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我在将 BLOB 上传到我的 MySQL 数据库时遇到问题，并收到以下错误：

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 '()*456789:CDEFGHIJSTUVWXYZcdefghijstuvwxyzƒ„…†‡ˆ‰Š’“”•–—˜™š¢£¤¥¦§¨©ª²³´µ¶·¸¹ºÂÃÄ' at line 1

我知道错误导致图像的文件内容，但我无法弄清楚语法有什么问题。有什么建议么？谢谢！

这是PHP：

    $file = $_FILES['image']['tmp_name'];

// If there's no file selected when button is pressed, echo out and tell the user to select an image to upload
if (!isset($file))
    echo "<p>Please select an image to upload.</p>";
else {
    //mysql escape string
    $image = file_get_contents($_FILES['image']['tmp_name']);
    //and here
    $image_name = $_FILES['image']['name'];
    $imagesize = getimagesize($_FILES['image']['tmp_name']);
}

// Checks that the file being uploaded is an image, i.e. has a size attribute with height & width dimensions
if ($imagesize == FALSE)
    echo "<p>Please upload only an image file such as .jpg or .png.</p>";
else {
    $sql = "INSERT INTO design (id, caption, image) VALUES ('', '$image_name', '$image')";
    $result = mysql_query($sql);
    if (!$result)
        echo "<p>Something went wrong.</p>" . mysql_error();
    else {
        echo "<p>Thank you for submitting your design.</p>";
    }
}
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1 回答 1

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显然，图像文件内容中有一个撇号。这并不奇怪。您需要正确转义输入（以及与此相关的所有输入）。

$image = mysql_real_escape_string($_FILES['image']['tmp_name']);

而不是使用ext/mysql，您应该使用带有 mysqli 或 PDO 的正确参数化查询。然后你不必明确地逃避。

于 2013-04-07T20:25:09.927 回答