Php Insert查询发出“无法获取mysqli_stmt”警告，但仍将数据发送到数据库
我有一张有两个下拉列表的表格。两者都使用数据库表中的值。 当访问者提交时，接下来的事情将发生： select查询将把选项id/POST值与数据库表中的值id进行比较。查询将在while循环中获取。while循环中有一个if-else，用于控制值是否相等。当它们运行时，则运行其他。 在else中，有一个insert查询，它将值id保存在一个新的数据库表中。 对于select和insert查询，我都使用准备好的语句。 获取select查询后，我在else$selControl->close中将其关闭；然后开始第二次查询插入 当我运行带有submit的网站时，我得到一个错误：无法获取select查询的mysqli_stmt。但它仍然可以工作并插入到DB表中。 当我写$selControl->close时；包含/独占$insKeuze->close；从第三个else的insert after}或}after while，我得到一个错误，第一个查询必须在一个新的prepare语句之前关闭，这就是逻辑。 如果没有close语句，也会出现“准备前关闭”错误。 我更新了我的代码。 我添加了绑定参数。它将数据发送到数据库，但给出了错误 我需要做什么来停止错误 如果有人能帮助我，请提前感谢 代码：insert.php 此结构会产生相同的错误：Php Insert查询发出“无法获取mysqli_stmt”警告，但仍将数据发送到数据库,php,mysqli,insert,prepared-statement,fetch,Php,Mysqli,Insert,Prepared Statement,Fetch,我有一张有两个下拉列表的表格。两者都使用数据库表中的值。 当访问者提交时，接下来的事情将发生： select查询将把选项id/POST值与数据库表中的值id进行比较。查询将在while循环中获取。while循环中有一个if-else，用于控制值是否相等。当它们运行时，则运行其他。 在else中，有一个insert查询，它将值id保存在一个新的数据库表中。 对于select和insert查询，我都使用准备好的语句。 获取select查询后，我在else$selControl->close中将其关闭
<?php
while ($selControl->fetch())
{
echo "<script>alert('". $kLand . $kGerecht . $lLand_id . $gGerecht_id . "')</script>";
// If selected land (land_id) is not the same as land_id in table landen
// or selected gerecht (gerecht_id) is not the same as gerecht_id in table gerechten --> send echo.
if ($kLand == $lLand_id && $kGerecht == $gGerecht_id)
{
$selControl->close();
// If query is prepared --> bind the columns and execute the query.
// Insert statement with placeholders for the values to database table landGerecht
if ($insKeuze = $mysqli->prepare("INSERT INTO lab_stage_danush . landGerecht (land_id, gerecht_id) VALUES ( ?, ?)"))
{
// Bind land_id and gerecht_id as integers with $landKeuze and $gerechtKeuze.
$insKeuze->bind_param('ii', $kLand, $kGerecht);
// If statement is not executed then give an error message.
if (!$insKeuze->execute())
{
echo "Failed to execute the query keuze: " . $mysqli->error;
}
$insKeuze->close();
} else // Else give an error message.
{
echo "Something went wrong in the insert query connection: " . $mysqli->error;
}
} else // Else insert the selected values in table landGerecht.
{
// Message when the combination is not correct
echo "<script>alert('Deze combinatie bestaat niet. Doe een nieuwe poging!');</script>";
}
}
?>
正如您在评论中所说，您认为您的mysql连接之间存在冲突。我已经创建了一个类来解决这个问题 landen.php
class landen{
//define our connection variable, this is set in __construct
private $mysqli;
//this function is called when you create the class. $l = new landen($mysqlconn)
public function __construct($mysqli){
$this->mysqli = $mysqli;
}
//setter for kLand
private $kLand = 0;
public function setKland($k){
$this->kLand = $k;
}
//setter for kGerecht
private $kGerecht = 0;
public function setKGerecht($k){
$this->kGerecht = $k;
}
//run is our main function here.
//if there was a return from select, it runs the insert.
//will return true if select + insert BOTH pass.
public function run(){
if($this->select()){
return $this->insert();
}
return false;
}
private function select(){
$q = "SELECT landen.land_id, gerechten.gerecht_id FROM landen INNER JOIN gerechten ON landen.land_id = gerechten.land_id WHERE landen.land_id = ? AND gerechten.gerecht_id = ?";
if($stmt = $this->mysqli->prepare($q)){
$stmt->bind_param("ii",$this->kLand,$this->kGerecht);
if($stmt->execute()){
while ($stmt->fetch()){
/*
In your original code, you had a check to see if
your post variable was the same as your returned query variables.
This was not required as you are selecting from your database with those.
They will **always** equal the post variables.
Line I'm disputing: if ($kLand != $lLand_id || $kGerecht != $gGerecht_id)
*/
return true;
}
}else{
print_r("Error in select execute: {$this->mysqli->error}");
}
}else{
print_r("Error in select prepare: {$this->mysqli->error}");
}
return false;
}
private function insert(){
$q = "INSERT INTO lab_stage_danush . landGerecht (land_id, gerecht_id) VALUES ( ?, ?)";
if($stmt = $this->mysqli->prepare($q)){
$stmt->bind_param("ii",$this->kLand,$this->kGerecht);
if($stmt->execute){
return true;
}else{
print_r("Error in insert execute {$this->myqsli->error}");
}
}else{
print_r("Error in insert prepare {$this->mysqli->error}");
}
return false;
}
}
<?php
require_once("dbConfig.php");
if(isset($_POST["submit"]) && !empty($_POST['landen']) && !empty($_POST['gerechten'])){
//Call the class code when we need it
require_once("landen.php");
//create the class when we need it.
$landen = new landen($mysqli);
//set our variables
$landen->setKland($_POST['landen']);
$landen->setKGerecht($_POST['gerechten']);
//landen run returns True if the select + insert statment passed.
//It will return false if they fail.
//if they fail, the page will not change and the errors will be outputted.
//Or it's failing because nothing has been returned by the select function.
if($landen->run()){
//send out header to go to Keuze page,
//you had a javascript location function here.
header("Location: keuze.php");
die();
}else{
//by default, let's say our queries are running fine and the only reason the select failed is because
//the database couldn't find anything.
echo "Nothing found with {$_POST['landen']} and {$_POST['gerechten']}. Please try again!";
}
}
正如您在评论中所说，您认为您的mysql连接之间存在冲突。我已经创建了一个类来解决这个问题 landen.php
class landen{
//define our connection variable, this is set in __construct
private $mysqli;
//this function is called when you create the class. $l = new landen($mysqlconn)
public function __construct($mysqli){
$this->mysqli = $mysqli;
}
//setter for kLand
private $kLand = 0;
public function setKland($k){
$this->kLand = $k;
}
//setter for kGerecht
private $kGerecht = 0;
public function setKGerecht($k){
$this->kGerecht = $k;
}
//run is our main function here.
//if there was a return from select, it runs the insert.
//will return true if select + insert BOTH pass.
public function run(){
if($this->select()){
return $this->insert();
}
return false;
}
private function select(){
$q = "SELECT landen.land_id, gerechten.gerecht_id FROM landen INNER JOIN gerechten ON landen.land_id = gerechten.land_id WHERE landen.land_id = ? AND gerechten.gerecht_id = ?";
if($stmt = $this->mysqli->prepare($q)){
$stmt->bind_param("ii",$this->kLand,$this->kGerecht);
if($stmt->execute()){
while ($stmt->fetch()){
/*
In your original code, you had a check to see if
your post variable was the same as your returned query variables.
This was not required as you are selecting from your database with those.
They will **always** equal the post variables.
Line I'm disputing: if ($kLand != $lLand_id || $kGerecht != $gGerecht_id)
*/
return true;
}
}else{
print_r("Error in select execute: {$this->mysqli->error}");
}
}else{
print_r("Error in select prepare: {$this->mysqli->error}");
}
return false;
}
private function insert(){
$q = "INSERT INTO lab_stage_danush . landGerecht (land_id, gerecht_id) VALUES ( ?, ?)";
if($stmt = $this->mysqli->prepare($q)){
$stmt->bind_param("ii",$this->kLand,$this->kGerecht);
if($stmt->execute){
return true;
}else{
print_r("Error in insert execute {$this->myqsli->error}");
}
}else{
print_r("Error in insert prepare {$this->mysqli->error}");
}
return false;
}
}
<?php
require_once("dbConfig.php");
if(isset($_POST["submit"]) && !empty($_POST['landen']) && !empty($_POST['gerechten'])){
//Call the class code when we need it
require_once("landen.php");
//create the class when we need it.
$landen = new landen($mysqli);
//set our variables
$landen->setKland($_POST['landen']);
$landen->setKGerecht($_POST['gerechten']);
//landen run returns True if the select + insert statment passed.
//It will return false if they fail.
//if they fail, the page will not change and the errors will be outputted.
//Or it's failing because nothing has been returned by the select function.
if($landen->run()){
//send out header to go to Keuze page,
//you had a javascript location function here.
header("Location: keuze.php");
die();
}else{
//by default, let's say our queries are running fine and the only reason the select failed is because
//the database couldn't find anything.
echo "Nothing found with {$_POST['landen']} and {$_POST['gerechten']}. Please try again!";
}
}
只是我从球棒上看到的东西。。hack_filter但真正的hack_filter是bind_param，您在选择中使用prepare后没有使用它。请删除该过滤器，并选择结束您准备好的语句；我是否必须在执行之前添加bind_参数，以使用变量绑定landen.land_id和gerechten.gerecht_id？使用bind_param将变量放入查询中，这必须在->执行之前完成。在运行->execute后，您可以使用bind_result来获取结果。您很可能会收到该错误，因为->prepare在选择时失败。请找到$selControl=$mysqli->prepareSELECT的结束if语句括号。。。然后抛出}else{print\r$mysqli->error；}来定位实际问题；和$selControl->error。这是错误的。语句/查询没有返回错误，实际上是mysqli返回了错误。这里正确的语句应该是$mysqli->error@WillParky93所以，如果我没弄错的话，trim、stripslashes和htmlspecialchar对于获得安全的post值来说是不必要的？当有人更改源代码F12中的值时，我使用它们来过滤这些值，这是我从bat中看到的。。hack_filter但真正的hack_filter是bind_param，您在选择中使用prepare后没有使用它。请删除该过滤器，并选择结束您准备好的语句；我是否必须在执行之前添加bind_参数，以使用变量绑定landen.land_id和gerechten.gerecht_id？使用bind_param将变量放入查询中，这必须在->执行之前完成。在运行->execute后，您可以使用bind_result来获取结果。您很可能会收到该错误，因为->prepare在选择时失败。请找到$selControl=$mysqli->prepareSELECT的结束if语句括号。。。然后抛出}else{print\r$mysqli->error；}来定位实际问题；和$selControl->error。这是错误的。语句/查询没有返回错误，实际上是mysqli返回了错误。这里正确的语句应该是$mysqli->error@WillParky93所以，如果我没弄错的话，trim、stripslashes和htmlspecialchar对于获得安全的post值来说是不必要的？当有人更改源代码F12中的值时，我使用它们来过滤这些值