dream02008 2014-01-28 15:01
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同时搜索三个区域,

this is how I have created search option does not appear on the page this thread and the fine I have to work but this is how I should look for 3 fields in the table, and it works in any way.

if($stmt = $this->mysqli->prepare('SELECT `navn`, `link`, `img`, `omrade` FROM `ordblindtest` WHERE `navn` LIKE "%" . $navn . "%" OR `omrade` LIKE "%" . $omrade . "%" OR `sogord` LIKE "%" . $sogord . "%"'))
    {
        $stmt->bind_param('sss', $navn, $omrade, $sogord);
        $navn = $_POST["sogord"];
        $omrade = $_POST["sogord"];
        $sogord = $_POST["sogord"];
        ...
    }

Problem is that I get this error:

Error on: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '. $navn . "%", OR omrade LIKE "%" . $omrade . "%", OR sogord LIKE "%" . $sog' at line 1

I do not understand why it appears with the error. Why am I receiving the error?

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1条回答 默认 最新

  • doutuo7156 2014-01-28 15:16
    关注

    Because you're using prepared statements you should be using the ? placeholder instead of the variables, and so your query should look like this:

    if ($stmt = $this->mysqli->prepare('
    
      SELECT
        `navn`,
        `link`,
        `img`,
        `omrade`
      FROM
        `ordblindtest`
      WHERE
        `navn` LIKE CONCAT("%", ?, "%")
        OR
        `omrade` LIKE CONCAT("%", ?, "%")
        OR
        `sogord` LIKE CONCAT("%", ?, "%")
    
    '))
    {
    
      $stmt->bind_param('sss', $navn, $omrade, $sogord);
    
      // ...
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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