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我遇到了一些问题,显示通过ajax请求从php文件传递的错误消息。jQuery Ajax Calculator问题
问题是,当我收到少于3个应该显示在无序列表中的错误消息时。我不想显示任何“未定义”变量 - 这就是为什么我没有通过一条语句将它们添加到无序列表中的原因。
有什么建议吗?
calculator.html
<div id="calculator">
<form>
<input type="text" name="years"/>
<label>How many years have you smoked for?</label><br />
<input type="text" name="smokes"/>
<label>How many cigarettes do you smoke each day?</label><br />
<input type="text" name="price"/>
<label>How much do you spend on each pack of cigarettes?</label><br />
<input type="submit" value="Show My Totals" name="submit" />
</form>
<div id="result"></div>
</div><!-- end calculator -->
calculator.js
$(function(){
$("#calculator form").submit(function(){
dataString = $("#calculator form").serialize();
$.ajax({
type: "POST",
url: "./calc.php",
data: dataString,
dataType: "json",
success: function(data) {
if(data.totalSmoked != null && data.totalSmoked != undefined &&
data.totalCost != null && data.totalCost != undefined &&
data.totalWeight != null && data.totalWeight != undefined)
{
$("#calculator #result").html("<ul><li>" + data.totalSmoked + "</li><li>" + data.totalCost + "</li><li>" + data.totalWeight + "</li></ul>");
}
else
{
$("#calculator #result").html("<ul>");
if(data.yearsError != null && data.yearsError != undefined) {
$("#calculator #result ul").html("<li>" + data.yearsError + "</li>");
}
if(data.smokesError != null && data.smokesError != undefined) {
$("#calculator #result ul").html("<li>" + data.smokesError + "</li>");
}
if(data.costError != null && data.costError != undefined) {
$("#calculator #result ul").html("<li>" + data.costError + "</li>");
}
$("#calculator #result").html("</ul>");
/**** If I replace the above content within the 'else' block, it works however - This will display an "undefined" variable if there are less than 3 errors ****/
//$("#calculator #result").html("<ul><li>" + data.yearsError + "</li><li>" + data.smokesError + "</li><li>" + data.costError + "</li></ul>");
}
}
});
return false;
});
});
Calculator.php中
$yearsSmoked = $_POST['years'];
$smokesPerDay = $_POST['smokes'];
$costPerPack = $_POST['price'];
$errors = array();
$results = array();
/**
*
*
* A bunch of error checking here...
*
*
*/
if(!empty($results)) {
echo json_encode($results);
}
elseif(!empty($errors)) {
echo json_encode($errors);
}
UPDATE
我想它不起作用,因为我在我的PHP文件中指定了数组键。把这个添加到我的else块中工作...有种丑陋的方式。
if(data['yearsError'] != undefined) {
$("#calculator #result ul").append("<li>" + data['yearsError'] + "</li>");
}
if(data['smokesError'] != undefined) {
$("#calculator #result ul").append("<li>" + data['smokesError'] + "</li>");
}
if(data['costError'] != undefined) {
$("#calculator #result ul").append("<li>" + data['costError'] + "</li>");
}
感谢您的提示马克。我已经尝试了for循环,并且只是再次尝试并在我的错误控制台中收到此消息:data.errors未定义。 我对json不太熟悉,似乎无法以这种方式访问数组。 – 2010-08-25 04:06:18
尝试用'alert()'或'console.log'吐出返回的json数据并查看实际返回的内容。 – 2010-08-25 13:55:50