duanlu8613 2012-06-20 10:01
浏览 198
已采纳

在PDO中实现LIKE查询

I am running problems in implementing LIKE in PDO

I have this query:

$query = "SELECT * FROM tbl WHERE address LIKE '%?%' OR address LIKE '%?%'";
$params = array($var1, $var2);
$stmt = $handle->prepare($query);
$stmt->execute($params);

I checked the $var1 and $var2 they contain both the words I want to search, my PDO is working fine since some of my queries SELECT INSERT they work, it's just that I am not familiar in LIKE here in PDO.

The result is none returned. Do my $query is syntactically correct?

  • 写回答

5条回答 默认 最新

  • douweicheng5532 2012-06-20 10:06
    关注

    You have to include the % signs in the $params, not in the query:

    $query = "SELECT * FROM tbl WHERE address LIKE ? OR address LIKE ?";
    $params = array("%$var1%", "%$var2%");
    $stmt = $handle->prepare($query);
    $stmt->execute($params);
    

    If you'd look at the generated query in your previous code, you'd see something like SELECT * FROM tbl WHERE address LIKE '%"foo"%' OR address LIKE '%"bar"%', because the prepared statement is quoting your values inside of an already quoted string.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(4条)

报告相同问题?

悬赏问题

  • ¥15 有了解d3和topogram.js库的吗?有偿请教
  • ¥100 任意维数的K均值聚类
  • ¥15 stamps做sbas-insar,时序沉降图怎么画
  • ¥15 unity第一人称射击小游戏,有demo,在原脚本的基础上进行修改以达到要求
  • ¥15 买了个传感器,根据商家发的代码和步骤使用但是代码报错了不会改,有没有人可以看看
  • ¥15 关于#Java#的问题,如何解决?
  • ¥15 加热介质是液体,换热器壳侧导热系数和总的导热系数怎么算
  • ¥100 嵌入式系统基于PIC16F882和热敏电阻的数字温度计
  • ¥15 cmd cl 0x000007b
  • ¥20 BAPI_PR_CHANGE how to add account assignment information for service line