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我在我的树莓派上运行 python/mysqldb 时遇到了一些麻烦。这是一个非常简单的脚本，所以我不确定我错过了什么。“SELECT * FROM ...”运行没有问题，但我似乎无法用新值更新表。该脚本运行时不会抛出错误，但是当我 ctrl-C 时，它给了我这个：

异常 _mysql_exceptions.ProgrammingError: (2014, "Commands out of sync; you can't run this command now") in bound method DictCursor.__del of MySQLdb.cursors.DictCursor object at 0x19dfd90

这是我的脚本：

    dhost = "localhost"
    duser = "root"
    dname = "rpi"
    dpass = "datPassword"

    import MySQLdb

    try:
        con = MySQLdb.connect(dhost,duser,dpass,dname);
        cur = con.cursor(MySQLdb.cursors.DictCursor)

    except MySQLdb.Error, e:
        print "Error %d: %s" % (e.args[0],e.args[1])
        sys.exit(1)


    def websiteToSensor():
        cur.execute("SELECT * FROM homeauto WHERE changedby = 'website'")
        rows = cur.fetchall()
        for row in rows:
            cur.execute("UPDATE homeauto SET changedby = 'script' WHERE id = '%s'",(row["id"]))
        return

    while True:
        websiteToSensor()

有谁知道为什么我的表没有更新？谢谢！

***编辑：解决方案* **

感谢 Martijn Pieters，这是我的新 websiteToSensor() 代码：

    def websiteToSensor():
        cur = con.cursor(MySQLdb.cursors.DictCursor)
        cur.execute("SELECT * FROM homeauto WHERE changedby = 'website'")
        rows = cur.fetchall()
        num = int(cur.rowcount)
        if num > 0:
            for row in rows:
                cur.execute("UPDATE homeauto SET changedby = 'script' WHERE id = '%s'",(row["id"]))
            con.commit()
            cur.close()
            con.commit()
        else:
            cur.close()
            con.commit()
        return
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1 回答 1

1

尝试提交您的更改：

def websiteToSensor():
    cur.execute("SELECT * FROM homeauto WHERE changedby = 'website'")
    rows = cur.fetchall()
    for row in rows:
        cur.execute("UPDATE homeauto SET changedby = 'script' WHERE id = '%s'",(row["id"]))
    con.commit()
    return
于 2012-09-22T21:38:29.800 回答