我有一个通过AJAX调用PHP的页面
:: 1 - - [24 / Oct / 2016:10:30:17 -0200]" POST /Media_Database/Pages/_copyFile.php HTTP / 1.1" 200 167 " http://localhost/Media_Database/Pages/menu.htm" " Mozilla / 5.0(Windows NT 10.0; WOW64)AppleWebKit / 537.36(KHTML,和Gecko一样) Chrome / 53.0.2785.148 Safari / 537.36 Vivaldi / 1.4.589.38"
...和Chrome开发者工具显示
点击" TEST 2"显示来自PHP的消息...但它不是我想要的,因为没有参数进入PHP
jquery lib是本地的,正在加载
<script src="_jquery-11.0.min.js"></script>
<script>
$(document).ready(function(){
$("#saveFavorites").click(function(){
var mp3src = document.getElementById("mp3player").src;
$.ajax({
type: "post",
url:"_copyFile.php",
data: { "varPath" : mp3src },
}).success(function(result){
console.log('AJAX success');
}).fail(function(jqXHR, textStatus, errorThrown){
console.log('AJAX failed' + textStatus);
});
});
});
</script>
<a href="#" id="saveFavorites"> TEST 1 </a>
<a href="_copyFile.php"> TEST 2 </a>
<?php echo '<script type="text/javascript"> alert("CALL OK");</script>'; ?>
使用Chrome作为客户端和XAMPP