The most probable reason for the error is that your SQL table does not contain a column named "active". You should also check if the query was successful (see the example code at PHP: mysqli::query - Manual[^] ) and rows are returned (calling mysqli_num_rows() before fetching).
Mar 13, 2018
Jul 10, 2015 · 1 Answer 1 ... You need to change: echo $row["userId"];. to: echo $row["userID"];. You're creating an index with userID from the database, so when ...
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Apr 25, 2023 · The solution simply is to check if it has been set before referencing. We can use the isset() function, which checks if the variable 'is set' or ...
Jun 15, 2020 · The output from the 2nd try is “Notice: undefined index: busName …”. Once the results from the query are put into an array ($DataRetrieved ...
Missing: mysqli | Show results with:mysqli
Nov 5, 2019 · Undefined Index errors come from referencing variables or indexes when they aren't created. You can't reference an index or a variable if you ...
Missing: mysqli | Show results with:mysqli
Returns the last error message for the most recent MySQLi function call that can succeed or fail. Parameters ¶. mysql. Procedural style only: A mysqli object ...
Jan 22, 2017 · To solve such error, you can use the isset() function, which will check whether the index or variable is set or not, If not then don't use it.
Aug 24, 2017 · I have a table with a column with UI: Single File, named "image" that has a related table of "directus_files". I would like to create multiple ...
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Posted: Feb 12, 2018
Posted: Feb 12, 2018
Mar 18, 2015 · Undefined index is likely referring to an element of an array that has not been defined yet. Example: $array['index1'] = 'Index 1';.