Python 函数使用循环生成查询字符串，只返回最后一个循环生成字符串

我有一个函数，用于为sql查询构建字符串

def create_text(tablelist=None, fieldlist=None):
alphalist = []
alpha = 'a'
for i in range(0, len(tablelist)): 
    alphalist.append(alpha) 
    alpha = chr(ord(alpha) + 1) 

for i,v in enumerate(alphalist):
    if i + 1 == len(alphalist):
        last=True
    else:
        last=False
    y=str(v)
    text ="SELECT "+', '.join([y+ '.' +x for x in fieldlist]) + ' FROM ' \
    +str(tablelist[i]) + " as " + str(y)  
    if not last:
        text = text + " UNION ALL"
    print(text)
return text

我这样称呼它

text = create_text(tablelist = ['table1', 'table2'], fieldlist = ['v1','v2'])

它在第二个循环后打印字符串时给我这个输出

SELECT a.v1, a.v2 FROM table1 as a UNION ALL
SELECT b.v1, b.v2 FROM table2 as b

但是，“text”本身的值是从函数返回的，如果我只是打印它的话

print(text)

是

这只是循环中的最后一个值，因为它覆盖了文本的值。我只是不知道如何通过循环并在每次迭代中连接文本，以便得到我想要的返回值，这是文本的值=

"SELECT a.v1, a.v2 FROM table1 as a UNION ALL SELECT b.v1, b.v2 FROM table2 as b"

---

谢谢

您的代码在循环中重新定义了

文本

，因此只会返回最后的循环迭代文本

您可以对代码进行大量清理，并得出以下结论：

def create_text(tablelist=None, fieldlist=None):
    # collects parts to be str.joined later
    parts = []

    # iterate the tablenames directly
    for i, table in enumerate(tablelist):
        # no neeed to precompute, simply use i to get a new alias
        alias = chr(ord('a')+i)  
        # string interpolation style formating for bonus points 
        fields = ', '.join( f"{alias}.{x}" for x in fieldlist)
        # collect in list
        parts.append( f"SELECT {fields} FROM {table} as {alias}")

    # returned joined list
    return " UNION ALL\n".join(parts)

text = create_text(tablelist = ['table1', 'table2'], fieldlist = ['v1','v2'])

print(text)

得到

SELECT a.v1, a.v2 FROM table1 as a UNION ALL
SELECT b.v1, b.v2 FROM table2 as b

每次需要连接文本时，将它们收集到一个列表中，然后使用

str.join（）

-这样连接字符串不需要中间字符串

SELECT a.v1, a.v2 FROM table1 as a UNION ALL
SELECT b.v1, b.v2 FROM table2 as b