更新mysqli錯誤

Question

有人請指出我出錯的地方。

我試着按照如何執行update-with-mysqli-prepare給出的建議？在網站上，但沒有運氣。

以下：

<?php 

//connection 
$con = new mysqli ("localhost","user","password","db"); 

$playno = "22"; 
$n1 = "4"; 
$n2 = "4"; 
$n3 = "4"; 

$stmt = $con -> prepare("UPDATE game SET no1 = ?, no2 = ?, no3 = ? WHERE id = ?"); 

$stmt -> bind_param ('iiii',"$n1","$n2","$n3","$playno"); 
$stmt -> execute(); 


?> 

在瀏覽器中給出了這樣的：

Fatal error: Cannot pass parameter 2 by reference in C:\xampp\htdocs... on line 13

提前非常感謝。

Answer 1

$stmt -> bind_param ('iiii',$n1,$n2,$n3,$playno); 

Answer 2

該錯誤表示bind_param中的第二個參數應該是對變量的引用。

$stmt -> bind_param ('iiii',"$n1","$n2","$n3","$playno"); 

從this：

Note that mysqli_stmt_bind_param() requires parameters to be passed by reference

它應該是：$stmt -> bind_param ('iiii',$n1,$n2,$n3,$playno);

Answer 3

當您嘗試通過PARAMS如下

$stmt -> bind_param ('iiii',"$n1","$n2","$n3","$playno"); 

的"$n1"等被視爲常數字符串，因此導致錯誤（而不是警告）。

此外，您需要傳遞整數值而不是字符串。

$stmt -> bind_param('iiii',$n1, $n2, $n3, $playno); 

Answer 4

$stmt -> bind_param ('iiii',"$n1","$n2","$n3","$playno"); 

應該

$stmt -> bind_param ('ssss',$n1,$n2,$n3,$playno); 

或

$playno = 22; 
$n1 = 4; 
$n2 = 4; 
$n3 = 4; 

$stmt -> bind_param ('iiii',$n1,$n2,$n3,$playno); 

Answer 5

你的問題很簡單：

這是一個字符串 $ playno = 「22」; // string $ n1 =「4」; // string $ n2 =「4」; // string $ n3 =「4」; //字符串

$stmt = $con -> prepare("UPDATE game SET no1 = ?, no2 = ?, no3 = ? WHERE id = ?"); 

這裏你的錯誤 $語句 - > bind_param（ 'IIII'， 「$ N1」， 「$ N2」， 「$ N3」， 「$ playno」）; //失敗，您必須使用s $ stmt - > bind_param（'ssss'，$ n1，$ n2，$ n3，$ playno）; //這是正確的

或變化：

$playno = 22; //integer 
$n1 = 4; //integer 
$n2 = 4; //integer 
$n3 = 4; //integer 
$stmt -> bind_param ('iiii',$n1,$n2,$n3,$playno); //this is correct 

然後：

$stmt -> execute();