我在MyFile.php中有这段代码
$db= mysqli_connect("host","user","pw","db");//connect to db
if (mysqli_connect_errno($con))//check connection
{echo "Failed to connect to MySQL: " . mysqli_connect_error();}
//Create a token for the unique link
$title= $_GET[apt_title];
$em
我正在尝试使用MySQLi创建一条准备好的语句,并在数据库上运行它,目前我正在获取带有结果元数据的列名,但是我没有将任何结果打印到我的HTML表中,并且mysqli_stmt_num_rows($statement)返回0。我已经运行了一些测试回显,并且$value确实有来自文本框的正确值,并且当我直接在数据库上运行查询时,替换?对于一个值,它确实会返回一些数据。你知道为什么我不能把数据传到MySQLi吗?
function Get_Data($value,$radio){
//gets data if the blank option was not selected
if
我正在尝试使用Flash Builder的php函数...但是,函数的返回值是意外的..空白对象( [] )而不是数组。该函数为:
public function getAllAuctions($search) {
$stmt = mysqli_prepare($this->connection, "SELECT * FROM $this->tablename WHERE name LIKE ?");
$this->throwExceptionOnError();
mysqli_stmt_bind
我正在用mysqli做一个PHP项目,需要通过$_POST从用户那里检索信息,我的问题是我想要像这样使用准备好的语句和mysqli->real_escape_string:
//$mysqli is the link to my database
if( isset($_POST['ID_name']) && isset($_POST['Nombre_name']) )
{
//just in case magic_quotes_gpc is on
$id = trim(htmlentities(my
我遇到的问题是,它在mysqli中给了我一个警告,声明如下:
Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn't match number of parameters in prepared statement in ...on line 89
我怎样才能摆脱警告?
$questionquery = "
SELECT q.QuestionId, q.QuestionContent, o.OptionType, q.NoofAnswers, GROUP_C
$stmt= mysqli_prepare($conn, "SELECT news_ttitle,pic_name,news_text,id FROM news order by rand()");
mysqli_stmt_bind_param($stmt, 'sssi');
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $tittle,$pic_name,$text,$id);
mysqli_stmt_fetch($stmt);
你好,我是php的新手,我正在制作一个基于网络的在线评分系统,我的问题是,当我作为一个学生登录时,我想要查看我的成绩,它显示了我上传的所有成绩的数据。
<?php
$con = mysqli_connect('localhost','root','','grade') or die(mysqli_error());
$query = "Select * from grades where
我使用mysqli来获取行,但它没有给我行,而且查询中也没有错误。
$query="select * from members where useremail='$user_email' and password='$password'";
$result=$db->query($query);
$row = $db->fetch_array($result);
echo $row['id'];
我的query函数
function query($query){
$result=mysqli_query
我正在尝试使用mysqli->prepare创建表,但在准备语句时遇到错误。我的声明如下:
$mysqli->prepare("CREATE TABLE ? (id INT AUTO_INCREMENT NOT NULL, PRIMARY KEY (id))");
然后,我只需将表名绑定到它
$mysqli->bind_param("s", $some_name);
并执行它。
$mysqli->execute();
我得到的错误是:
You have an error in your SQL syntax; check the manu
我有以下代码:
$id = $_GET['id'];
// get the recod from the database
if($stmt = $mysqli->prepare("SELECT * FROM date WHERE id=?"))
{
$stmt->bind_param("isssssssss", $id, $mtcn, $amount, $currency, $sender_name, $sender_country, $receiver_name, $comment, $support, $ema
我一天中的大部分时间都在尝试让它工作,但都没有效果。
基本上,我试图在foreach()循环中执行bind_param(),但是,当我这样做时,我得到了以下错误:
Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn't match number of parameters in prepared statement in /nfs/domains/domain.com/html/v2/includes/classes/class.Database.php on li
这是我的函数,它给出了错误,不知道为什么它根本不能执行,这个if语句是第86行
if($stmt = $mysqli->prepare("SELECT time FROM login_attempts WHERE user_id=? AND time > '$valid_attempts'"))
{
$stmt->bind_param('i', $user_id);
$stmt->execute();
$stmt->store_result();
if($stmt->num_r