标签: mysqli

码:

$stmt->bind_param("s", md5($input['user'] . $config['salt']));

PHP错误消息:

只应通过引用传递变量

我一直在研究这个项目,但现在我被困住了.我是PHP的新手.该怎么办？

我的要求是在这个链接要求.如果我们运行查询,我们将分别得到Chennai和Pune与Count 3,因为chennai和pune的regionid存在于表格中.但我需要将班加罗尔和德里显示为0.我怎么做

Plz帮助我不知道这个功能有什么问题....

$gsql = "SELECT * FROM posts WHERE group='$group_name' ORDER BY postdate DESC LIMIT 0,20";
$gquery = mysqli_query($db_conx, $gsql);
$gstatusnumrows = mysqli_num_rows($gquery);

while ($grow = mysqli_fetch_array($gquery, MYSQLI_ASSOC)) {

它一直在说这个错误: -

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in D:\group.php on line 3

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in D:\group.php on line 5

我有功能:

function dbConnect($usertype, $connectionType = 'mysqli') {

   // some code hare

   return new mysqli($host, $user, $pwd, $db) or die ('Cannot open database');
}

但是当我尝试这个:

$conn = dbConnect('read');
$result = $conn->query('SELECT * FROM images');

函数不返回任何东西,它说:

致命错误:在第10行的C:\ xampp\htdocs\phpsols\mysql\mysqli.php中调用非对象的成员函数query()

但它以这种方式工作(没有die())

return new mysqli($host, $user, $pwd, $db);

我singer id在另一个表中插入数据后尝试生成,当我使用prepare语句时会出现问题:

警告:mysqli_insert_id()期望参数1为mysqli,object>在第22行的/Applications/XAMPP/xamppfiles/htdocs/admin/singer.php中给出警告:mysqli :: prepare()[mysqli.prepare]:couldn'在第24行的/Applications/XAMPP/xamppfiles/htdocs/admin/singer.php中获取> mysqli>

<?php
include('../db_inc.php');    
$singer_name =$_POST['singer_name'];
$singer_gender=$_POST['singer_gender'];
$singer_des=$_POST['singer_description'];
$singer_genre=$_POST['genre_list'];

        if($stmt=$connection->prepare("INSERT INTO singers(singer_name,singer_gender,singer_description) VALUES (?,?,?)")){
            $stmt->bind_param('sss',$singer_name,$singer_gender,$singer_des);
            $result1=$stmt -> execute();
            $singer_id=mysqli_insert_id($stmt);
            $stmt->close();
            }

  if($stmt2=$connection->prepare("INSERT INTO genre_singer(f_singer_id,f_genre_id) VALUES (?,?)")){
      $stmt2->bind_param('fs',$singer_id,$singer_genre);
            $result2=$stmt2 -> execute();
            $stmt2->close();
            $connection->close();
      }
if($result1 & $result2){
    echo "insert successfully"; 
    };
?>

经过多次编辑和检查教程网站.代码当前未从数据库调用信息,单击"批准"按钮时,不编辑数据库.我有一个名为Reg_ID的列标识符,可以指定您选择编辑的数据列.表单正在提交,只清除我输入的信息而不存储数据.

此文件名为Approve Deny Prayer Request.

<?php
$DB_HOST = "XXXXXXX";
$DB_NAME = "XXXXXXX";
$DB_PASS = "XXXXXXX";
$DB_USER = "XXXXXXX";

$link = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($link->connect_errno > 0) {
die('Connection failed [' . $db->connect_error . ']');
}

$query = "SELECT * FROM Request";
$result = mysqli_query($link,$query); //<----- Added link
$row = mysqli_fetch_array($result);

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

$id = mysqli_real_escape_string($link,$_POST['id']);
$firstname = mysqli_real_escape_string($link,$_POST['first']);
$lastname = mysqli_real_escape_string($link,$_POST['last']);
$phone = mysqli_real_escape_string($link,$_POST['phone']);

$query2=mysqli_query($link,"UPDATE Request SET Reg_F_Name='$firstname',     Reg_L_Name='$lastname',Reg_Request='$phone' WHERE id='$id'" );

if($query2){
header("Location: fbcaltusprayerorg.ipagemysql.com");
}

} // brace …

我正在做一个教程,其中作者没有更新他的内容以反映PHP文档中的更改.无论如何,我需要知道我要提供的参数是什么.我已经检查过所有事情都是有序的,但我确实不知道我应该提供什么.这就是我所拥有的:

Connects.php

<?php
$connect_error = 'Sorry, we\'re experiencing connection issues.';
$con = mysqli_connect('localhost', 'root', 'PwdSQL5');
mysqli_select_db('phpcadet') or die($connect_error);
?>

但我得到错误:

编辑:在确定解决Connects.php问题之后,这就是为什么我修复它的原因...更多错误,这是我的代码.记住我是PHP的新手,并且正在完成一个做得不好的教程.

警告: mysqli_real_escape_string()期望参数1为mysqli,第4行的C:\ vhosts\phpcadet\core\functions\general.php中给出的字符串

General.php

<?php
function sanitize($data)
{
return mysqli_real_escape_string($data, 'What goes here?');
}
?>

然后这个:警告:mysqli_query()期望参数1是mysqli,第7行的C:\ vhosts\phpcadet\core\functions\users.php中给出的字符串

Users.php

<?php
require 'core/database/connects.php';

function user_exists($username)
{
  $username = sanitize($username);
  $query = mysqli_query($_POST['username'], "SELECT * FROM users");
  $row = mysqli_fetch_array($query);

  if($row['username']==$username)
  {
     echo "Welcome, $username!";
     return true;   
  }
  else echo "Please enter a valid username and password";
  return false;
} …

好像我PHP试图MySQL用我没有提供的用户名登录数据库.

我得到的错误是:

Warning: mysql_query(): Access denied for user 'radiocaf'@'localhost' (using password: NO) in /home/radiocaf/public_html/layout.php on line 16
Warning: mysql_query(): A link to the server could not be established in /home/radiocaf/public_html/layout.php on line 16

我肯定提供了一个密码,并没有使用"radiocaf"作为我的连接文件中的用户名,所以经过3个小时的盯着,我仍然无法解决我出错的地方.

这是我的代码:

PSL-config.php文件:

define("HOST", "localhost");

define("USER", "carl");

define("PASSWORD", "xxxxxxxxx");

define("DATABASE", "wlist");

db_connect.php

include_once 'psl-config.php';   // As functions.php is not included

$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);

然后最后,我收到错误的主页面(我已经删除了PHP之间的HTML:

ini_set('display_errors',1);
error_reporting(E_ALL);
//Include Connection PHP and connect
include_once('includes/db_connect.php');

//Check Connection
if ($mysqli->connect_error) {
    die('Connection failed: …

我正在使用PHP更新数据库中的记录.这个字符串中有一些单引号.我想知道如何使用sql update在数据库中更新此字符串.

PHP代码:

$content = $_POST['content'];

$query = "UPDATE about SET content = '" . $content . "' WHERE about_id = 1";
$result = mysql_query($query) or die(mysql_error());

例如,内容是:

嗨,我叫Sam,我住在荷兰.我24岁,我在'DiscoFever'工作.

由于文中有许多单引号,我无法更新我的记录.解决这个问题的方法有哪些

MySQLi不将NULL作为空值进行比较.

就我而言:

if($Id == 1) {
    $cond = " AND my_field != '' ";
} elseif ($Id == 2){
    $cond = " AND my_field = '' ";
}

在执行具有条件的mysql语句时!='',WHERE结果集与=''WHERE条件中的结果集不完全相反.这里!=''声明完美正常.

但是当我把这个条件放在elseif它的工作正常时.

elseif ($Id == 2){
   $cond = " AND my_field IS NULL ";
}