我有以下 Javascript 代码:
function checkIfValid(){
var form = document.createuserform;
if(form.fName.value == "" || form.lName.value == "" || form.pass.value == "" || !isValidEmail()){
alert("Please Ensure All Fields Are Filled In Correctly");
}else{
//submit form
form.submit();
}
}
function isValidEmail(){
var x=document.forms["createuserform"]["email"].value
var atpos=x.indexOf("@");
var dotpos=x.lastIndexOf(".");
if (atpos<1 || dotpos<atpos+2 || dotpos+2>=x.length)
{
return false;
}
return true;
}
这是创建用户页面。 javascript 检查输入的数据是否有效，然后将表单提交给 php 应用程序，然后将数据存储在数据库中...
这是 HTML:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Create User</title>
<script src="http://dev.speechlink.co.uk/David/js/createuser.js" type="text/javascript"></script>
</head>
<body>
<h1>Create User</h1>
<p>Please fill in the details below to create a new user account</p>
<form name = "createuserform" action = "http://dev.speechlink.co.uk/David/createuser.php" method = "post">
<p>First Name: <input name = "fName" type="text" size="25"></p>
<p>Last Name: <input name = "lName" type="text" size="25"></p>
<p>E-mail: <input name = "email" type="text" size="25"></p>
<p>Password: <input name = "pass" type="text" size="25"></p>
<p> Gender: <input type="radio" name="gender" value="M" checked> Male
<input type="radio" name="gender" value="F"> Female</p>
<p>User Type: <select name="usertype">
<option>Athlete</option>
<option>Support Staff</option>
</select></p>
<p><input type ="button" value="Create Account" name = "createuser" onclick="checkIfValid()"></p>
</form>
</body>
</html>
php代码如下:
<?php
session_start();
$dbh = connect();
//connect to database
function connect() {
$dbh = mysql_connect ("localhost", "abc123", "def567") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db("PDS", $dbh);
return $dbh;
}
//insert into database
if (isset($_POST['createuser'])){
//prevent SQL injections (this version of php does not support mysqli functions)
$fn = mysql_real_escape_string($_POST['fName']);
$ln = mysql_real_escape_string($_POST['lName']);
$email = mysql_real_escape_string($_POST['email']);
$gender = mysql_real_escape_string($_POST['gender']);
$usertype = mysql_real_escape_string($_POST['usertype']);
if($usertype == 'Athlete'){
$usertype = 'Athletes';
}else{
$usertype = 'SupportStaff';
}
$pass = md5($_POST['pass']);
if(eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$", $email)) {
$query= "SELECT * FROM $usertype WHERE email = '" . $email . "'";
$result = mysql_query($query);
if(mysql_num_rows($result) < 1 || !mysql_num_rows($result)) {
$query= "INSERT INTO $usertype (fName, lName, gender, email, password) VALUES ('$fn', '$ln', '$gender', '$email', '$pass')";
mysql_query($query);
mysql_close($dbh);
echo ("&result=true&");
} else {
mysql_close($dbh);
echo "&result=userexists&";
}
}
else {
mysql_close($dbh);
echo "&result=false&"; //invalid e-mail address
}
}
?>
当我使用 HTML 表单中的提交按钮直接向它发送 POST 时，php 工作。但是，当我使用 javascript 时，该页面是有针对性的，但它是空白的……没有回显……有什么想法吗？
我有一种感觉，表单没有捕捉到触发 javascript 的 HTML 按钮的“createuser”名称...因为我的 php 文件期望设置 $_POST['createuser']...也许它不是...？
最佳答案
因为您没有按下名称为“createuser”的按钮，“createuser”不是发送到服务器的数据的一部分。
因此，您不应检查“createuser”，而应在您的表单中插入一个隐藏字段并检查它。
关于php - JavaScript 通过 POST 提交 HTML 表单，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/6847864/