SQLite3::prepare

(PHP 5 >= 5.3.0, PHP 7)

SQLite3::prepare — Prepares an SQL statement for execution

Description

public SQLite3Stmt SQLite3::prepare ( string $query )

Prepares an SQL statement for execution and returns an SQLite3Stmt object.

Parameters

query: The SQL query to prepare.

Return Values

Returns an SQLite3Stmt object on success or FALSE on failure.

Examples

Example #1 SQLite3::prepare() example


<?php
unlink('mysqlitedb.db');
$db = new SQLite3('mysqlitedb.db');

$db->exec('CREATE TABLE foo (id INTEGER, bar STRING)');
$db->exec("INSERT INTO foo (id, bar) VALUES (1, 'This is a test')");

$stmt = $db->prepare('SELECT bar FROM foo WHERE id=:id');
$stmt->bindValue(':id', 1, SQLITE3_INTEGER);

$result = $stmt->execute();
var_dump($result->fetchArray());
?>

User Contributed Notes

Venkata Subbaraju

3 months ago


Without checking the return value of "prepare" if "exec" is used then it will cause Fatal Error.

"PHP Fatal error:  Call to a member function execute() on a non-object "

To avoid this error,need to check return value as following:

<?php
 $db = new SQLite3('school.db');
 if($stmt = $db->prepare('SELECT id,student_name FROM classTen '))
 {
         $result = $stmt->execute();
         $names=array();
         while($arr=$result->fetchArray(SQLITE3_ASSOC))
         {
          $names[$arr['id']]=$arr['student_name'];
         }
}
?>

pavel dot lisa at centrum dot cz

5 years ago


SQLite3::prepare() will fail, if you pass in a WHERE clause (in the $query) referring to a non-existent column.