php 在 Wordpress 中使用 ajax 提交表单
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Submitting a form with ajax in Wordpress
提问by MariaZ
I am trying to get the results of an ajax request in wordpress, but I am getting result of '0' in an alert box of javascript, so the form looks like this:
我正在尝试在 wordpress 中获取 ajax 请求的结果，但我在 javascript 的警报框中获取结果为“0”，因此表单如下所示：
<form class="form" id="ajax-contact-form" action="#">
<input type="text" name="name" id="name" placeholder="Name" required="">
<button type="submit" class="btn">Submit</button>
</form>
The javascript looks like this:
javascript 看起来像这样：
$('#ajax-contact-form').submit(function(e){
$.ajax({
data: {action: 'contact_form'},
type: 'post',
url: ajaxurl,
success: function(data) {
alert(data); // This prints '0', I want this to print whatever name the user inputs in the form.
}
});
})
And the PHP:
和 PHP：
add_action('wp_ajax_contact_form', 'contact_form');
add_action('wp_ajax_nopriv_contact_form', 'contact_form');
function contact_form()
{
echo $_POST['name'];
}
Does anyone know if the code above is correct, I have also tried $_REQUEST['name'] and it doesnt work.
有谁知道上面的代码是否正确，我也试过 $_REQUEST['name'] 但它不起作用。
Thanks soo much,
非常感谢，
采纳答案by Sterling Archer
Try something like this, you did not add the nameparameter you are expecting in your PHP contact_formfunction, so you must add it to the dataattribute in the jQuery ajax function call.
尝试这样的事情，您没有name在PHPcontact_form函数中添加您期望的参数，因此您必须将其添加到datajQuery ajax函数调用中的属性中。
$('#ajax-contact-form').submit(function(e){
var name = $("#name").val();
$.ajax({
data: {action: 'contact_form', name:name},
type: 'post',
url: ajaxurl,
success: function(data) {
console.log(data); //should print out the name since you sent it along
}
});
});
回答by Shashank Shah
You should add an attribute for name too in your javascript.
您也应该在 javascript 中为 name 添加一个属性。
It may look like this........
它可能看起来像这样......
$('#ajax-contact-form').submit(function(e){
$.ajax({
data: {action: 'contact_form', name:name},
type: 'post',
url: ajaxurl,
success: function(data) {
alert(data);
}
});
})
})
