我使用SQLite3为我的一个web应用程序。一切正常。我可以访问数据库，创建和所有东西。
但. .当我使用以下代码检索信息时:
function db_select($username, $password, $database = 'Railway') {
$db = db_connect($database);
$result = $db->query("SELECT * FROM users ") or die('Query failed');
$rowss = array();
while ($row = $result->fetchArray()) {
$rowss[] = $row;
}
print_r($rowss);
}
Array (
[0] => Array ( [0] => John123 [username] => John123
[1] => foo [password] => foo
[2] => John [fname] => John
[3] => Smith [lname] => Smith
[4] => 21 [age] => 21
[5] => john@gmail.com [email] => john@gmail.com
[6] => Male [sex] => Male ) )
这个数组没有意义，如何单独检索信息，如fname, lname, age, username等?
是否有办法将其转换为以下形式?
Array (
username => John123,
password => foo,
fname => John
lname => Smith
)
Try
$result->fetchArray(SQLITE3_ASSOC)
既然你似乎对列号不感兴趣，
从PHP文档中找到这个答案。
$row = array();
$i = 0;
while($res = $result->fetchArray(SQLITE3_ASSOC)){
if(!isset($res['username'])) continue;
$row[$i]['username'] = $res['username'];
$row[$i]['fname'] = $res['fname'];
$row[$i]['lname'] = $res['lname'];
$row[$i]['age'] = $res['age'];
$row[$i]['email'] = $res['email'];
$row[$i]['sex'] = $res['sex'];
$row[$i]['phone'] = $res['phone'];
$i++;
}
print_r($row);
输出这个漂亮的数组。
Array (
[0] => Array (
[username] => John123
[fname] => John
[lname] => Smith
[age] => 21
[email] => john@gmail.com
[sex] => Male
)