Question

我正在做简单的表单，在数据库表中插入值，这是代码

new_subject，php

        <form action="create_subject.php" method="post">
          <p>Menu name:
          <input type="text" name="menu_name" value="" />
          <p>Position:
          <select name="position">
              <?php  
               $subject_set= find_all_subjects();
               $subject_count=mysqli_num_rows($subject_set);
                for($count=1; $count<=$subject_count+1; $count++){
                echo "<option value='{$count}' >{$count}</option>";
               }
              ?>
               </select>
          </p>
          <p>Visible
              <input type="radio" value="0" name="visible" />No &nbsp;
              <input type="radio" value="1" name="visible" />Yes
          </p>
          <input type="submit" value="Submit" name="submit" />
          <br/>          <br/>
          <a href="manage_content.php">Cancel</a>
     </form>

create_subject.php

 <?php  
    if(isset($_POST['submit'])){
        $menu_name= $_POST["menu_name"]; 
        $position = (int) $_POST["position"];
        $visible= (int)$_POST["visible"]; 
        $query="INSERT INTO subjects (menu_name , position , visible)  VALUES($menu_name, $position, $visible)";        
        $result= mysqli_query($connection, $query); 
        if($result){
            header("Location : manage_content.php");
            exit();
        }
        else {
            echo 'There was some problem ' ;
            echo mysqli_error($connection);
        }
    }
    else {
        header('Location: new_subject.php');
        exit(); 
    }

 ?>

现在这是错误有一些问题未知字段'name'// name是我在文本框中输入的值。有人请帮我解决这个问题。如果我没有输入文本框的值，则其他值已成功输入数据库

Answer 1

作为使易于阅读和维护的sql语句的提示使用sprintf()：

$query= sprintf("INSERT INTO subjects (menu_name , position , visible)  VALUES('%s', '%s', '%s')",$menu_name, $position, $visible);

或

$sql = "INSERT INTO subjects (menu_name , position , visible)  VALUES('%s', '%s', '%s')";
$query = sprintf($sql, $menu_name, $position, $visible);

Answer 2

将单引号括在它将解决问题的值

$query="INSERT INTO subjects (menu_name , position , visible)  VALUES('$menu_name', '$position', '$visible')";

PHP / mysql：插入语句不按预期工作

2 个答案: