所以我想用AJAX给出一个简单的建议,任务工作正常,但它不会用POST读取输入的值...我可以验证如果isset,索引正在工作......但我甚至不能_POST输入......
ajax的JQ post方法:
$(document).ready(function(){
$("input").keyup(function(){
var name = $("input").val();
$.post("suggestions.php", {
suggestion: name
}, function(data, status){
$("#test").html(data);
});
});
});
输入名称和#test paraghaph在那里,他们工作
suggestions.php:
$existingNames = array("Daniel","Dennis", "Alex");
if (isset($_POST['suggestion']))
{
$name = $_POST['suggestion'];
//$name = "D" ( if i uncomment this it will show Daniel and Denis)
foreach ($existingNames as $exista) {
if(strpos($exista , $name) !== false){
echo $exista;
echo "<br>";
}
}
}
仅仅因为它的设定,并不意味着它不是empty
if (isset($_POST['suggestion'] && ! empty($_POST['suggestion'])) {
// your code here
}