Question

我尝试创建没有真实提交按钮的表单，但是使用了样式的href和javascript，但我无法自行发布表单。

如果我var_dump($_POST);，那么我得到一个包含4个项目（3个输入和1个文本区域）的数组。表单本身不存在，因此if (isset($_POST['submit']))的值为false。我也试过了if (isset($_POST['ContactForm']))，但它不起作用

表单代码：

<form name='ContactForm' id='ContactForm' action='verzendvraag.php' method='POST'>
    <input type="text" class="col-md-6 col-xs-12 name" id="naam" name='naam' placeholder='Naam *' value="<? echo $_SESSION['naam']; ?>" required/>
    <input type="email" class="col-md-6 col-xs-12 Email" id="email" name='email' placeholder='Email *' value="<? echo $_SESSION['email']; ?>" required/>
    <input type="text" class="col-md-12 col-xs-12 Subject" name='Subject' id='Subject' placeholder='Onderwerp'/>
    <textarea type="text" class="col-md-12 col-xs-12 Message" id="vraag" name="vraag" placeholder='Bericht *' style="left: 0px; top: 0px" required><? echo $_SESSION['vraag']; ?></textarea>
    <div class="cBtn col-xs-12">
        <ul>                                
            <li class="clear"><a href="#" onclick="resetForm(event)"><i class="fa fa-times"></i>ledig vakjes</a></li>
            <li class="send"><a href="#" onclick="submitForm(event)"><i class="fa fa-share"></i>Verstuur bericht</a></li>
        </ul>
    </div>
</form>

的javascript：

function submitForm(e) {
    if (validateForm())
    {
        document.getElementById("ContactForm").submit();
    }
    e.preventDefault();
}

PHP：

session_start();    
//Validatie serverside => bescherming inbouwen voor als iemand zijn javascript uitschakelt. Javascript echter wel nodig om de server en netwerktraffic te ontlasten. 
if (isset($_POST['submit']))
{ 
    $naam  = $_POST['naam'];
    //$bedrijf = $_POST['bedrijf'];
    $subject = $_POST['Subject'];
    $email = $_POST['email'];
    $vraag = $_POST['vraag'];

    //In session steken zodat we bij een eventuele terugkeer naar de form de reeds ingevulde gegevens kunnen terug zetten.
    $_SESSION['naam']  = $_POST['naam'];
    //$_SESSION['bedrijf'] = $_POST['bedrijf'];
    $_SESSION['Subject'] = $_POST['Subject'];
    $_SESSION['email'] = $_POST['email'];
    $_SESSION['vraag'] = $_POST['vraag'];

    $isOK = true;
    $error = '0';

    $atpos = strrpos($email,"@");
    $dotpos = strrpos($email,".");

    if(is_null($naam) || $naam == '')
    {
        $isOK = false;
        echo('NAAM NIET OK ');
        $error .= ',1';
    }

    if(is_null($email) || $email == '')
    {
        $isOK = false;
        echo('EMAIL NIET OK '.$email);
        $error .= ',2';
    }
    else
    {
        //checken geldig e-mailadres
        if ($atpos === false || $dotpos === false || $dotpos < $atpos+2 || $dotpos+2 >= strlen($email))  //=== moet => http://php.net/manual/en/function.strpos.php en http://www.php.net/manual/en/language.operators.comparison.php
        {
            $isOK = false;
            echo("Ongeldig e-mailadres ");
            $error .= ',3';
        }
    }

    if(is_null($vraag) || $vraag== '')
    {
        $isOK = false;
        echo('VRAAG NIET OK ');
        $error .= ',4';
    }
}

if(!($isOK))
{
    echo("<script>location.href='index.php?error=$error'</script>");
    exit;
}
else
{
.....
}

我查了Javascript Submit is not Sending POST to php file，他说他的代码有效，所以我不明白我的赢了。

Answer 1

如果（isset（$ _ POST [＆＃39; submit＆＃39;]））if（（__ POST [＆＃39; naam＆＃39;]）

，你也应该改变这个

Answer 2

由于您没有任何名为submit的按钮或输入，因此它不存在于$ _POST变量中。我认为您应该检查表格中的内容，以便使用。

if (isset($_POST['naam'])){
// your code 
}

而不是

if (isset($_POST['submit'])){
// your code 
}

Answer 3

使用像这样的ajax请求

$.ajax({//create an ajax request to load_page.php
        type: "POST",
        url: "receive.php",
        data: $("#ContactForm").serialize(),
        success: function(response) {
            if (response) {
                alert('Successfully posted.');
            }
            else {
                alert('Successfully not posted.');
            }
        }
    });

Answer 4

只需使用ajax提交表单

<!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=iso-8859-1" />
<title>Untitled Document</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
</head>

<body>
<form name='ContactForm' id='ContactForm' action='verzendvraag.php' method='POST'>
<input type="text" class="col-md-6 col-xs-12 name" id="naam" name='naam' placeholder='Naam *' value="<? echo $_SESSION['naam']; ?>" required/>
<input type="email" class="col-md-6 col-xs-12 Email" id="email" name='email' placeholder='Email *' value="<? echo $_SESSION['email']; ?>" required/>
<input type="text" class="col-md-12 col-xs-12 Subject" name='Subject' id='Subject' placeholder='Onderwerp'/>
<textarea type="text" class="col-md-12 col-xs-12 Message" id="vraag" name="vraag" placeholder='Bericht *' style="left: 0px; top: 0px" required><? echo $_SESSION['vraag']; ?></textarea>
            <div class="cBtn col-xs-12">
                <ul>                                
                    <li class="clear"><a href="#" onclick="resetForm(event)"><i class="fa fa-times"></i>ledig vakjes</a></li>
                    <li class="send"><a href="#" onclick="submitForm(event)"><i class="fa fa-share"></i>Verstuur bericht</a></li>
                </ul>
            </div>
        </form>
        <script>        
        function submitForm(e) {
        alert('working');
        var naam = $("#naam").val();
        var email = $("#email").val();
        var Subject = $("#Subject").val();

        $.ajax({//create an ajax request to load_page.php
        type: "POST",
       url: "verzendvraag.php",
       data: {"naam":naam,"email":email,"Subject":Subject},
       success: function(response) {
        if (response) {
        alert(response);
            alert('Successfully posted.');
        }
        else {
            alert('Successfully not posted.');
        }
    }
});
}
</script>

</body>
</html>

verzend vraag.php

 <?php session_start();    
  //Validatie serverside => bescherming inbouwen voor als iemand zijn javascript uitschakelt. Javascript echter wel nodig om de server en netwerktraffic te ontlasten. 


  if (isset($_POST))
   { 
    $naam  = $_POST['naam'];
  //$bedrijf = $_POST['bedrijf'];
   $subject = $_POST['Subject'];
   $email = $_POST['email'];
  //$vraag = $_POST['vraag'];

  //In session steken zodat we bij een eventuele terugkeer naar de form de reeds ingevulde gegevens kunnen terug zetten.
$_SESSION['naam']  = $_POST['naam'];
//$_SESSION['bedrijf'] = $_POST['bedrijf'];
$_SESSION['Subject'] = $_POST['Subject'];
$_SESSION['email'] = $_POST['email'];
//$_SESSION['vraag'] = $_POST['vraag'];

$isOK = true;
$error = '0';

$atpos = strrpos($email,"@");
$dotpos = strrpos($email,".");

if(is_null($naam) || $naam == '')
{
    $isOK = false;
    echo('NAAM NIET OK ');
    $error .= ',1';
}

if(is_null($email) || $email == '')
{
    $isOK = false;
    echo('EMAIL NIET OK '.$email);
    $error .= ',2';
}
else
{
    //checken geldig e-mailadres
    if ($atpos === false || $dotpos === false || $dotpos < $atpos+2 || $dotpos+2 >= strlen($email))  //=== moet => http://php.net/manual/en/function.strpos.php en http://www.php.net/manual/en/language.operators.comparison.php
    {
        $isOK = false;
        echo("Ongeldig e-mailadres ");
        $error .= ',3';
    }
}

   /* if(is_null($vraag) || $vraag== '')
{
    $isOK = false;
    echo('VRAAG NIET OK ');
    $error .= ',4';
}*/
}

if(!($isOK))
 {
echo("<script>location.href='index.php?error=$error'</script>");
exit;
}
else
{
echo "SUCCESS";
}
 ?>

Javascript提交不发送SUBMIT POST到php文件

4 个答案: