PHP 5.5.37 is released

SQLite3::prepare

(PHP 5 >= 5.3.0, PHP 7)

SQLite3::preparePrepares an SQL statement for execution

Descrizione

public SQLite3Stmt SQLite3::prepare ( string $query )

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

Elenco dei parametri

query

The SQL query to prepare.

Valori restituiti

Returns an SQLite3Stmt object on success o FALSE in caso di fallimento.

Esempi

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'1SQLITE3_INTEGER);

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

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User Contributed Notes 2 notes

up
1
Venkata Subbaraju
5 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'];
         }
}
?>
up
-28
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.
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