Question

使用python 3尝试烧瓶中的mysql查询

SELECT title FROM Book WHERE title LIKE '%booktitlehere%';

从我的搜索表单的输入中，这是尝试烧瓶实现

search_string = search.data['search']
cur = mysql.connection.cursor()
likeString = "'%" + search_string + "%'"
cur.execute('''SELECT title FROM Book WHERE title LIKE %s;''',likeString)

当转到包含此表单的页面

时，它无法解决此错误

_mysql_exceptions.ProgrammingError：并非在字符串格式化期间转换所有参数

我尝试添加%%以逃避文字％

likeString = "'%%" + search_string + "%%'"
# print(like)
cur.execute("""SELECT title FROM Book WHERE title LIKE %s;""",likeString)
rv = cur.fetchall()

但仍然出现错误

_mysql_exceptions.ProgrammingError: not all arguments converted during string formatting

还尝试使用select

附近的双引号

cur.execute("SELECT title FROM Book WHERE title LIKE %s;",likeString)

Answer 1

Double it：

likeString = "'%%" + search_string + "%%'"

cur.execute("""SELECT title FROM Book WHERE title LIKE %s;""", (likeString))

Answer 2

示例：

cari = "%" + request.form['cari'] +"%"

conn.execute("SELECT nama, harga FROM barang where nama LIKE %s OR harga LIKE %s", (cari, cari))

Answer 3

这对我有用，我使用%%，喜欢后应该有“”

cur.execute（'从名称中选择*，例如“ %%% s %%”'，[名字]）

Answer 4

在sql命令中使用搜索字符串之前

.col-style-3 { padding: 0px !important; background-color: #fff; border: solid 1px #e0e0e0; border-radius: 5px; text-align: center; color: #000; transition: 0.5s; } .col-style-3 > a { color: #FFFFFF; text-decoration: none; } .col-style-3:hover { border: solid 1px #bebebe; box-shadow: 0 5px 10px #adadad; transition: 0.3s; }

然后在sql命令中应用

search_string= f"%{search_string}%"

另外还可以降低它
cur.execute("SELECT * FROM books WHERE title LIKE :searching_string ORDER BY id ",{"searching_string":search_string})

search_string= f"%{search_string}%".lower()

使用带有LIKE %%的mysql查询的Flask

4 个答案: