致命错误:在非对象上调用成员函数prepare() [英] Fatal error: Call to a member function prepare() on a non-object

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本文介绍了致命错误:在非对象上调用成员函数prepare()的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

您好我有这段代码





$ sql =INSERT INTO tblLecturer(lec_id,lec_lastname,lec_firstname)VALUES( ?,?,?);



if(!($ stmt = $ mysqli-> prepare($ sql))){

die(准备失败:。$ mysqli-> errno);

}



if(!$ stmt-> ; bind_param('sss',$ id,$ lname,$ fname)){

die(Binding parameters failed:。$ stmt-> errno);

}



if(!$ stmt-> execute()){

die(插入注册表失败:。$ stmt-> errno);

}



我在这里遇到致命错误:if(!($ stmt = $ mysqli- >准备($ sql))){

我不知道这段代码有什么问题..

任何帮助plzz

解决方案

sql =INSERT INTO tblLecturer(lec_id,lec_lastname,lec_firstname)VALUES(?,?,?);



if(! (


stmt =


mysqli->制备(

Hello i have this piece of code


$sql = "INSERT INTO tblLecturer (lec_id,lec_lastname,lec_firstname) VALUES (?, ?, ?)";

if (!($stmt = $mysqli->prepare($sql))) {
die("Prepare failed: ".$mysqli->errno);
}

if (!$stmt->bind_param('sss', $id, $lname, $fname)){
die("Binding parameters failed: ".$stmt->errno);
}

if (!$stmt->execute()) {
die("Insert registration table failed: ".$stmt->errno);
}

I am getting a fatal error here: if (!($stmt = $mysqli->prepare($sql))) {
I don't know whats wrong with this code..
Any help plzz

解决方案

sql = "INSERT INTO tblLecturer (lec_id,lec_lastname,lec_firstname) VALUES (?, ?, ?)";

if (!(


stmt =


mysqli->prepare(


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