Bug #55001 [Opn->Fbk]: Mysql explain command with prepared statement
Edit report at http://bugs.php.net/bug.php?id=55001&edit=1
ID: 55001
Updated by: johannes@php.net
Reported by: enrico dot triolo at gmail dot com
Summary: Mysql explain command with prepared statement
-Status: Open
+Status: Feedback
Type: Bug
Package: MySQLi related
Operating System: Ubuntu 11.04
PHP Version: Irrelevant
-Assigned To:
+Assigned To: mysql
Block user comment: N
Private report: N
New Comment:
Are you using mysqlnd or libmysql. If libmysql which version? (check phpinfo() output or php
--ri mysqli from command line)
Previous Comments:
------------------------------------------------------------------------
[2011-06-06 16:30:24] enrico dot triolo at gmail dot com
Description:
------------
Hi, I found out an anomaly executing an "explain" command using prepared
statements functions. Using "standard" functions works as expected.
In short, if I execute an "explain" on a query with a subquery and fetch the
resulting "type" field, I get "unique_subq" instead of
"unique_subquery".
Please view the attached code snippet.
I'm using PHP 5.3.5-1ubuntu7.2, but other versions have the same behaviour too.
The attached script uses a table created with this instruction:
CREATE TABLE IF NOT EXISTS mytest_table (
id int(11) NOT NULL AUTO_INCREMENT,
idParent int(11) NOT NULL,
PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
Test script:
---------------
$sql = 'explain SELECT id FROM mytest_table WHERE idParent <> -1 AND idParent NOT IN (
SELECT id FROM mytest_table)';
$link = mysqli_connect("localhost", 'user', 'password',
'dbName');
printf("Using prepared statement functions...\n");
$stmt = mysqli_stmt_init($link);
mysqli_stmt_prepare($stmt, $sql);
mysqli_stmt_execute($stmt);
mysqli_stmt_store_result($stmt);
$result = mysqli_stmt_result_metadata($stmt);
printf("Fields:\n");
while($field = mysqli_fetch_field($result))
printf("\t%s(%d)\n", $field->name, $field->length);
mysqli_free_result($result);
mysqli_stmt_bind_result($stmt, $id, $select_type, $table, $type, $possible_keys, $key, $key_len,
$ref, $rows, $extra);
while(mysqli_stmt_fetch($stmt))
printf("Type field value: %s\n", $type);
printf("\nUsing mysqli_query...\n");
$result = mysqli_query($link, $sql);
while($row = mysqli_fetch_array($result))
printf("Type field value: %s\n", $row['type']);
mysqli_free_result($result);
mysqli_close($link);
Expected result:
----------------
Using prepared statement functions I'd expect the second row value for the "type"
field being "unique_subquery".
Actual result:
--------------
I get "unique_subq" instead of "unique_subquery". Using mysqli_query (i.e.
without prepared statement functions), I get the expected result:
Using prepared statement functions...
...
Type field value: unique_subq
Using mysqli_query...
...
Type field value: unique_subquery
------------------------------------------------------------------------
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Edit this bug report at http://bugs.php.net/bug.php?id=55001&edit=1
Thread (4 messages)