2013-06-11 84 views
0

我的代码插入一个空的记录到MySQL表“激活”,而不是获取数据activate.html。它调用我剥去的activate.php。我也应该补充说我是PHP的新手,但是我知道注入攻击。我原来有一些解决的安全问题,但正如我所说,已经剥离了代码来解决问题的根源。另外,当我回显表单字段时,它们会填充,而不是放入MySql表中。任何想法为什么?先谢谢你。Php将空白数据插入表

<?php 
$host = "host"; // Host name 
$username = "user"; // Mysql username 
$password = "pass"; // Mysql password 
$db_name = "db"; // Database name 
$tbl_name = "activate"; // Table name 

// Connect to server and select database. 

mysql_connect("$host", "$username", "$password") or die("cannot connect"); 
mysql_select_db("$db_name") or die("cannot select DB"); 

// Get values from form 

if (isset($_POST['submit'])) { 
    $esn = mysql_real_escape_string($_POST['esn']); 
    $esnverify = mysql_real_escape_string($_POST['esnverify']); 
    $zip = mysql_real_escape_string($_POST['zip']); 
    $comments = mysql_real_escape_string($_POST['comments']); 
} 

// Insert data into mysql 

$sql = "INSERT INTO $tbl_name (esn, esnverify, zip, comments) VALUES ('$esn', '$esnverify', '$zip', '$comments')"; 
$result = mysql_query($sql); 

// if successfully insert data into database, displays message "Successful". 

if ($result) { 
    echo "Successful"; 
    echo "<br />"; 
    echo $_POST['esn']; 
    echo "<br />"; 
    echo $_POST['esnverify']; 
    echo "<br />"; 
    echo $_POST['zip']; 
    echo "<br />"; 
    echo $_POST['comments']; 
    echo "<br />"; 
    echo "<a href='thankyou.html'>Back to main page</a>"; 
} 
else { 
    echo "ERROR"; 
} 

?> 

    <?php 

// close connection 

mysql_close(); 
?> 

Activate.html

<form method="post" action="activate.php"> 
<p><b>ESN:</b> <input type="text" id="esn" name="esn" maxlength="50"><br/> 
<b>Confirm ESN:</b> <input type="text" name="esnverify" id="esnverify" maxlength="50"><br/> 
<b>Zip:</b> <input type="text" name="zip" id="zip" maxlength="5"><br/> 

<p>Your comments:<br /> 
<textarea name="comments" rows="10" cols="40" id="comments" maxlength="500"></textarea></p> 

<p><input type="submit" value="Send it!"></p></form> 
+0

从哪个页面收到'$ _POST'变量? – Kevin

+1

由于您的提交按钮没有'name'属性,这个条件'if(isset($ _ POST ['submit']))'永远不会被满足。它应该像''。 – Lion

+0

谢谢你狮子!与移动我的大括号相结合,它被纠正!再次感谢大家,我非常感谢帮助。 – JuStin

回答

1
<?php 
$host = "host"; // Host name 
$username = "user"; // Mysql username 
$password = "pass"; // Mysql password 
$db_name = "db"; // Database name 
$tbl_name = "activate"; // Table name 

// Connect to server and select database. 

mysql_connect("$host", "$username", "$password") or die("cannot connect"); 
mysql_select_db("$db_name") or die("cannot select DB"); 

// Get values from form 

if (isset($_POST['submit'])) { 
    $esn = mysql_real_escape_string($_POST['esn']); 
    $esnverify = mysql_real_escape_string($_POST['esnverify']); 
    $zip = mysql_real_escape_string($_POST['zip']); 
    $comments = mysql_real_escape_string($_POST['comments']); 


// Insert data into mysql 

$sql = "INSERT INTO $tbl_name (esn, esnverify, zip, comments) VALUES ('$esn', '$esnverify', '$zip', '$comments')"; 
$result = mysql_query($sql); 

// if successfully insert data into database, displays message "Successful". 

if ($result) { 
    echo "Successful"; 
    echo "<br />"; 
    echo $_POST['esn']; 
    echo "<br />"; 
    echo $_POST['esnverify']; 
    echo "<br />"; 
    echo $_POST['zip']; 
    echo "<br />"; 
    echo $_POST['comments']; 
    echo "<br />"; 
    echo "<a href='thankyou.html'>Back to main page</a>"; 
} 
else { 
    echo "ERROR"; 
} 

} 

?> 

    <?php 

// close connection 

mysql_close(); 
?> 

另外,在HTML代码更改此:

添加名称= '提交' 到输入字段/提交按钮。

+0

我改变了大括号,但现在页面什么都不做,甚至没有回声输入? – JuStin

+0

@JuStin,编辑回复 - 另外,我看到狮子已经回答。 :) – sinisake

1

即使$_POST['submit']未设置(在这种情况下,您的变量将没有价值),你的代码仍然会尝试将数据插入到表中。这就是为什么有空白行。

正如Lion在他的评论中所说,由于您的提交按钮没有名称属性,所以$_POST['submit']将永远不会被设置,因此您将始终插入空白数据。