以下python代码片段中的错误是什么?

时间:2015-01-01 18:03:55

标签: python-2.7 machine-learning

我正在阅读关于编程集体智慧的搜索引擎的章节,并且遇到了以下代码片段并尝试在IPython中实现它。但是,我遇到了错误。请帮助。

def getmatchrows(self,q):
    fieldlist='w0.urlid'
    tablelist=''
    clauselist=''
    wordids=[]

    words=q.split()
    tablenumber=0

    for word in words:
        wordrow=self.con.execute("select rowid from wordlist where word='%s'" % word).fetchone()
        if wordrow!=None:
            wordid=wordrow[0]
            wordids.append(wordid)
            if tablenumber>0:
                tablelist+=','
                clauselist+=' and '
                clauselist+='w%d.urlid=w%d.urlid and '&(tablenumber-1,tablenumber)
            fieldlist+=',w%d.location'%tablenumber
            tablelist+='wordlocation w%d'%tablenumber
            clauselist+='w%d.wordid=%d'%(tablenumber,wordid)
            tablenumber+=1
    fullquery="select %s from %s where %s"%(fieldlist,tablelist,clauselist)
    cur=self.con.execute(fullquery)
    rows=[row for row in cur]
    return row,wordids

我收到以下错误:

e.getmatchrows('the')
---------------------------------------------------------------------------
OperationalError                          Traceback (most recent call last)
<ipython-input-4-65d6f760f6c8> in <module>()
----> 1 e.getmatchrows('the')

C:\Users\Blue\Anaconda\searchengine.py in getmatchrows(self, q)
    134                                 tablenumber+=1
    135                 fullquery="select %s from %s where %s"%(fieldlist,tablelist,clauselist)
--> 136                 cur=self.con.execute(fullquery)
    137                 rows=[row for row in cur]
    138                 return row,wordids

OperationalError: near "where": syntax error

1 个答案:

答案 0 :(得分:0)

您有SQL错误。首先正确格式化SQL。 Here is a quick tutorial。这是我首先要做的语法更改。

fullquery="SELECT %s FROM %s WHERE %s"%(fieldlist,tablelist,clauselist)

换句话说,考虑在SQL关键字上使用大写并重新运行它。