Question

我正在开发一个电子商务项目，该项目目前位于该项目的后端，您可以在该项目中添加或删除类别的菜单或子菜单。

我已检查dll已启用

这是我得到的错误

致命错误：在第15行的C：\ xampp \ htdocs \ project \ admin \ categories.php中调用未定义的函数sanitize（）

这是代码

<?php
 require_once $_SERVER['DOCUMENT_ROOT'].'/project/core/init.php';
 include 'includes/head.php';
 include 'includes/navigation.php';

 $sql = "SELECT * FROM categories WHERE parent = 0";
$result = $db->query($sql);
$errors = array();
// when the form is cick process

$Category = '';
$parent= '';

if(isset($_POST) && !empty($_POST)){
  $parent = sanitize($_POST['parent']);
  $category = sanitize($_POST['category']);
  $sqlform = "SELECT * FROM categories WHERE category ='$category' AND parent = '$parent'";
  $fresult = $db->query($sqlform);
  $count = mysqli_num_rows($fresult);
 //if category is blank

 if($category == ''){
  $errors[] .= 'The category cannot be left blank';
  }

 //if it already exixt in database
 if($count > 0){
  $errors[] .= $category.' Already exits please choose a new category'; 
 }
 //display error or update database
 if(!empty($errors)){
 $display = display_errors($errors);?>
 <script>
    jQuery('document').ready(function(){
  jQuery ('#errors').html('<?php $display;?>');   

    });

 </script>


<?php } else{
 //update database
 $updatesql = "INSERT INTO categories(category,parent)VALUES ('$category','$parent')";
 $db->query($updatesql);
 header('location:categories.php');
}
}


?>

Answer 1

在我看来，$parent = mysqli_real_escape_string($_POST['parent']); $category = mysqli_real_escape_string($_POST['category']);并未在范围内定义。它不是标准函数，因此它应该在代码中的某个位置。您最有可能直接用mysqli_real_escape_string代替它：

hash -r
which linter

您应该使用PDO代替，这会更好！

希望这有帮助！

调用未定义的函数sanitize（）

1 个答案: