Question

所以，我在检查空值时遇到了一些麻烦。我该如何解决这个问题？我想检查category_name中是否存在category_name（它的工作），我想检查输入是否为空（这不起作用）。这是我的代码：

PHP：

 <?php
    error_reporting(E_ERROR);
    include("../db_config.php");
    $category = mysqli_real_escape_string($conn, $_POST['category']);
    $result = "SELECT * FROM category WHERE category_name='$category'";
    $rs = mysqli_query($conn,$result);
    $data = mysqli_fetch_array($rs);
    if($data > 1 || $category === NULL)
    {
        echo "<script>
        alert('Category name already exist or the value is empty.');
        window.location.href='category.php';
        </script>";
    }

    else
    {   
        $sql[0] = "INSERT INTO category (category_name) VALUES ('$category')";
        if(mysqli_query($conn, $sql[0]))
        { 
                header("Location:category.php");
                exit();
        } 
        else
        {
        echo "ERROR: Could not able to execute $sql[0]. " . mysqli_error($conn);
        }
    $conn->close();
    }
?>

HTML：

 <form action="category_insert.php" method="post" enctype="multipart/form-data">
        <input type="hidden" name="mode" value="add">
        Name:<br />
        <input type="text" name="category" placeholder="Category name"><br><br>
        <input type="submit" value="Submit"><br><br>
    </form>

Answer 1

函数mysqli_real_escape_string返回转义字符串（正如您在此处的文档http://php.net/manual/en/mysqli.real-escape-string.php中所读到的那样），因此$category变量中无法预期NULL。替换代码的这一部分

if($data > 1 || $category === NULL)

用这个

if($data > 1 || empty($category))

检查空值有问题

1 个答案: