java.sql.sqlexception：几何图形字节字符串必须是little endian

Question

我想使用contains函数获取其title或question包含某个值的所有值。然而,我得到了这个错误：

java.sql.sqlexception：几何图形字节字符串必须是小尾序。在com.mysql.jdbc.sqlerror.createsqlexception（sqlerror.Java:1074)在com.mysql.jdbc.mysqlio.checkerrorpacket（mysqlio.Java:4120)在com.mysql.jdbc.mysqlio.nextrowfast（mysqlio.Java:2076)在com.mysql.jdbc.mysqlio.nextrow（mysqlio.Java:1932)在com.mysql.jdbc.mysqlio.readsinglerowset（服务(httpservlet.java:790)在org.apache.jasper.servlet.jspserver.service(jspservlet.java:411)在org.apache.jasper.servlet.jspserver.service(jspservlet.java:473)在org.apache.jasper.servlet.jspserver.service(jspservlet.java:377)在org.apache.jatalina.core.standardwrapper.service(httpservlet.java:790)在eAdapter.Service(coyoteAdapter.java:283)位于com.sun.enterprise.v3.services.impl.containerMapper$HttpHandlerCallable.call(containerMapper.java:459)位于com.sun.enterprise.v3.services.impl.containerMapper.Service(containerMapper.java:167)位于org.glassfish.grizzly.http.server.httpandler.runservice(httphandler.java:206)位于oreXecutor.execute(processoreXecutor.java:77)在org.glassfish.grizzly.nio.transport.tcpniotransport.fireioevent(tcpniotransport.java:536)在org.glassfish.grizzly.strategies.abstractionostrategy.fireioevent(abstractionostrategy.java:112)在

这是代码：

public String getRecentQuestions(String search) throws SQLException {
        PreparedStatement ps=con.prepareStatement("select title, question_id from questions where contains(title, ?) or contains(question, ?) limit 20");
        ps.setString(1, search);
        ps.setString(2, search);
        ResultSet rs=ps.executeQuery();
        StringBuilder sb=new StringBuilder("");
        while (rs.next()) {
            sb.append("<a href='Question.jsp?question=").append(rs.getInt(2)).append("'><h2>");
            sb.append(rs.getString(1));
            sb.append("</h2></a>");
        }
        return sb.toString();
    }

My PreparedStatement有效，title列数据类型为varchar，question列数据类型也为varchar。提前道谢。

Answer 1

徐尚说得对。您可以使用“instr（str,substr）”对文本进行模糊查询。例如：如果INSTR（content,“test”）>0，则“INSTR”比“like”更有效。

Answer 2

mysql使用instr（str,substr）而不是contains

也许你可以试试instr或like