Php ajax allows you to get the weather status without refreshing

First, we need to understand that we cannot do the weather forecast function by ourselves. Here we only need to call the data returned by the api, the following is an example of how to call the api of www.weather.com.cn. Weather has become an indispensable topic in our life and is closely related to our life. On the right side of my blog, I used php + ajax to create a weather query module.

The ideal status should be that the user automatically obtains local weather information based on different access locations, but the current technology is limited, and only manual query can be completed. This is much simpler, and too many technologies are not used. The main reason is that ajax is used to call an open interface and then process the returned json data.

Interface address: http://www.weather.com.cn/data/cityinfo/101200101.html

Returned value: {"weatherinfo": {"city": "Wuhan", "cityid": "101200101", "temp1": "28 ℃", "temp2 ": "36 ℃", "weather": "clear to cloudy", "img1": "n0.gif", "img2": "d1.gif", "ptime "}}

The interface address part is "101200101". The ID string is the city id. The ID corresponding to my Baidu city is encapsulated into an array and can be called directly when used. There are not many core code, mainly because the city -- the ID is relatively large, so I will not post the source code, just package it and share it directly. A friend in need can download it directly!

Some code

The Code is as follows:

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<Meta http-equiv = "Content-Type" content = "text/html; charset = UTF-8"/> <Script type = "text/javascript" src = "jquery. js"> </script> <Script type = "text/javascript"> $ (Function (){ $ ("# Submit"). click (function (){ // Send an ajax request Var city = $ ("# city"). val (); $. Post ("getweather. php", {city: city}, function (data ){ If (data. weatherinfo. city ){ Var city = data. weatherinfo. city; // city name Var temp1 = data. weatherinfo. temp1; // maximum temperature Var temp2 = data. weatherinfo. temp2; // minimum temperature Var weather = data. weatherinfo. weather; // weather description ("clear to cloudy ") Alert (city + ":" + weather + "," + temp2 + "-" + temp1 ); Return; } Else { Alert ("the city is not found "); } }, "Json "); }); }); </Script>

Getweather. php file

The Code is as follows:

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<Form method = "post"> Enter the city: <input type = "text" name = "city" id = "city" value = "Wuhan"/> <Input type = "button" name = "sub" id = "submit" value = "view weather"/> </Form> Is, <? Php Include "citycode. php "; $ City = $ _ POST ['city']; $ Citycode = @ $ citycode [$ city]; // Echo "shibushi "; If (empty ($ citycode )){ Echo "the city you entered is out of the range "; } Else { Echo file_get_contents ("http://www.weather.com.cn/data/cityinfo/". $ citycode. ". html "); } ?>

Test Results

Source code download: php ajax implements refreshing to get the weather status Source Code download: