mysqli的bind_param（）预计为基准，给出的价值 [英] mysqli bind_param() expected to be a reference, value given

问题描述

想不通，什么导致错误的参数3 mysqli_stmt :: bind_param（）预计为参考，值...给出

Can't figure out, whats causing error Parameter 3 to mysqli_stmt::bind_param() expected to be a reference, value given in...

PDO
$query = "INSERT INTO test (id,row1,row2,row3) VALUES (?,?,?,?)";
$params = array(1,"2","3","4");
$param_type = "isss";
$sql_stmt = mysqli_prepare ($mysqli, $query);
call_user_func_array('mysqli_stmt_bind_param', array_merge(array($sql_stmt, $param_type), $params));
mysqli_stmt_execute($sql_stmt);

也试过OOP

OOP
$insert_stmt = $mysqli->prepare($query);
array_unshift($params, $param_type);
call_user_func_array(array($insert_stmt, 'bind_param'), $params);
$insert_stmt->execute();

但同样的错误，只是现在参数2引起的问题。

But same error, only that now Parameter 2 is causing problem.

那么，什么是错$ PARAMS？我需要$ PARAMS是值的数组。

So, what's wrong with $params? I need $params to be an array of values.

推荐答案

从PHP实况：

护理必须与call_user_func_array结合使用mysqli_stmt_bind_param（）时，应考虑（）。需要注意的是mysqli_stmt_bind_param（）需要的参数按引用传递，而call_user_func_array（）可以作为参数接受，可以重新present引用或值的变量列表。

Care must be taken when using mysqli_stmt_bind_param() in conjunction with call_user_func_array(). Note that mysqli_stmt_bind_param() requires parameters to be passed by reference, whereas call_user_func_array() can accept as a parameter a list of variables that can represent references or values.

在页面上的mysqli-stmt.bind-参数你有不同的解决方案

And on the page mysqli-stmt.bind-param you have different solutions:

例如：

call_user_func_array(array($stmt, 'bind_param'), refValues($params));

function refValues($arr){
    if (strnatcmp(phpversion(),'5.3') >= 0) //Reference is required for PHP 5.3+
    {
        $refs = array();
        foreach($arr as $key => $value)
            $refs[$key] = &$arr[$key];
        return $refs;
    }
    return $arr;
}

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