我是phone gap的新手，尝试调用在php中创建的rest web服务，并在我的phone gap android应用程序中使用application/x-www-form-urlencoded中的post参数，但没有得到响应。以下是调用该服务的代码：
$.ajax({
type: "POST",
url: "URL.php",
contentType: "application/x-www-form-urlencoded",
data: dataString,
success: function(response) {
var resp = response.responseText;
var jsonObj = JSON.parse(resp);
console.log("Success: " + jsonObj);
},
error: function(request, status, error) {
console.log("Error status " + status);
console.log("Error request status text: " + request.statusText);
console.log("Error request status: " + request.status);
console.log("Error request response text: " + request.responseText);
console.log("Error response header: " + request.getAllResponseHeaders());
}
});
我得到[信息：控制台（1）]“未捕获的语法错误：意外的令牌u”，来源：file:///android_asset/www/a.html (1).
如果有好的教程/例子，请告诉我。
谢谢
因为您只是在Phonegap中开发和应用程序，所以在页面中的div标记中打印响应，而不是使用控制台。有关如何使用jquery发送请求，您可以查看以下示例：http://labs.jonsuh.com/jquery-ajax-php-json/
终于我完成了！
$.ajax({
type: "POST",
url: "http://www.url.php",
contentType: "application/x-www-form-urlencoded",
data: dataString,
success: function(response) {
//entered in the success block means our service call is succeeded properly
var resp = JSON.stringify(response.text); // we are accessing the text from the json object(response) and then converting it in to the string format
console.log(JSON.stringify(response)); // print the response in console
alert(resp); // alert the response
},
error: function(request, status, error) {
console.log("Error status " + status);
console.log("Error request status text: " + request.statusText);
console.log("Error request status: " + request.status);
console.log("Error request response text: " + request.responseText);
console.log("Error response header: " + request.getAllResponseHeaders());
}
});
