在我的登录表单上,我正在进行服务器端验证,如果发生错误,我想在验证的控件下方显示这些错误。现在为此,我试图调用javascript函数在PHP代码中显示验证消息但无法调用。
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
if($_POST['txtUsername']=='')
{
//here i want to call javascript function to display message
}
}
?>
<form action="login.php" method="POST">
Username <input type="text" size="30" name="txtUsername" id="user" /><br />
Password <input type="password" size="30" name="txtPassword" id="pass" /><br />
<input type="submit" value="Login" name="loginSubmit"/>
</form>
<script type="text/javascript">
function showMessage(value)
{
document.getElementById(value).innerHTML= value+"can not be empty.";
}
</script>
请告诉我如何在表单中的验证控件下方显示服务器端验证。
答案 0 :(得分:3)
像这样的东西
<html>
<head>
<script type="text/javascript">
function showMessage(value)
{
document.getElementById(value).innerHTML= value+"can not be empty.";
}
</script>
</head>
<body>
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
if($_POST['txtUsername']=='')
{
echo '<script> showMessage("txtUsername"); </script>';
}
}
?>
<form action="login.php" method="POST">
Username <input type="text" size="30" name="txtUsername" id="txtUsername" /><br />
Password <input type="password" size="30" name="txtPassword" id="txtPassword" /><br />
<input type="submit" value="Login" name="loginSubmit"/>
</form>
</body>
</html>
答案 1 :(得分:3)
使用此
if($_POST['txtUsername']=='')
{
echo '<script> showMessage("txtUsername"); </script>';
}
答案 2 :(得分:2)
你可以将你的php代码放在你喜欢的任何地方,让我们说在body like属性中。您可以尝试以下代码:
<body <?php
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
if($_POST['txtUsername']=='')
{
echo "onload = 'showMessage("VALUE")'";
}
}
?> > // end of body start tag
<form action="login.php" method="POST">
Username <input type="text" size="30" name="txtUsername" id="user" /><br />
Password <input type="password" size="30" name="txtPassword" id="pass" /><br />
<input type="submit" value="Login" name="loginSubmit"/>
</form>
</body>
<script type="text/javascript">
function showMessage(value)
{
document.getElementById(value).innerHTML= value+"can not be empty.";
}
</script>
如果验证成功,php代码将不会回显任何内容,并且不会调用javascript函数。适合我:)。告诉我这是否有帮助。
答案 3 :(得分:1)
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
if($_POST['txtUsername']=='')
{
?>
<script>
//Define the function somewhere in the top or in external js and include it.
callyourfunction();
</script>
<?php
}
}
?> //Its not working