【发布时间】:2014-12-02 09:39:58
【问题描述】:
index.php
<form id="myForm" action="test2.php" method="POST">
<input type="submit" value="Print 1" name="submit" id="submit">
<input type="submit" value="Print 2" name="submit2" id="submit2">
</form>
<div id="ack"></div>
这是我的脚本
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script type="text/javascript" src="scripts/login_script.js"></script>
test2.php
<?php
if(isset($_POST['submit'])){
echo "1"; // i want to show this everytime i click submit (not working)
} // or what is the proper way to do this..
if(isset($_POST['submit2'])){
echo "2"; // i want to show this everytime i click submit2 (not working)
} // or what is the proper way to do this..
?>
login_script.js
$("#submit, #submit2").click( function() {
$.post( $("#myForm").attr("action"),
$("#myForm :input").serializeArray(),
function(data) {
$("#ack").empty();
$("#ack").html(data);
});
$("#myForm").submit( function() {
return false;
});
});
如果我删除 if(isset($_POST['submit'])).... 它的工作... 那么这样做的正确方法是什么..? 抱歉,我是 ajax 和 jquery 中的菜鸟
【问题讨论】:
-
确保第二个输入的
name=也是submit2 -
仍然无法正常工作,先生,我认为问题出在 if(isset($_POST['submit2'])){...?你有什么想法吗?
标签: javascript php jquery ajax