Why does my PHP script fail to create a file when executing a specific query that returns \"unknown table status: TABLE_TYPE\"?

PHP file unable to execute part of the code

The provided PHP script serves as a function to handle check-in requests from an Android application and interacts with a database. However, the program encountered an issue where it could not execute a specific section of code, which prevented it from creating a file named 'file4.txt'.

Tracing the Problem

The developer identified that the script only failed to create the file when executing the query $query. Upon further investigation, it was discovered that MySQL was responding with the message "unknown table status: TABLE_TYPE" when attempting to execute the query. This behavior was inconsistent, as the function worked correctly sometimes but not others.

Proposed Solutions

To resolve this issue, the developer suggested a divide and conquer approach by breaking down the function into smaller sections. Specifically, the file writing section was factored out into a separate function:

<code class="php">function file_put($number, $data)
{
    $path = sprintf("C:/temp/wamp/www/file%d.txt", $number);
    file_put_contents($path, $data);
}</code>

This simplified the code and made it easier to debug.

Additionally, the developer proposed reorganizing the database operations into a separate class called MySql. This would isolate the error handling and provide a more efficient way to connect to the database and retrieve data:

<code class="php">class MySql
{
    // ... (Database connection and query methods) ...
}</code>

Updated Code Example

Combining the proposed solutions, the updated checkin() function became much cleaner and easier to maintain:

<code class="php">function checkin(MySql $DB, $TechID, $ClientID, $SiteID)
{
    $query = sprintf("SELECT `Type` FROM `Log` WHERE `TechID` = '%d' ORDER BY LogTime DESC LIMIT 1", $TechID);
    file_put(5, $query);

    $result1 = $DB->query("SELECT COUNT(*) FROM Log");
    $result2 = $DB->query($query);

    foreach ($result1 as $row1) {
        list($count) = $row1;
        $data = "ClientID:$ClientID TechID:$TechID SiteID:$SiteID Count:$count";
        file_put(3, $data);
        foreach ($result2 as $row2) {
            file_put(4, $data);
        }
    }
}</code>

The full code example, including the MySql class, was provided, demonstrating how to divide and conquer the problem to achieve a more manageable and efficient solution.

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