[英]SQLite3 PHP “Update” -result of prepared query
Does somebody know how i can check in PHP if an SQLite3 update query (prepared) was succesful or not? 有人知道我如何检查SQLite3更新查询(准备好的)是否成功吗?
Her my code... 她是我的密码...
$stmt = $project->prepare( 'UPDATE tasks set title=:title WHERE rowid=:rowid' );
$stmt->bindValue(':title', rtrim($_POST['title'],'<br>'), SQLITE3_TEXT);
$stmt->bindValue(':rowid', (int)$_POST['rel'], SQLITE3_INTEGER);
$result = $stmt->execute();
var_dump( $result );
This code is upodating my table. 这段代码正在更新我的表。 But "var_dump( $result )" return everytime an empty object.
但是“ var_dump($ result)”每次返回一个空对象。 Even if i force an error by passing "rowid=non-existing-rowid".
即使我通过传递“ rowid = non-existing-rowid”来强制执行错误。
Any ideas, how i can check my update query? 任何想法,我如何检查我的更新查询?
An UPDATE
statements updates as many record as match the WHERE
condition; UPDATE
语句更新与WHERE
条件匹配的记录。 this could be zero, one, or many records. 可以是零个,一个或多个记录。 On the SQL level, all of this is considered a success.
在SQL级别,所有这些都被认为是成功的。
If you want to find out how many records were affected, you can use the changes
method of the database connection object. 如果要查找受影响的记录数,可以使用数据库连接对象的
changes
方法 。 In your case: 在您的情况下:
...
$result = $stmt->execute();
if ($result) {
echo 'Updated rows: ', $project->changes();
}
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