PHP $_POST 为空——不适用于使用 MAMP 的本地服务器

Question

所以目前，这是我拥有的代码

index.php:

        <form action="insert.php" method="post">
            Comments:
            <input type="text" name="comment">
            <input type="submit">
        </form>

插入.php：

<?php

include ('index.php');

$user = 'x';
$password = '';
$db = 'comment_schema';
$host = 'localhost';
$port = 3306;

/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */

$link = mysqli_connect("localhost", "x", "", "comment_schema");

// Check connection
if($link === false){
   die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Escape user inputs for security
$comment = mysqli_real_escape_string($link, $_POST["comment"]);
$sql = "INSERT INTO parentComment(ID, Comment) VALUES('','$comment')";

// attempt insert query execution
if(mysqli_query($link, $sql)){
   echo $comment;
} else{
 echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// close connection
mysqli_close($link);

当我回显 $comment 时，没有任何内容被打印出来。但是，如果我执行类似 echo "hi" 的操作，它会起作用。我认为由于某种原因 $_POST 未被识别。任何使这项工作的建议，或者如果我做错了。

我的目标是获取用户输入并插入到 phpmyadmin 上的数据库中。目前，它能够插入，但是它插入一个空值。我的数据库中只有两列。一个 ID 和一个 Comment 列。 ID 会自动递增。评论是我从用户那里得到的。

Answer 1

检查一下您在数据库中的评论数据类型是什么。（我更喜欢Varchar()）。 试试这个：-

<form action="insert.php" method="POST">
                Comments:
                <input type="text" name="comment"/>
                <input type="submit" name="submit" value="submit">
            </form>

<?php

include ('index.php');

$user = 'x';
$password = '';
$db = 'comment_schema';
$host = 'localhost';
$port = 3306;


$link = mysqli_connect("localhost", "x", "", "comment_schema");


if($link === false){
   die("ERROR: Could not connect. " . mysqli_connect_error());
}


$comment = mysqli_real_escape_string($link, $_POST["comment"]);
$sql = "INSERT INTO parentComment(ID, Comment) VALUES('','$comment')";


if(mysqli_query($link, $sql)){
   echo $comment;
} else{
 echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// close connection
mysqli_close($link);

Answer 2

试试这个

<?php
    $user = 'x';
    $password = '';
    $db = 'comment_schema';
    $host = 'localhost';

    $link = mysqli_connect($host, $user, $password, $db);

    if($link === false){
       die("ERROR: Could not connect. " . mysqli_connect_error());
    }

    // Escape user inputs for security
    $comment = mysqli_real_escape_string($link, $_POST["comment"]);
    $sql = "INSERT INTO parentComment(ID, Comment) VALUES(NULL,'$comment')";

// attempt insert query execution
if(mysqli_query($link, $sql)){
   echo $comment;
} else{
 echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// close connection
mysqli_close($link);

Answer 3

一些调试建议：

• var_dump($_POST); // 在 mysqli_real_escape_string 之前
• var_dump($comment); // 在 mysqli_real_escape_string 之后

mysqli api 可能无法正常工作！像这样修复查询

• $sql = "INSERT INTO parentComment(ID, Comment) VALUES('','".$comment."')";
• echo $sql; // 检查你的 sql 语法
• echo mysqli_error($link); // 你有错误吗
• 查看您的 .htaccess 文件，看看您是否有 <Limit POST> 标签
• 删除此行include ('index.php');。假设这两个文件在一个文件夹中。所以只需运行 index.php 。尝试了没有该行的代码，它对我有用。

Answer 4

使用下面的代码：-

include ('index.php');

$user = 'x';
$password = '';
$db = 'comment_schema';
$host = 'localhost';
$port = 3306;    

$link = mysqli_connect("localhost", "x", "", "comment_schema");

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

if(!empty($_POST["comment"])){
    $comment = mysqli_real_escape_string($link, $_POST["comment"]);    
    // Escape user inputs for security
    $sql = "INSERT INTO parentComment(ID, Comment) VALUES('','$comment')";
    $result = mysqli_query($link, $sql);
    // attempt insert query execution
    if($result){
       echo $comment;
    } else{
        echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
    }
    // close connection
    mysqli_close($link);

}else{
    die('comment is not set or not containing valid value');
}

希望对你有帮助:-)