PHP建议表格

时间:2015-08-23 20:48:41

标签: php forms validation

我一直试图在我的网站上提出建议表格,但我不能成功! 这是我的代码

<div class="suggest" id="suggest">
<center>
<span class="title">Suggest a Game</span>

<br> 
<br>
<form action="suggest.php" method="get">
<input type="text" class="suggestion" id="suggestion" name="suggestion">   
<input onclick="validate()" type="submit" value="Send" class="send">  
</form>
<br>
<center>
<span id="validator"></span>
</center>
</div> 

和PHP(suggest.php):

<?php

if (isset($_REQUEST['suggestion']))  {

//Email information
$admin_email = "greglypet@gmail.com";
$suggest = $_REQUEST['suggestion'];
$subject = "Game Suggestion:";
//send email
mail($admin_email, "$subject", "$suggest" );

echo }

else  { 

?>

Javascript(验证):

function validate() {  
var text = document.getElementById("suggestion").innerHTML;
if (text.length > 4) {
document.getElementById("validator").style.color = "lightgreen" ; 
document.getElementById("validator").innerHTML = "Thanks for the suggestion" ;  
}
 if (text.length < 4) { 
document.getElementById("validator").style.color = "red" ; 
document.getElementById("validator").innerHTML = "Input is too sort" ;  

}     

}

我做错了什么!我是php的总菜鸟,所以请帮忙!

我想要的是使用validatetrue()和validatefalse()验证文本输入长度,但它似乎无法正常工作!

1 个答案:

答案 0 :(得分:0)

在您的js-function validate()中,您现在拥有:

var text = document.getElementById("suggestion").innerHTML;

但是输入没有这个属性/属性,所以它总是返回undefined(因此小于4)。

要获取输入文本,请使用value

var text = document.getElementById("suggestion").value;