如何在iphone iOS中使用参数在SQLITE中形成LIKE语句？

Question

说,我有一个名为name的搜索域.如果用户在搜索字段中键入"John",我应该在"名称"字段中检查值"John"并选择所有相应的字段值.'John'将是来自searchBar.text的NSString,它必须作为参数添加到查询中.我构建了一个像这样的sql语句,从来没有用过.

NSString *query = [NSString stringWithFormat:@"select * from Employee where EmployeeName LIKE '%%@%'", searchText.text];
const char *sqlStatement = [query UTF8String];

我也尝试了这个

NSString *query = [NSString stringWithFormat:@"select * from Employee where EmployeeName LIKE %'%@'%", searchText.text];
const char *sqlStatement = [query UTF8String];

没有任何效果.

Answer 1

诀窍是获得那里的百分比.这很尴尬,但你可以这样做:

@"select * from Employee where EmployeeName LIKE '%%%@%%'"

%%是百分比,%@是替换字符串.