如何在PHP中将2个参数传递给array_map函数？

Question

$sql="select * from question_test where test_id='".$test_id."'   and  difficulty_level   BETWEEN  ".$_SESSION['difficulty_start']." and  ".$_SESSION['difficulty_end']." and question_id NOT IN ('".implode("', '", array_map('mysqli_real_escape_string', $_SESSION['question_attempt']))."')  order by rand()*favourability_level  desc";

在运行上面的代码时，我得到一个错误：

警告：mysqli_real_escape_string（）恰好需要2个参数，第359行的C：\\ xampp \\ htdocs \\ exam.php中给出了1个参数

Answer 1

改用回调函数

function array_map_callback($a)
{
  global $con;
  return mysqli_real_escape_string($con, $a);
}

array_map('array_map_callback', $_SESSION['question_attempt']);

其中$con是连接变量。

因此，您的$sql变量为：

$sql="select * from question_test where test_id='".$test_id."' and  difficulty_level BETWEEN  ".$_SESSION['difficulty_start']." and  ".$_SESSION['difficulty_end']." and question_id NOT IN ('".implode("', '", array_map('array_map_callback', $_SESSION['question_attempt']))."')  order by rand()*favourability_level  desc";

或者你可以选择array_walk

array_walk($_SESSION['question_attempt'], function(&$string) use ($con) { 
  $string = mysqli_real_escape_string($con, $string);
});