


Why does my PHP script fail to create a file when executing a specific query that returns \'unknown table status: TABLE_TYPE\'?
Oct 31, 2024 pm 04:31 PMPHP 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.
The above is the detailed content of Why does my PHP script fail to create a file when executing a specific query that returns \'unknown table status: TABLE_TYPE\'?. For more information, please follow other related articles on the PHP Chinese website!

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