通过AJAX传递JSON加按钮Dilema
问题描述:
所以我试图传递一些JSON对象的信息,并有一个PHP页面插入数据到数据库。但是,我遇到了一些麻烦。 “更新”按钮存在于一个弹出窗口中。然后用户点击“更新”,输入的数据应该被相应地处理。但是,我担心我甚至没有达到我的.click功能。我的警报似乎没有被触发。下面我会指出问题的发生地点。谢谢!通过AJAX传递JSON加按钮Dilema
<script>
function updateTable()
{
document.getElementById("testLand").innerHTML = "Post Json";
//echo new table values for ID = x
}
$('#update').click(function() {
alert("help!");
var popupObj = {};
popupObj["Verified_By"] = $('#popupVBy').val();
popupObj["Date_Verified"] = $('#popupDV').val();
popupObj["Comments"] = $('#popupC').val();
popupObj["Notes"] = $('#popupN').val();
var popupString = JSON.stringify(popupObj);
alert(popupString);
#.ajax({
type: "POST",
dataType: "json",
url: "popupAjax.php",
//data: 'popUpString = '+ popupString,
data: popupObj,
cache: false,
success: function(data) {
updateTable();
alert("testing tests");
}
});
});
</script>
<html>
<button onClick="openPopup(<?php echo $row['ID'];?>);"><?php echo $row['ID'];?></button> <!--opens a popup with input options-->
<button id="update">Update</button> <!-- this button is supposed to cause the javascript above to run when clicked, however none of my alerts seem to be reached.-->
</html>
谢谢你看!
看到代码在这里工作:http://jsbin.com/zocekohalo/1/edit?html,js,output – 2014-10-30 22:48:08