Question

目前，这是我的代码

的index.php：

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

insert.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);

当我回复$ comment时，没有任何内容被打印出来。但是，如果我做一些像回声＆＃34; hi＆＃34;有用。我认为由于某种原因，$ _POST没有得到认可。任何建议使这项工作或我做错了。

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

Answer 1

检查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($_POST); //在mysqli_real_escape_string
var_dump($comment); //在mysqli_real_escape_string

mysqli api可能效果不好！修复像

这样的查询

$sql = "INSERT INTO parentComment(ID, Comment) VALUES('','".$comment."')";
echo $sql; //检查您的sql语法
echo mysqli_error($link); //您有错误
查看.htaccess文件，看看您是否有<Limit POST>标记
删除此行include ('index.php');。据说，这两个文件位于一个文件夹中。所以只需运行index.php。尝试没有那条线的代码，它对我有用。

Answer 4

使用以下代码： -

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');
}

希望它会对你有所帮助： - ）

PHP $ _POST为空 - 使用MAMP不适用于本地服务器

4 个答案: