Javascript 防止验证失败时引导模式消失
我试图阻止引导模式消失,以防用户输入任何无效数据 我的php验证代码是:Javascript 防止验证失败时引导模式消失,javascript,php,jquery,twitter-bootstrap,Javascript,Php,Jquery,Twitter Bootstrap,我试图阻止引导模式消失,以防用户输入任何无效数据 我的php验证代码是: if (isset($_POST['submit'])) { $conn = new mysqli("localhost", "ahmesmat", "ZainMalek3110", "SignUpsIroners"); if ($conn->connect_error) { die("Connection failed: " . $conn->connect_er
if (isset($_POST['submit'])) {
$conn = new mysqli("localhost", "ahmesmat", "ZainMalek3110", "SignUpsIroners");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else
if (!$_POST['lname'])
$error="</br>Your first name.";
if (!$_POST['fname'])
$error.="</br>Your last name.";
if (!$_POST['email'])
$error.="</br>Your email address.";
if (!(isset($_POST['day']) && isset($_POST['month']) && isset($_POST['year'])))
$error.="</br>Your full date of birth.";
if (!$_POST['phone'])
$error.="</br>Your phone number.";
if (!$_POST['ssn'])
$error.="</br>Your social security number.";
if (!$_POST['staddress'])
$error.="</br>Your street address.";
if (!$_POST['city'])
$error.="</br>Your city.";
if (!$_POST['state'])
$error.="</br>Your state.";
if (!$_POST['zcode'])
$error.="</br>Please enter your zip code.";
if (!$_POST['country'])
$error.="</br>Your country.";
if (!$_POST['radio'])
$error.="</br>Tell us if you have an iron or planning to get one.";
}
if (isset($error)) {
$flag=1;
}
if(isset($\u POST['submit'])){
$conn=newmysqli(“localhost”、“ahmesmat”、“ZainMalek3110”、“SignUpsIroners”);
如果($conn->connect\u错误){
die(“连接失败:”.$conn->connect\U错误);
}否则
如果(!$\u POST['lname'])
$error=“您的名字。”;
如果(!$\u POST['fname'])
$error.=“您的姓。”;
如果(!$\u POST['email'])
$error.=“您的电子邮件地址。”;
如果(!(isset($发布日期)&&isset($发布月份)&&isset($发布年份))
$error.=“您的完整出生日期。”;
如果(!$_POST['phone']))
$error.=“您的电话号码。”;
如果(!$_POST['ssn']))
$error.=“您的社会保险号码。”;
如果(!$\u POST['staddress'])
$error.=“您的街道地址。”;
如果(!$_POST['city'])
$error.=“您所在的城市。”;
如果(!$_POST['state']))
$error.=“您的状态。”;
如果(!$\u POST['zcode'])
$error.=“请输入您的邮政编码。”;
如果(!$_POST['country'])
$error.=“您的国家/地区。”;
如果(!$_POST['radio']))
$error.=“告诉我们您是否有熨斗或计划购买熨斗。”;
}
如果(isset($error)){
$flag=1;
}
在验证之后,如果验证失败,php应该返回一个标志。以下脚本应读取标志并检索模式,包括用户先前输入的数据:
<script>
//this will launch the modal the first time
$(window).load(function(){
$('#myModal').modal('show');
});
//this was suppose to retrieve the modal
$.ajax({
url:"signupstore.php"
}).done(function() {
var flag='<?php echo json.encode($flag); ?>';
if (flag==1) {
$('#myModal').modal('show');
}
});
//这将在第一次启动modal
$(窗口)。加载(函数(){
$('myModal').modal('show');
});
//假设这是为了检索模态
$.ajax({
url:“signupstore.php”
}).done(函数(){
var标志=“”;
如果(标志==1){
$('myModal').modal('show');
}
});
这显然不起作用。有什么建议吗?服务器端代码
<?php
$error = '';
if (isset($_POST['submit'])) {
$conn = new mysqli("localhost", "ahmesmat", "ZainMalek3110", "SignUpsIroners");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}else{
if (!$_POST['lname'])
$error="</br>Your first name.";
if (!$_POST['fname'])
$error.="</br>Your last name.";
if (!$_POST['email'])
$error.="</br>Your email address.";
if (!(isset($_POST['day']) && isset($_POST['month']) && isset($_POST['year'])))
$error.="</br>Your full date of birth.";
if (!$_POST['phone'])
$error.="</br>Your phone number.";
if (!$_POST['ssn'])
$error.="</br>Your social security number.";
if (!$_POST['staddress'])
$error.="</br>Your street address.";
if (!$_POST['city'])
$error.="</br>Your city.";
if (!$_POST['state'])
$error.="</br>Your state.";
if (!$_POST['zcode'])
$error.="</br>Please enter your zip code.";
if (!$_POST['country'])
$error.="</br>Your country.";
if (!$_POST['radio'])
$error.="</br>Tell us if you have an iron or planning to get one.";
}
}
if ($error != '') {
echo $error;
}else{
return true;
}
?>
<script>
//this will launch the modal the first time
$(document).ready(function(){
$('#myModal').modal('show');
$.ajax({
url: 'signupstore.php',
type: 'POST',
data: {email: 'your_email', country: 'your country'}, // pass your data here
success: function(data) {
//called when successful
if(data != ''){
$('.error').html(data); // add a div to display the return errors
}else{
$('#myModal').modal('hide');
}
}
});
});
</script>
客户端代码
<?php
$error = '';
if (isset($_POST['submit'])) {
$conn = new mysqli("localhost", "ahmesmat", "ZainMalek3110", "SignUpsIroners");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}else{
if (!$_POST['lname'])
$error="</br>Your first name.";
if (!$_POST['fname'])
$error.="</br>Your last name.";
if (!$_POST['email'])
$error.="</br>Your email address.";
if (!(isset($_POST['day']) && isset($_POST['month']) && isset($_POST['year'])))
$error.="</br>Your full date of birth.";
if (!$_POST['phone'])
$error.="</br>Your phone number.";
if (!$_POST['ssn'])
$error.="</br>Your social security number.";
if (!$_POST['staddress'])
$error.="</br>Your street address.";
if (!$_POST['city'])
$error.="</br>Your city.";
if (!$_POST['state'])
$error.="</br>Your state.";
if (!$_POST['zcode'])
$error.="</br>Please enter your zip code.";
if (!$_POST['country'])
$error.="</br>Your country.";
if (!$_POST['radio'])
$error.="</br>Tell us if you have an iron or planning to get one.";
}
}
if ($error != '') {
echo $error;
}else{
return true;
}
?>
<script>
//this will launch the modal the first time
$(document).ready(function(){
$('#myModal').modal('show');
$.ajax({
url: 'signupstore.php',
type: 'POST',
data: {email: 'your_email', country: 'your country'}, // pass your data here
success: function(data) {
//called when successful
if(data != ''){
$('.error').html(data); // add a div to display the return errors
}else{
$('#myModal').modal('hide');
}
}
});
});
</script>
//这将在第一次启动modal
$(文档).ready(函数(){
$('myModal').modal('show');
$.ajax({
url:'signupstore.php',
键入:“POST”,
数据:{email:'your_email',country:'your country'},//在此处传递数据
成功:功能(数据){
//成功时调用
如果(数据!=''){
$('.error').html(数据);//添加一个div以显示返回错误
}否则{
$('#myModal').modal('hide');
}
}
});
});
我希望这有帮助。对不起,如何使用data:{}传递数据?!以前从未做过。我用一个小例子更新了答案,你也可以参考:模式现在闪烁,这意味着我不能输入任何东西来测试重新显示。但是,对于传递的数据,我是否可以输入包含数据的变量,或者必须是静态文本?您可以使用jQuery示例检索数据:$('#inputId').val();此外,您还需要在“提交”单击时发送数据(用户默认设置)。由于模式仍在闪烁,因此不确定我是否以正确的方式发送数据<代码>//这将在第一次启动模式$(document).ready(function(){$(''#myModal').modal('show');event.preventDefault();$.ajax({url:'signupstore.php',type:'POST',var data=$('.#form#val').val();success:function(data){//如果(data!='')成功,则调用函数(data){$('.error').html(data);//添加一个div以显示返回错误}else{$('#myModal').modal('hide');}}}}}});