Question

我写了以下HTML代码：

  <form action="index.php" method="POST">    
      <input type="text" name="title" required>
      <input type="text" name="brief_text" required>
      <textarea name="text" required></textarea>
      <input type="submit" name="add" value="Add">
  </form>

我的PHP代码：

<?php
require_once('db.php');

if(isset($_POST['add'])){
  $title = $_POST['title'];
  $brief_text = $_POST['brief_text'];
  $text = $_POST['text'];
  $blog_cat_id = $_POST['blog_cat_id'];

  if($title AND $brief_text AND $text AND $blog_cat_id){
    $insert_blog = "insert into blog values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
    $run_insertion = mysqli_query($con, $insert_blog);

    if($run_insertion){
      echo "Blog has been added!";
     }
    else{
      echo "Error adding blog!!!";
    }
 }
 else{
   echo "All fields are required!";
 }
}
else{
echo "GOODBYE";
}

?>

每次刷新页面时，它只显示表单和＆＃34; GOODBYE＆＃34;甚至不会将数据插入数据库表。请帮帮我。

Answer 1

你几乎没有错，

1）应该是：$_POST['add']而不是$_POST['add_blog']

2）没有$_POST['blog_cat_id']没有形式

修改复制了您的代码并进行了一些更改：

<强>码

if(isset($_POST['add'])){ print_r($_POST); $title = $_POST['title']; $brief_text = $_POST['brief_text']; $text = $_POST['text']; $blog_cat_id = $_POST['blog_cat_id']; if($title AND $brief_text AND $text AND $blog_cat_id){ echo "inside condition"; $insert_blog = "insert into blog values ('','$title','$brief_text','$text','$blog_cat_id',NOW())"; $run_insertion = mysqli_query($con, $insert_blog); if($run_insertion){ echo "Blog has been added!"; } else{ echo "Error adding blog!!!"; } } else{ echo "All fields are required!"; } } else{ echo "GOODBYE"; }

<强> HTML：

<form action="index.php" method="POST"> <input type="text" name="title" required> <input type="text" name="brief_text" required> <input type="text" name="blog_cat_id" required> <textarea name="text" required></textarea> <input type="submit" name="add" value="Add"> </form>

<强>输出

Array ( [title] => test [brief_text] => test [blog_cat_id] => 1 [text] => testing [add] => Add ) inside condition

现在，检查您的查询是否无效。

希望这会对你有所帮助。

Answer 2

您的查询错误，未指定列，并且您对sql注入开放，您应该学会使用参数化查询。但是这次你可以使用以下内容。

试试这个：

<强> htmlcode

<form action="index.php" method="POST">    
  <input type="text" name="title" required>
  <input type="text" name="brief_text" required>
  <input type="text" name="blog_cat_id" required>
  <textarea name="text" required></textarea>
  <input type="submit" name="add" value="Add">
</form>

<强>的index.php

<?php
require_once('db.php');
if(isset($_POST['add'])){
  $title = $_POST['title'];
  $brief_text = $_POST['brief_text'];
  $text = $_POST['text'];
  $blog_cat_id = $_POST['blog_cat_id'];

  if($title AND $brief_text AND $text AND $blog_cat_id){
    $insert_blog = "insert into blog('col1','col2','col3','col4','col5','col6') values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
    $run_insertion = mysqli_query($con, $insert_blog);

    if($run_insertion){
      echo "Blog has been added!";
     }
    else{
      echo "Error adding blog!!!";
    }
 }
 else{
   echo "All fields are required!";
 }
}
?>

注意： col1, col2, col3, col4, col5 and col6将是您的列名。

Answer 3

它是否仍在展示＆＃39; GOODBYE＆＃39;现在你已经改变了

$_POST['add_blog']

到

$_POST['add']

点击提交时？

为什么我的表单不能使用POST方法？

3 个答案: