問題描述

現(xiàn)在遇到的問題是：得到用戶輸入的數(shù)據(jù)后，怎么根據(jù)用戶的數(shù)據(jù)去結(jié)合公式，從而計算出距離，返回結(jié)果。目前有兩個文件，一個是py

一個是html.

我在anaconda-jupyter運行過這個公式，是可以得出距離的。可是jupyter運行是因為上面有運行過的，所以有數(shù)據(jù)可以運行。

然而現(xiàn)在我不知道要在哪里填寫，填寫什么才可以將機場的數(shù)據(jù)輸入到公式，距離的結(jié)果可以返回。

以下是部分ppy代碼

def searchcities() -> ’html’:airportone = request.form[’user_airportone’]airporttwo = request.form[’user_airporttwo’]distanceone = calcDistance[’distance’]return render_template(’results.html’, the_title = ’以下是您選取的機場：’, the_airportone = airportone, the_airporttwo = airporttwo, the_distance = distanceone )

以下是部分html代碼

from math import radians, cos, sin, atan, acos,tandef calcDistance(a1,a2):ra = 6378.140 # 赤道半徑 (km)rb = 6356.755 # 極半徑 (km)Lat_A = airportone[’latitude’]Lng_A = airportone[’longitude’]Lat_B = airporttwo[’latitude’]Lng_B = airporttwo[’longitude’]flatten = (ra - rb) / ra # 地球扁率rad_lat_A = radians(Lat_A)rad_lng_A = radians(Lng_A)rad_lat_B = radians(Lat_B)rad_lng_B = radians(Lng_B)pA = atan(rb / ra * tan(rad_lat_A))pB = atan(rb / ra * tan(rad_lat_B))xx = acos(sin(pA) * sin(pB) + cos(pA) * cos(pB) * cos(rad_lng_A - rad_lng_B))c1 = (sin(xx) - xx) * (sin(pA) + sin(pB)) ** 2 / cos(xx / 2) ** 2c2 = (sin(xx) + xx) * (sin(pA) - sin(pB)) ** 2 / sin(xx / 2) ** 2dr = flatten / 8 * (c1 - c2)distance = ra * (xx + dr)return distanceairportone = request.form[’user_airportone’]airporttwo = request.form[’user_airporttwo’]calcDistance(airportone,airporttwo)

懇求幫助，謝謝。

問題解答

def searchcities() -> ’html’:airportone = request.form[’user_airportone’]airporttwo = request.form[’user_airporttwo’]distanceone = calcDistance[’distance’]return render_template(’results.html’, the_title = ’以下是您選取的機場：’, the_airportone = airportone, the_airporttwo = airporttwo, the_distance = distanceone )

這一部分不就是獲取 機場 然后計算結(jié)果，并且返回距離的么距離顯示在 result.html 頁面，這不都寫完了么