类型定义字符串中的元素数量与第8行的/home/a3598479/public_html/Register.php中的绑定变量数量不匹配 [英] Number of elements in type definition string doesn't match number of bind variables in /home/a3598479/public_html/Register.php on line 8

问题描述

请帮助我.我实际上并没有使用PHP，但是需要在我的Login/Register项目中使用.

Please help me. I don't actually use PHP but I need to use in my Login/Register project.

$con = mysqli_connect("***", "***", "***", "***");

$name = $_POST["name"];
$username = $_POST["username"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO user (name, username, password) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement, "siss", $name, $username, $password);
mysqli_stmt_execute($statement);

$response = array();
$response["success"] = true;  

echo json_encode($response);

但它说

Warning: mysqli_stmt_bind_param() [function.mysqli-stmt-bind-param]: Number of elements in type definition string doesn't match number of bind variables in /home/a3598479/public_html/Register.php on line 8

我该如何解决?

推荐答案

我认为以下语句存在问题:

I think problem in following statement:

mysqli_stmt_bind_param($statement, "siss", $name, $username, $password);

您正在传递siss，这意味着应该有4个params，其类型为string，integer，string和string，这是错误的，因为只有three parameters都是字符串

You are passing siss which means it there should be 4 params with types string, integer, string and string which is wrong as you have only three parameters which are all string.

所以声明应该像这样:

 mysqli_stmt_bind_param($statement, "sss", $name, $username, $password);

这篇关于类型定义字符串中的元素数量与第8行的/home/a3598479/public_html/Register.php中的绑定变量数量不匹配的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！