This code works with SQlite version 3 databases.
<?php
function sqlite_field_names($dbfile, $tablename){
try
{
$db = new PDO("sqlite:".$dbfile);
$sql=$db->query("SELECT sql FROM sqlite_master WHERE tbl_name='".$tablename."'");
foreach($sql as $row)
{
$fields = $row["sql"];
}
$db = NULL;
}
catch(PDOException $e)
{
print "Exception : ".$e->getMessage();
}
$cut=strtok($fields,"(");
while($fieldnames[]=strtok(",")) {};
array_pop($fieldnames);
foreach($fieldnames as $no => $field)
if (strpos($field, "PRIMARY KEY")){
strtok($field,"(");
$primary=strtok(")");
unset($fieldnames[$no]);
} else
$fieldnames[$no]=strtok($field, " ");
return $fieldnames;
}
print_r(array_values(sqlite_field_names($dbfile, $tablename)));
?>
sqlite_field_name
SQLiteResult::fieldName
SQLiteUnbuffered::fieldName
(PHP 5 < 5.4.0, PECL sqlite >= 1.0.0)
sqlite_field_name -- SQLiteResult::fieldName -- SQLiteUnbuffered::fieldName — 特定のフィールドの名前を返す
説明
string sqlite_field_name
( resource
$result
, int $field_index
)オブジェクト指向型 (メソッド):
string SQLiteResult::fieldName
( int
$field_index
)
string SQLiteUnbuffered::fieldName
( int
$field_index
)
Given the ordinal column number, field_index,
sqlite_field_name() returns the name of that field in
the result set result.
パラメータ
-
result -
SQLite 結果リソース。 このパラメータは、 オブジェクト指向言語型メソッドを使用する場合は不要です。
-
field_index -
結果セットにおけるオリジナルのカラム番号
返り値
与えられたオリジナルのカラム番号での SQLite
結果セット中のフィールド名を返します。エラーの場合は、FALSE を返します。
SQLITE_ASSOC および SQLITE_BOTH で
返されるカラム名は、設定オプション
sqlite.assoc_case の値に基づき、
大文字小文字が変換されます。
Jan Holeek
11-Jan-2011 10:27
admin at psychonautical dot org
09-Nov-2010 04:58
It's not the most elegant way, but it works:
<?php
function sqlite_field_names($db, $tablename){
$sq5=sqlite_single_query($db, "SELECT sql FROM sqlite_master WHERE tbl_name='".$tablename."'", SQLITE_NUM);
$kommando=$sq5[0];
$cut=strtok($kommando,"(");
while($fieldnames[]=strtok(",")) {};
array_pop($fieldnames);
foreach($fieldnames as $no => $field)
if (strpos($field, "PRIMARY KEY")){
strtok($field,"(");
$primary=strtok(")");
unset($fieldnames[$no]);
} else
$fieldnames[$no]=strtok($field, " ");
return $fieldnames;
}
?>
rrf5000 at psu dot edu
21-Jun-2007 07:03
While working with SQLite using its object-oriented mode, I found need to display a column/field name without knowing what it was in advance. I couldn't find any examples on the Internet, just this document. So, for anyone who happens to need to do this, here's an example.
<?php
$db = "db/database.sqlite";
// create new database (OO interface)
$dbo = new SQLiteDatabase("$db");
// create table foo and insert sample data
$dbo->query("
CREATE TABLE foo(id INTEGER PRIMARY KEY, name CHAR(255));
INSERT INTO foo (name) VALUES('Ilia1');
INSERT INTO foo (name) VALUES('Ilia2');
INSERT INTO foo (name) VALUES('Ilia3');
");
$query = "SELECT * FROM foo;";
$result = $dbo->query($query) or die("Error in query");
echo "
<table border='1' cellpadding='10'>
<tr>
<td>".$result->fieldName(0)."</td>
<td>".$result->fieldName(1)."</td>
</tr>";
// iterate through the retrieved rows
while ($result->valid()) {
// fetch current row
$row = $result->current();
echo "
<tr>
<td>".$row[0]."</td>
<td>".$row[1]."</td>
</tr>";
// proceed to next row
$result->next();
}
echo "</table>";
?>
