传递了无效的参数join() [英] Invalid arguments passed join()

问题描述

因此，我制作了该脚本，应使用mysqli_fetch_array和mysqli_query来获取数组，所有脚本都可以正常工作，并且不会显示任何错误，但是在我更改了某些内容之后，使用了另一个函数，例如，如果我更改了心情，昵称或其他任何内容，均显示错误

So I made this script that should fetch an array using mysqli_fetch_array and mysqli_query it all works fine, and no error shows up, but then after I change something, using another function, for example, If I change the mood, nickname or whatever, it shows an error

Warning: join(): Invalid arguments passed in Main.php on line 526

我一直在尝试使用while loop修复此问题，并且发生了相同的事情

Iv'e been trying to fix this using a while loop and the same thing happens

$info = mysqli_fetch_array(mysqli_query($con, "SELECT id, nickname, mood, credits, colour, curhead, curface, curneck, curbody, curhands, curfeet, curflag, curphoto, rank * 146 FROM `sync_users` WHERE id='" . mysqli_real_escape_string($con, $raw[5]) . "';"), MYSQLI_ASSOC);
$client->sendPacket("%xt%gp%-1%" . join("|", $info) . "%");
print str_replace('Array', '', print_r($info, true));

我试图问其他几个了解php的朋友，他们也无法给我解决方案.

I tried asking a few other friends who knew php well and they were unable to give me a solution as well.

推荐答案

以下是有关基本调试的课程:

如果PHP函数抱怨输入参数，验证这些参数将是一个好主意.因此，添加

var_dump($info);

您的代码很可能会输出(bool)false

to your code will most likely output (bool)false

这使我们相信SQL查询不返回任何行，因此是时候对该查询进行调试了.

Which lead us to believe that SQL query returned no rows, and so it's time to debug the query.

希望有帮助

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