Question

我知道这个问题之前已被多次询问过，我已经搜索过并试图弄明白我哪里出错了，我似乎无法做到。

任何帮助都会被贬低。提前谢谢。

我已经设置了dreamweaver cc更新查询，效果很好。我想现在将图像添加到我的数据库。

这是我的问题：

    $image = $_FILES['speakerimage']['tmp_name'];
    $imgContent = addslashes(file_get_contents($image));
    $updateSQL = sprintf("UPDATE Speakers SET speaker=%s,     chkSportsCeleb=%s, chkEntertainment=%s, chkEmcee=%s, chkCFLAA=%s, chkKeynote=%s, chkAutioneer=%s, speakerImg=%s WHERE speakerID=%s",
                   GetSQLValueString($_POST['speaker'], "text"),
                   GetSQLValueString(isset($_POST['SportsCeleb']) ? "true" : "", "defined","1","0"),
                   GetSQLValueString(isset($_POST['Entertainment']) ? "true" : "", "defined","1","0"),
                   GetSQLValueString(isset($_POST['Emcee']) ? "true" : "", "defined","1","0"),
                   GetSQLValueString(isset($_POST['CFLAA']) ? "true" : "", "defined","1","0"),
                   GetSQLValueString(isset($_POST['Keynote']) ? "true" : "", "defined","1","0"),
                   GetSQLValueString(isset($_POST['Autioneer']) ? "true" : "", "defined","1","0"),
                   $imgContent,
                   GetSQLValueString($_POST['speakerID'], "int"));

当我在浏览器中预览我的代码时，这就是我得到的：

   query = UPDATE Speakers SET speaker='John Doe', chkSportsCeleb=1, chkEntertainment=0, chkEmcee=0, chkCFLAA=1, chkKeynote=1, chkAutioneer=0, speakerImg= WHERE speakerID=59  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 'WHERE speakerID=59' at line 1

我认为图片内容应该显示为胡言乱语。我无法弄清楚如何重写这个。

由于

罗兰

Answer 1

尝试将图片加载到变量中，然后在UPDATE语句中直接指定它：

$image = $_FILES['speakerimage']['tmp_name'];
$imgContent = addslashes(file_get_contents($image));
$updateSQL = sprintf("UPDATE Speakers SET speaker=%s,     chkSportsCeleb=%s, chkEntertainment=%s, chkEmcee=%s, chkCFLAA=%s, chkKeynote=%s, chkAutioneer=%s, $imgContent WHERE speakerID=%s",
// ...

Dreamweaver cc SQL查询

1 个答案: