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Embedding PHP variable into Sqlite statement

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1 answer
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I have been struggling with this for hours. I am pretty new to Sqlite and have been trying to write a prepared statement, everything works up until I try to get my variable in. The following works as does a direct SELECT - entering the string straight in the statement. prepare('SELECT * FROM Users WHERE Name = ?'); $statement->bindValue(1, 'aName'); $result = $statement->execute(); echo("Get the 1st row as an associative array:\n"); print_r($result->fetchArray(SQLITE3_ASSOC)); echo("\n"); $result->finalize(); However as soon as I bring a variable in to play with bindParam , I do not get results. like so: prepare('SELECT * FROM Users WHERE Name = ?'); $statement->bindParam('s', $myusername); $result = $statement->execute(); echo("Get the 1st row as an associative array:\n"); print_r($result->fetchArray(SQLITE3_ASSOC)); echo("\n"); $result->finalize(); I have checked the $_POST['user'] variable and it is pulling through. Any help that solves my issue or shows how I might debug would be most welcome. I am working in PHP Storm just in case there is something specific to that IDE. I have also tried with Name = :Name and binding with that. Thanks in advance.
Asked by fcreav (99 rep)
Mar 2, 2024, 01:38 AM
Last activity: Mar 3, 2024, 12:25 AM