Question

这是我的表单，在我使用input type = submit之前，但现在我使用验证状态，这就是为什么我需要打开输入类型=按钮，我的问题是如何提交这个表格使用按钮？

// HTML CODE

<div class="panel-body" style="width:700px;">
    <form class="form-group" method="post" id="question_form">
        <div class="form-group">
            <label>Your email address<span style="color:#FF0000">*</span></label>
            <input type="text" name="email" size="80" class="form-control" id="emailq" placeholder="Enter you email address">
        </div>
    <div class="form-group">
        <label>Message<span style="color:#FF0000">*</span></label>
        <textarea rows="8" cols="50" class="form-control" name="message" id="messageq" placeholder="What is your concern?"></textarea>
    </div>
    <div class="form-group">
        <label style="width:170px;">Enter the code you see<span style="color:#FF0000">*</span></label>
        <span style="background:#CCC; font-size:20px; color:#000; padding-top:5px; padding-bottom:5px; padding-left:30px; padding-right:30px"><?=$random_text?></span> 
        <input type="text" size="80" name="uIcaptcha" class="form-control" id="codeq" placeholder="This code is for security purposes only.">
    </div>
    <div class="form-group">
        <input type="hidden" name="captcha">
        <input type="button" id="question_submit" name="submit" value="Send Request" class="btn btn-primary">
    </div>
    </form>
</div>

// JS CODE

function validateText(id){
if($("#"+id).val()==null || $("#"+id).val()=="")
{
    var div = $("#"+id).closest("div");
    div.removeClass("has-success");
    $("#gly"+id).remove();
    div.addClass("has-error has-feedback");
    div.append('<span id="gly'+id+'" class="glyphicon glyphicon-remove form-control-feedback"></span>');
    return false;
}else{
    var div = $("#"+id).closest("div");
    div.removeClass("has-error");
    div.addClass("has-success has-feedback");
    $("#gly"+id).remove();
    div.append('<span id="gly'+id+'" class="glyphicon glyphicon-ok form-control-feedback"></span>');
    return true;
}

}

$(document).ready(
function()
{
    $("#question_submit").click(function()
    {
        validateText("emailq");

        validateText("messageq");

        validateText("codeq");


        $("form#question_form").submit();

    });
}

）;

// PHP代码

 if(isset($_POST['submit'])){
 $email = $_POST['email'];
 $message = $_POST['message'];
 $captcha = $_POST['captcha'];
 $user_Input = $_POST['uIcaptcha'];

$exists = mysqli_query($con,"SELECT * FROM users WHERE email = '".$email."'");
if(!empty($email) and !empty($message)and !empty($user_Input)){    //check if empty

    if(mysqli_num_rows($exists) > 0){ //check if email is valid

        if($captcha == $user_Input){ //check if captcha is correct
            echo 'Query Here';

        } else{
            $error_code = 'Please enter correct code.'; //close if captcha checking
        }

    }else{
        $error_email = 'Please enter the valid email address'; // close if email validation         
    }
}else{
    $error = 'Please fill up all information!';  //close if empty checking
}

}

Answer 1

function getStates(value) {   //this my code that show result 
    $.post("search.php", {name:value},function(data){
      $("#results").html(data); 
      if(!data) {
        alert("i am blank and hiding result text box");
        $("#results").hide();
      }
   }); 
 }

如果这不起作用，请告诉我。

Answer 2

您只需在每个表单中插入一个隐藏按钮：例如：

<div class="form-group">
    <input type="hidden" name="captcha">
    <input type="button" id="question_submit" name="fakeSubmit" value="Send Request" class="btn btn-primary">
     <input type="submit" name="submit" id="submit" style="display:none;"/>
</div>

然后当你点击假一个

时，只需点击实际提交按钮上的点击

jQuery('#question_submit').click(function(){jQuery('#submit).toggle('click')});

多数民众赞成我如何做到这一点然后你可以简单地在点击真实的隐藏提交之前添加if（条件）

希望我能帮到你

如何使用input type = button提交表单

2 个答案: