Question

*更新我试图为我的儿子制作一个游戏，它有2个振动传感器，代表球和罢工（棒球）。当一个传感器被击中时，我有一个循环，为击球或球加上一个勾号。当传感器被击中时，我也希望它进入我的sqlite数据库。然后我有一个php文件，它读取sqlite数据库并显示实时发生的球和罢工。我使用的是php5，sqlite3和apache2服务器。我有服务器设置，python脚本正确检测传感器并更新2个变量，球和罢工。我似乎无法将py文件链接到php文件，因此当我导航到basesloaded.php时，它会执行传感器.py。如果我首先在空闲状态下运行py脚本然后导航到它正确更新的php页面，只是不会从php文件运行py脚本。

#sensor.py file
#!/usr/bin/env python
import RPi.GPIO as GPIO
import time
import sqlite3

KnockPin = 12 
ShockPin = 13 

BtnPin = 15

inning = 1
runs = 0
strikes = 0
balls = 0

print ("Welcome to Bases Loaded!")

def setstr():
    global strikes
    strikes = 0
    global balls
    balls = 0

def setup():
    GPIO.setmode(GPIO.BOARD)         
    GPIO.setup(KnockPin, GPIO.IN, pull_up_down=GPIO.PUD_UP)
    GPIO.setup(ShockPin, GPIO.IN, pull_up_down=GPIO.PUD_UP)
    GPIO.setup(BtnPin, GPIO.IN, pull_up_down=GPIO.PUD_UP)   

def knock(ev=None):

    with sqlite3.connect('basesLoaded.db') as conn;

        global strikes
        strikes += 1

        if strikes ==0:
            pass
        elif strikes == 1:            
            conn.execute("UPDATE bl set STRIKE = 1 WHERE ID =1");
            print ("Strike 1 :", conn.total_changes);

        elif strikes == 2:
            conn.execute("UPDATE bl set STRIKE = 2 WHERE ID=1");
            print ("Strike 2 :", conn.total_changes);

        elif strikes == 3:
            conn.execute("UPDATE bl set STRIKE = 0 WHERE ID=1");
            print ("Strike 3 Your Out :", conn.total_changes);

            global inning
            inning +=1
            conn.execute("UPDATE bl set INNING = inning WHERE ID=1");
            setstr()
            print ("Inning Number :", inning);

def shock(ev=None):
    with sqlite3.connect('basesLoaded.db') as conn;
        print ("Opened database successfully");
        global balls
        balls += 1

        if balls ==0:
            pass
        elif balls == 1:          
            conn.execute("UPDATE bl set BALL = 1 WHERE ID=1");
            print ("Ball 1 :", conn.total_changes);

        elif balls == 2:
            conn.execute("UPDATE bl set BALL = 2 WHERE ID=1");
            print ("Ball 2 :", conn.total_changes); 

        elif balls == 3:
            conn.execute("UPDATE bl set BALL = 3 WHERE ID=1");
            print ("Ball 3 :", conn.total_changes);

        elif balls == 4:

            global runs
            runs += 1
            conn.execute("UPDATE bl set BALL = 0 WHERE ID=1");
            conn.execute("UPDATE bl set RUN = 'runs' WHERE ID=1");
            print ("Run Scored! score is :", runs); 
            setstr()

def loop():

    GPIO.add_event_detect(ShockPin, GPIO.FALLING, callback=shock, bouncetime=2500)

    GPIO.add_event_detect(KnockPin, GPIO.FALLING, callback=knock, bouncetime=2500) 

if __name__ == '__main__':     # Program start from here
    setup()
    try:
        loop()

    except KeyboardInterrupt: 
        destroy()

这是我的php脚本来检索sqlite数据 -

#basesLoaded.php file
<?php
$try = exec('python sensor.py');
print_r ($try);
$db = new SQLite3('basesLoaded.db');

echo '<br>';
echo '<br>';
$strikes = ($db->querySingle('SELECT STRIKE FROM bl WHERE ID = 1'));
$balls = ($db->querySingle('SELECT BALL FROM bl WHERE ID = 1'));
$outs = ($db->querySingle('SELECT OUT FROM bl WHERE ID = 1'));
$runs = ($db->querySingle('SELECT RUN FROM bl WHERE ID = 1'));
$inning = ($db->querySingle('SELECT INNING FROM bl WHERE ID = 1'));


echo 'Strikes: ' . $strikes . '<br>';
echo 'Balls: ' . $balls . '<br>';
echo 'Outs: ' . $outs . '<br>';
echo 'Runs: ' . $runs . '<br>';
echo 'Inning: ' . $inning . '<br>';
?>

Answer 1

如@Parfait的评论中所述，您需要提交更新以使其持久化。您的PHP代码没有看到更新，因为它们没有被提交。

您可以在代码中调用conn.commit()，但是，保证更新已提交（或在出现错误时回滚）的简单方法是使用context manager打开数据库，即使用with语句，例如：

def knock(ev=None):
    with sqlite3.connect('basesLoaded.db') as conn:
        conn.execute("UPDATE bl set STRIKE = 1 WHERE ID =1")

现在，当with语句终止时，您的更新将自动提交，在这种情况下，当函数knock()返回时，或者在发生异常时回滚。

<强>更新

正如评论中所讨论的，事件处理似乎存在问题。我怀疑你的代码需要显式继续运行，否则，一旦主线程终止，整个程序将终止。你可以这样做：

WAIT_PIN = 14

def register_callbacks():
    GPIO.add_event_detect(ShockPin, GPIO.FALLING, callback=shock, bouncetime=2500)
    GPIO.add_event_detect(KnockPin, GPIO.FALLING, callback=knock, bouncetime=2500) 

def wait():
    GPIO.wait_for_edge(WAIT_PIN, GPIO.BOTH)    # wait for level change

#    # Or wait in a while loop        
#    while True:
#        time.sleep(0.1)

#    # Or wait for keyboard input
#    input('Press <ENTER> to quit: ')

if __name__ == '__main__':
    try:
        setup()
        register_callbacks()
        wait()    # block the main thread
    except KeyboardInterrupt: 
        destroy()

您需要始终运行此Python脚本。它独立于PHP脚本，该脚本仅在通过Web服务器响应HTTP请求时运行。您可以将Python脚本作为后台进程运行：

$ python sensor.py &

此外，从PHP脚本中删除exec()调用。

使用python将数据添加到sqlite3，然后php5使用python，php5，sqlite3动态检索apache2服务器上的数据

1 个答案: