Php文件
<?php global $ajax_recieve;
if(isset($_POST['values'])){
$ajax_recieve = $_POST['values'];
}
$result = mysqli_query($read_info,"SELECT Namn FROM information WHERE Namn LIKE ('".$ajax_recieve.'%")";
$resultSet = array();
while($row = mysqli_fetch_array($result)) {
$resultSet[] = array( 'Name'=> $row['Namn']);
}
$json = json_encode($resultSet);
echo $json;
?>
JS文件
$(document).ready(function(){
$('#search').on('input', function() {
var inputSearch = $('#search').val();
$.ajax({
type: "POST",
contentType: "application/json; charset=utf-8",
datatype: 'json',
url: "createJSON_list.php",
data: {values: inputSearch},
success: function(response){
$.each(response, function(){
$('#json_datalist').append(response[0]);
console.log(response[0]);
});
}
});
});
});
我在console.log中看不到值。或者我只从数据库中获取最后的数据。我需要一个必须作为数组追加到html datalist的数组。
解决了
contentType:“application / json; charset = utf-8”,删除此行并更改此内容
$.each(response, function(key,Val){$('#json_datalist').append(“<value=‘“+Val.Name”’+”>”);