PHP submitting a form with multiple submit buttons

2019-08-06 04:06发布

站内问答 / PHP
570 3 5

I have the following HTML code:

<form method="post" action="the_file.php" id="the-form">
    <input type="text" name="the_input" value="<?php if ( isset( $_POST['the_input'] ) echo $_POST['the_input]; ?>">
    <button type="submit" name="eat_something" value="TRUE">Eating</button>
    <button type="submit" name="eat_something" value="FALSE">Don't Eat</button>
</form>
<textarea id="result"></textarea>

Followed by this JS:

$('#the-form').bind('submit', submitForm);
function submitForm(evt) {
    jQuery.post(
        $(this).attr('action'),
        $(this).serialize(),
        function(data) {
            $('#result').empty().append(data).slideDown();
        });
    evt.preventDefault();
}

I also have a PHP script that receives the $_POST value from the input on submit and runs a conditions to test which submit button was clicked.

Like this:

$input = $_POST['the_input'];
$eating = $_POST['eat_something'];
if ( $eating == 'TRUE' ) {
    // Do some eating...
} else {
    // Don't you dare...
}

If I don't use the jQuery.post() function the submit values from the button are posted. However, for some reason, I can't manage to pass the button value to PHP $_POST with the jQuery.post() function. If I don't use jQuery.post() the output doesn't get appended to the textarea but rather in a text format on a separate page, like a document. I've also tried calling the submit function on the button $('button[type=submit]').bind('submit', submitForm); but this doesn't solve my problem either.

Thanks in advance for you help.

3条回答
Emotional °昔
2楼-- · 2019-08-06 04:32

Simplest answer would be:

<form method="post" action="the_file.php" id="the-form">
    <input type="text" name="the_input" value="<?php if ( isset( $_POST['the_input'] ) echo $_POST['the_input']; ?>">
    <input type="submit" name="eat_something" value="Eating" />
    <input type="submit" name="eat_something" value="Don't Eat" />
</form>

Of course, this won't give exactly what you're looking for.

You can also do something like this:

<form method="post" action="the_file.php" id="the-form">
    <input id="theInput" type="text" name="the_input" value="<?php if ( isset( $_POST['the_input'] ) echo $_POST['the_input']; ?>">
    <button type="button" name="eat_something" data-value="TRUE">Eating</button>
    <button type="button" name="eat_something" data-value="FALSE">Don't Eat</button>
</form>

$('#the-form button').bind('click', function(){
    jQuery.post($('#the-form').attr('action'),
    {
        the_input: $('#theInput').val(),
        eat_something: $(this).attr('data-value')
    },
    function(data) { $('#result').empty().append(data).slideDown() },
    'json'
});
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可以哭但决不认输i
3楼-- · 2019-08-06 04:45

Change the buttons to this (change w/your form names).. 

<input type="hidden" name="eat_something" value="" />
<input type="button" onclick="$('input[name=eat_something]').val('TRUE')" />
<input type="button" onclick="$('input[name=eat_something]').val('FALSE')" />

The Javascript function and the PHP are fine.

Thanks!

@leo

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乱世女痞
4楼-- · 2019-08-06 04:48

You forget signle quote and ) in the_name input.

<input type="text" name="the_input" value="<?php if (isset($_POST['the_input'])) { echo $_POST['the_input'] ; } ?>">

and for getting form button pressed value you need to append it value manually to serialize.

$form.serialize() + "&submit="+ $('button').attr("value")

Example

<script type="text/javascript">
    $(function()
    {
        $('button[type="submit"]').on('click',function(e)
        {
            e.preventDefault();
            var submit_value = $(this).val();
            jQuery.post
            (
                $(this).attr('action'),
                $(this).serialize()+ "&submit="+ submit_value,
                function(data)
                {
                    $('#result').empty().append(data).slideDown();
                }
            );
        });
    });
</script>

Complete Tested Code

<?php
    if(isset($_POST['the_input']))
    {
        $input = $_POST['the_input'];
        $eating = $_POST['eat_something'];
        exit;
    }
?>
<!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" xml:lang="en">
<head>
    <meta http-equiv="Content-Type" content="text/html;charset=UTF-8" />
    <title>StackOverFlow</title>
    <script type="text/javascript" src="js/jquery.js"></script>
    <script type="text/javascript">
        $(function()
        {
            $('button[type="submit"]').on('click',function(e)
            {
                e.preventDefault();
                var submit_value = $(this).val();
                jQuery.post
                (
                    $('#the-form').attr('action'),
                    $('#the-form').serialize()+ "&eat_something="+ submit_value,
                    function(data)
                    {
                        $('#result').empty().append(data).slideDown();
                    }
                );
            });
        });
    </script>
</head>
<body>
    <form method="post" action="random.php" id="the-form">
        <input type="text" name="the_input" value="<?php if (isset($_POST['the_input'])) { echo $_POST['the_input'] ; } ?>">
        <button type="submit" name="eat_something" value="TRUE">Eating</button>
        <button type="submit" name="eat_something" value="FALSE">Don't Eat</button>
    </form>
    <textarea id="result"></textarea>
</body>
</html>
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