Question

我有一个简单的表单，通过POST消息将数据发送到服务器。但是，我收到了错误＃34;无法执行查询＆＃34;每当我点击提交按钮。这是我的实施：

sample.html

<!DOCTYPE html>
<html>
<head>
<script>
        function pullMore(){
            var xmlhttp;
            if (window.XMLHttpRequest){ // code for IE7+, Firefox, Chrome,etc.
                xmlhttp = new XMLHttpRequest();
            }else{ // code for IE6, IE5
                xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xmlhttp.onreadystatechange = function() {
                if (xmlhttp.readyState==4 && xmlhttp.status==200) {
                    document.getElementById("news_mesgs").innerHTML = xmlhttp.responseText;
                }
            }

            var name = document.getElementById("name");
            var email = document.getElementById("email");
            var comments = document.getElementById("comment");
            var parameters="name"+name.value+"&email="+email.value+"&comments="+comments.value;

            xmlhttp.open("POST", "reviews.php", true);
            xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
            xmlhttp.send(parameters);
        }
</script>
</head>
<body style="background-color : #e9e9e9;">
<div> Hello there </div>
<form>
    Name: <input type="text" id="name" name="name">
    Comment: <input type="text" id="comment" name="comment">
    Email: <input type="text" id="email" name="email">
    <input type="button" value="Submit" onclick="pullMore()">
</form>


<div id="news_mesgs"> come here </div>
</body>
</html>

reviews.php

mysql_connect($host,$username,$password);
mysql_select_db($database) or die( "Unable to select database");


$query = 'INSERT INTO Reviews (Name, Email, Review) VALUES ('.$_POST['name'].','.$_POST['email'].','.$_POST['comments'].');';
$result = mysql_query($query) or die( "Unable to execute query");

更新：出于某种原因，$_POST["name"]显示为空。我尝试print var_dump($_POST);获取一些示例数据，这就是我得到的： array（2）{[＆＃34; email＆＃34;] =＆gt; string（11）＆＃34; abc@abc.com" [＆＃34;注释＆＃34;] =＆GT; string（5）＆＃34;你好＆＃34;无法执行查询

Answer 1

您缺少任务操作员。您没有正确发送名称，这就是查询失败的原因。

尝试使用

var parameters="name="+name.value+"&email="+email.value+"&comments="+comments.value;

你也缺少引号。使用

$query = 'INSERT INTO Reviews (Name, Email, Review) VALUES ("'.$_POST['name'].'","'.$_POST['email'].'","'.$_POST['comments'].'");';

或更好一点

$query = 'INSERT INTO Reviews (Name, Email, Review) VALUES ("'.mysql_escape_string($_POST['name']).'","'.mysql_escape_string($_POST['email']).'","'.mysql_escape_string($_POST['comments']).'");';

并且不要只将post变量插入查询中。这是SQL注入的直接方式。

检查How can I prevent SQL-injection in PHP?

HTTP POST请求问题

1 个答案: