如何使用javascript和php检查符号，像这样

How to check sign using javascript and php , Like this?

本文关键字：符号 像这样 检查 php 何使用 javascript      更新时间：2023-09-26

如何使用javascript和php检查符号，像这样？

首先，用户将数据填写到输入username并password

如果username和password正确，它将保持警惕SUCCESS

但是username或password不正确，它将保持警惕FAIL

我测试了我的代码，但不起作用。我该怎么做？

.HTML

<form method="post" action="" ENCTYPE = "multipart/form-data" onsubmit="return checkform(this);" id="sign_in_fid">
    <label>
        Your Username
    </label>
    <input type="text" name="username" id="username">
    <br> 
    <label>
        Your Password
    </label>
    <input type="password" name="password" id="password">
    <br>       
    <br> 
<input name="submit" type="submit" value="Sign in"/>
</form>

JAVASCRIPT

<script language="JavaScript" type="text/javascript">
function checkform ( form )
{
  var username_val = document.getElementById("username").value;
  var password_val = document.getElementById("password").value;         
</script>

<?PHP
    include("connect.php");
    $strUsername = "<script>document.writeln(username_val);</script>";
    $strPassword = "<script>document.writeln(password_val);</script>";
    $sql = "SELECT * FROM av8_users WHERE BINARY username = '$strUsername' and password = '$strPassword'";
    $result=mysql_query($sql);
    $row=mysql_fetch_array($result);
    $active=$row['active'];
    $count=mysql_num_rows($result);
    if($count==1)
        {
?>
            <script>
            alert("SUCCESS");
            </script>
<?PHP
        }
    else 
        {
?>
            <script>
            alert("FAIL");
            </script>
<?PHP
        }
?>
<script>
return true ;
}
</script>

你的代码失败的原因是你在php代码中使用了JavaScript。尝试更改您的 

$strUsername = "<script>document.writeln(username_val);</script>";
$strPassword = "<script>document.writeln(password_val);</script>";

自 

$strUsername = $_POST['username'];
$strPassword = $_POST['password'];
相关文章：