php – 构建mysqli查询？

如果我有查询选择users.user_id,users.fname,users.lname,bios.bio,groups.groupid来自用户LEFT JOIN bios on users.user_id = bios.userid那么我想在条件上添加另一个表然后最后添加where语句.问题是因为当我绑定param时,它说“变量的数量与prepare语句中的变量数量不匹配”.我该如何解决这个问题？干杯.例：

    $info = "select users.user_id, users.fname, users.lname, bios.bio, groups.groupid from users LEFT JOIN bios on users.user_id = bios.userid";
    $content = $members->prepare($info);
    if ($_GET['where'] == 'requests') $info .= "LEFT JOIN requests on users.user_id = requests.receiver";
    else if ($_GET['where'] == 'referrals') $info .= "LEFT JOIN referrals on users.user_id = referrals.receiver";
    $info .= "where users.user_id = ?";
    $content->bind_param('s', $_SESSION['token'][1]);
    $content->execute();

解决方法:

您正在准备它之后更改sql字符串.不要那样做.这样做是这样的：

$info = "select users.user_id, users.fname, users.lname, bios.bio, groups.groupid from users LEFT JOIN bios on users.user_id = bios.userid";
if ($_GET['where'] == 'requests') $info .= " LEFT JOIN requests on users.user_id = requests.receiver";
else if ($_GET['where'] == 'referrals') $info .= " LEFT JOIN referrals on users.user_id = requests.receiver";
$info .= " where users.user_id = ?";
$content = $members->prepare($info);
$content->bind_param('s', $_SESSION['token'][1]);
$content->execute();

编辑：另外,确保您的sql片段在必要时用空格分隔; .=运算符不会自动为您添加空格.