如何修复关于admin-ajax.php的400(错误请求)错误

Question

我想通过JS传递html值到PHP处理，但我仍然得到错误，400(错误请求)。

这是HTML:

</div>
<form class="row g-3" id="my_form">
<div class="col-md-12">
<label for="url" class="form-label">Input URL</label>
<input type="text" class="form-control my_info" id="url" required>

<div class="col-12">
<button class="btn btn-primary submit" type="button">Submit form</button>
</div>
</form>
</div>

这是JavaScript:

function getBaseUrl() {
var re = new RegExp(/^.*//);
var my_local_url = re.exec(window.location.href)[0];
return my_local_url;
}
console.log(getBaseUrl())
jQuery(document).ready(function($) {
console.log('test');
// We'll pass this variable to the PHP function example_ajax_request
var my_data = {
'action':'crawler_info',
'my_info' : document.getElementById('url').value
};
$('.submit').click(function (){
// This does the ajax request
$.ajax({
url: getBaseUrl() + 'admin-ajax.php',
data: my_data,,
success:function(data) {
// This outputs the result of the ajax request
console.log(data);
},
error: function(errorThrown){
console.log(errorThrown);
}
}); 
})
});

这是PHP:

function get_info() {
// The $_REQUEST contains all the data sent via ajax 
if ( isset($_REQUEST) ) {
$infos = $_REQUEST['my_info'];
// Now we'll return it to the javascript function
// Anything outputted will be returned in the response
echo $infos;
// If you're debugging, it might be useful to see what was sent in the $_REQUEST
print_r($_REQUEST);
}
// Always die in functions echoing ajax content or it will display 0 or another word
die();
}
add_action( 'wp_ajax_crawler_info', 'get_info' );
add_action( 'wp_ajax_nopriv_crawler_info', 'get_info' );

我尝试了很多方法来修复它，但它仍然显示这个错误。有没有人知道如何解决我的问题，谢谢。

Answer 1

我已经修复了我的问题，一些代码放在错误的地方，我发现