

<?php
    include_once('ligacao.php');
    if (mysqli_connect_errno())
    {
    echo "Falha ao conectar à Base de Dados: ".mysqli_connect_error();
    }

        $pesq = $_POST['id'];
        echo $pesq; 

        $sqlveri = "SELECT * FROM distrito WHERE iddistrito LIKE '12'";
        $result = mysql_query ($ligar,$sqlveri);
        echo $resul;
        mysqli_close ($ligar);
        ?>

我提交时唯一出现的是保留在变量上的数字$pesq.帮帮我这些家伙我连续两天都在努力寻找答案,没有运气

谢谢您的帮助.