Question

我正在研究我的大学项目，因为我有一个应该显示城市和温度的网页，我正在尝试使用AJAX在某些时间间隔更新温度。这是代码

对于ajax请求Java脚本，我在这个答案https://stackoverflow.com/a/18243161/4341530

中看到了这段代码

function postAjaxRequest(qry){
alert("Ajax response on " +qry);
$.ajax({
type: "POST",
url: "ajax_resp.php",
data: {q: qry},
dataType:'JSON', 
success: function(response){
    dispatchCallBack(qry,response);
}});}

alert("Ajax response on " +qry);工作正常，所以我知道这段代码正在运行。但是，当我把这种＆＃34; alert（）＆＃34;就在dispatchCallBack(qry,response);之前它不起作用，所以我在此推断请求不成功。

此处ajax_resp.php

<?php

if(isset($_POST['q'])&&!empty($_POST['q'])
{
    if($_POST['q']=="c_temp")
    {
        $api_key = "Here's my api key, working alright"; 
        $city_id = "here's city id, also working correct";
        $url = "http://api.openweathermap.org/data/2.5/forecast/city?id=".$city_id."&APPID=".$api_key;

        $json = file_get_contents($url);
        $json_o = json_decode($json);

        $city_name =  $json_o->city->name;
        $temp_c =  $json_o->list[0]->main->temp-273;
        echo json_encode(array("city"=>$city_name,"temp"=>$temp));
//I have checked JSON response seperately its correct. I was able to get city name and temperature variable properly initialised.
    }
}?>

所以我无法弄清楚出了什么问题。我是新手：p可能是一些愚蠢的错误，如果指出会很棒。

Answer 1

请查看以下代码并更正

[1]使用单引号或双引号从$_POST变量中获取具有字符串索引的数据

[2] $qry之前未定义。在阅读问题之后，$ qry的最可能值可能是$ _POST ['q']。因此，将$qry替换为$_POST['q']或指定$qry = $_POST['q']

if(isset($_POST['q'])&&!empty($_POST['q'])
{
if($_POST['q']=="c_temp") <==  $_POST['q'] instead of $qry
{
    $api_key = "Here's my api key, working alright"; 
    $city_id = "here's city id, also working correct";
    $url = "http://api.openweathermap.org/data/2.5/forecast/city?id=".$city_id."&APPID=".$api_key;

    $json = file_get_contents($url);
    $json_o = json_decode($json);

    $city_name =  $json_o->city->name;
    $temp_c =  $json_o->list[0]->main->temp-273;
    echo json_encode(array("city"=>$city_name,"temp"=>$temp));
    //I have checked JSON response seperately its correct. I was able to get city name and temperature variable properly initialised.
}
}?>

为什么php ajax没有给出任何回复？

1 个答案: