drqja5919276 2017-10-24 18:37
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Ajax PHP表单按钮[重复]

This question already has an answer here:

Can anyone tell me what I am missing here please?

Here is the basic HTML file with a simple "Create" and "Destroy" button that will eventually call an API. The button click with Ajax seems to work, but the PHP function is not being called.

testapi.html

<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
    $('.button').click(function(){
        var clickBtnValue = $(this).val();
        var ajaxurl = 'ajax.php',
        data =  {'action': clickBtnValue};
        $.post(ajaxurl, data, function (response) {
            alert("Action performed successfully!");
        });
    });

});
</script>
</head>
<body>
    <input type="submit" class="button" name="create" value="Create" />
    <input type="submit" class="button" name="destroy" value="Destroy" />
</body>
</html>

ajax.php

if (isset($_POST['action'])) {
    switch ($_POST['action']) {
        case 'create':
            create();
            break;
        case 'destroy':
            destroy();
            break;
    }
}
function create() {
    echo "Create function.";
    exit;
}
function destroy() {
    echo "Destroy function.";
    exit;
}
</div>
  • 写回答

1条回答 默认 最新

  • dreamer1231 2017-10-24 19:23
    关注

    Your case is wrong: the value in your input is Create, while in your code you check for create.

    Change

    switch ($_POST['action']) {
            case 'create':
                create();
                break;
            case 'destroy':
                destroy();
                break;
        }
    

    to

    switch ($_POST['action']) {
            case 'Create':
                create();
                break;
            case 'Destroy':
                destroy();
                break;
        }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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