Invalid arguments passed join()

问题

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

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));

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

回答1:

One or more elements of the array isn't a string. Either cast the offending fields in the query, or pass all the fields through strval() first.

回答2:

Here goes a lesson on basic debugging:

If a PHP function complains for the input parameters, it would be a good idea to verify these parameters. So, adding

var_dump($info);

to your code will most likely output (bool)false

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

Hope it helps

回答3:

First of all don't use mysql_* functions because are deprecated, try PDO for example

I think that if mysqli_fetch_array returns an empty resultset then the join will fail because mysqli returns a FALSE

Perhaps it will be your problem

来源：https://stackoverflow.com/questions/18374881/invalid-arguments-passed-join

标签

php

join

mysqli