Jquery提交并发布一个无法返回数据的表单

i have problem with jquery submit and post. I would like to submit a form and post it to my php for a check, and return the php result back to the form. However, I have problem with the returning of data, it basically ignores the result.

我有jquery提交和发布的问题。我想提交一个表单并将其发布到我的php进行检查，并将php结果返回给表单。但是，我有数据返回的问题，它基本上忽略了结果。

This is my html form:

这是我的html表单：

<form id="form" method="post">
    <p id="status">Status:</p>
    <p class="text">
        <label for="name" class="label">Name:</label>
        <input type="text" id="name" name="name" value="" size="30" />
    </p>
    <p class="text">
        <label for="email" class="label">Email:</label>
        <input type="text" id="email" name="email" value="" size="30" />
    </p>
    <p class="submit">
        <input type="submit" name="send_btn" id="send_btn" value="Send" />
    </p>
</form>

This is my javascript to do the submit and post:

这是我的javascript做提交和发布：

$('#form').submit(function(e) {
    e.preventDefault();

        var name = $('#name').val();
        var email = $('#email').val();

        $.post('notify.php', {name: name, email: email}, function(data) {
            $('#status').html(data);
        });

});

This is the php that does the check and return the data:

这是检查并返回数据的php：

<?php 
if (isset($_POST['name'], $_POST['email']))
{
    $name = htmlentities($_POST['name']);
    $email = htmlentities($_POST['email']);

    if ($name == "myname")
    {
        $output = 'It matches!';
    }
    else
    {
        $output = 'No matches!";
    }
}
?>

Can please highlight what has gone wrong? Thank you.

可以请突出显示出错的地方吗？谢谢。

4 个解决方案

#1

You need to echo or die in your php script, so your function can get the results.

你需要在php脚本中回显或消亡，这样你的函数才能得到结果。

So, change your script to this:

所以，将脚本更改为：

<?php 
if (isset($_POST['name'], $_POST['email']))
{
    $name = htmlentities($_POST['name']);
    $email = htmlentities($_POST['email']);

    if ($name == "myname")
    {
        $output = 'It matches!';
    }
    else
    {
        $output = 'No matches!';
    }

    echo $output;
}
?>

Notice the third to last line, where I am calling echo $output - whatever you echo will be returned from your ajax call. If you want to get more complex, you should return a JSON object.

注意第三行到最后一行，我调用echo $ output - 无论你回应什么，都会从你的ajax调用中返回。如果你想变得更复杂，你应该返回一个JSON对象。

$results = array("result" => "It Matches!", "foo" => "bar");
echo json_encode($results);

EDIT: You also need to change the " to a ' at the end of your else.

编辑：您还需要在其他人的末尾更改“到a”。

#2

Print the answer on PHP as comment suggest echo $output;

在PHP上打印答案作为评论建议echo $ output;

at the end of the line

在行尾

#3

In your php code, you aren't actually writing $output to the page.

在您的PHP代码中，您实际上并没有将$ output写入页面。

<?php
    // your code

    echo $output;
?>

#4

dont forget

别忘了

return false;

返回虚假;

If u forget it, the form when submit will refesh and ajax will fail ;

如果你忘了它，提交时的表格将会改写，而ajax将会失败;

$('#form').submit(function(e) {
    e.preventDefault();

        var name = $('#name').val();
        var email = $('#email').val();

        $.post('notify.php', {name: name, email: email}, function(data) {
            $('#status').html(data);
        });
    return false;       
});

#1