mysqli::$field_count

mysqli_field_count

(PHP 5)

mysqli::$field_count -- mysqli_field_count — 直近のクエリのカラムの数を返す

説明

オブジェクト指向型

int $mysqli->field_count;

手続き型

int mysqli_field_count ( mysqli $link )

link が指す接続における直近のクエリのカラムの数を返します。この関数は、クエリが空でない結果セットを生成すべきなのかそうではないのかを判断するために mysqli_store_result() 関数を使用する際に有用です。

パラメータ

link: 手続き型のみ: mysqli_connect() あるいは mysqli_init() が返すリンク ID。

返り値

結果セットのフィールド数を整数で返します。

例

例1 $mysqli->field_count の例

オブジェクト指向型


<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "test");

$mysqli->query( "DROP TABLE IF EXISTS friends");
$mysqli->query( "CREATE TABLE friends (id int, name varchar(20))");

$mysqli->query( "INSERT INTO friends VALUES (1,'Hartmut'), (2, 'Ulf')");


$mysqli->real_query("SELECT * FROM friends");

if ($mysqli->field_count) {
    /* これは select/show あるいは describe クエリです */
    $result = $mysqli->store_result();

    /* 結果セットを処理します */
    $row = $result->fetch_row();

    /* 結果セットを開放します */
    $result->close();
}

/* 接続を閉じます */
$mysqli->close();
?>

手続き型


<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "test");

mysqli_query($link, "DROP TABLE IF EXISTS friends");
mysqli_query($link, "CREATE TABLE friends (id int, name varchar(20))");

mysqli_query($link, "INSERT INTO friends VALUES (1,'Hartmut'), (2, 'Ulf')");

mysqli_real_query($link, "SELECT * FROM friends");

if (mysqli_field_count($link)) {
    /* これは select/show あるいは describe クエリです */
    $result = mysqli_store_result($link);

    /* 結果セットを処理します */
    $row = mysqli_fetch_row($result);

    /* 結果セットを開放します */
    mysqli_free_result($result);
}

/* 接続を閉じます */
mysqli_close($link);
?>

mysqli::get_charset

mysqli::$error

[edit] Last updated: Fri, 10 Feb 2012

add a note User Contributed Notes mysqli::$field_count

Jonathan 06-Mar-2007 03:43


Some corrections ;o)



$mysqli_type = array();

$mysqli_type[0] = "DECIMAL";

$mysqli_type[1] = "TINYINT";

$mysqli_type[2] = "SMALLINT";

$mysqli_type[3] = "INTEGER";

$mysqli_type[4] = "FLOAT";

$mysqli_type[5] = "DOUBLE";



$mysqli_type[7] = "TIMESTAMP";

$mysqli_type[8] = "BIGINT";

$mysqli_type[9] = "MEDIUMINT";

$mysqli_type[10] = "DATE";

$mysqli_type[11] = "TIME";

$mysqli_type[12] = "DATETIME";

$mysqli_type[13] = "YEAR";

$mysqli_type[14] = "DATE";



$mysqli_type[16] = "BIT";



$mysqli_type[246] = "DECIMAL";

$mysqli_type[247] = "ENUM";

$mysqli_type[248] = "SET";

$mysqli_type[249] = "TINYBLOB";

$mysqli_type[250] = "MEDIUMBLOB";

$mysqli_type[251] = "LONGBLOB";

$mysqli_type[252] = "BLOB";

$mysqli_type[253] = "VARCHAR";

$mysqli_type[254] = "CHAR";

$mysqli_type[255] = "GEOMETRY";

Typer85 at gmail dot com 02-Jan-2007 08:33


For those interested and to clarify the Manual Entry.



For query statements that are DESIGNED to return a result set of some sort, this function will always return the number of fields in the table that was queried.



I said DESIGNED because the return value has no effect on whether or not the actual query matched any rows or not.



For example, say I have a table that has 2 fields and only 10 rows. I issue the following query:



<?php



// Assume Connection Blah Blah.



mysqli_query( $connObject , "Select * From `table` Where `Id` > 1000");



// Get Number Of Fields.



mysqli_field_count( $connObject );



// Will Return 2 --> The Number of fields in the table!



?>



It is quite clear that the query itself will never return a result set because I asked it to return rows which have an Id over 1000 and there are only 10 rows.



But because the nature of the query itself is to return a result set, the field count is always returned no matter what.



In contrast, if the query does anything that does not return a result set by nature, such as an insert or update, the field count will always be 0.



Hence, you can easily determine the nature of this query dynamically using these return values.



Good Luck,



?>

dedlfix 18-Jul-2006 10:26


There are MYSQLI_TYPE_* constants for the type property (listed in http://php.net/manual/en/ref.mysqli.php).



e.g.

<?php

if ($finfo->type == MYSQLI_TYPE_VAR_STRING)

  // a VARCHAR

jakerosoft at hotmail dot com 16-Aug-2005 09:15


<?


$fieldinfo = $result->fetch_field();


if ($fieldinfo & MYSQLI_NOT_NULL_FLAG)  {


  print "not null flag is set";


} else {


  print "not null flag is NOT set";


}


?>

Marc-Andr� 07-Jul-2005 03:56


The "type" property will return a numerical representation of a field type instead of a "meaningful" string.



Here is an array that may help you:



<?php 

$mysqli_type = array();

$mysqli_type[0] = "decimal";

$mysqli_type[1] = "tinyint";

$mysqli_type[2] = "smallint";

$mysqli_type[3] = "int";

$mysqli_type[4] = "float";

$mysqli_type[5] = "double";

$mysqli_type[7] = "timestamp";

$mysqli_type[8] = "bigint";

$mysqli_type[9] = "mediumint";

$mysqli_type[10] = "date";

$mysqli_type[11] = "time";

$mysqli_type[12] = "datetime";

$mysqli_type[13] = "year";

$mysqli_type[252] = "blob"; // text, blob, tinyblob,mediumblob, etc...

$mysqli_type[253] = "string"; // varchar and char

$mysqli_type[254] = "enum";

?>

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