這裏的中間就是我有這麼遠 $awards_sql_1 = mysql_query('SELECT * FROM categories WHERE section_id = 1') or die(mysql_error());
$awards_sql_2 = mysql_query('SELECT * FROM categories WHERE section_id = 2') or die(my
好難倒了,所以我也跟着在這個問題上的意見:Stumped in the middle of a PHP loop 但現在,我有比以前更多的問題。這是我到目前爲止有: $sql = "SELECT section_name, category_name, result_level, url, winner FROM 2009_RKR_bestof INNER JOIN categories ON 2
我有一個使用一個簡單的while循環顯示進度條,但似乎並不奏效,因爲我預料的腳本: count = 1
maxrecords = len(international)
p = ProgressBar("Blue")
t = time
while count < maxrecords:
print 'Processing %d of %d' % (count, maxrecords
我是一名Python編程初學者。我寫了下面的程序,但沒有按照我的要求執行。這裏是代碼: b=0
x=0
while b<=10:
print 'here is the outer loop\n',b,
while x<=15:
k=p[x]
print'here is the inner loop\n',x,
x=x+1
b=b