mysqli::get_charset

mysqli::$error

[edit] Last updated: Fri, 17 May 2013

mysqli::$field_count

mysqli_field_count

(PHP 5)

mysqli::$field_count -- mysqli_field_count — Devuelve el número de columnas para la consulta más reciente

Descripción

Estilo orientado a objetos

int $mysqli->field_count;

Estilo por procedimientos

int mysqli_field_count ( mysqli $link )

Devuelve el numero de columnas para la consulta más reciente en la conexión representada por el parámetro link. Esta función puede ser útil cuando se utiliza la función mysqli_store_result() para determinar si la consulta ha producido un resultado no vacío o no, sin saber la naturaleza de la consulta.

Parámetros

link: Sólo estilo por procediminetos: Un identificador de enlace devuelto por mysqli_connect() o mysqli_init()

Valores devueltos

Un entero que representa el número de campos en un set de resultado. An integer representing the number of fields in a result set.

Ejemplos

Ejemplo #1 Ejemplo de $mysqli->field_count

Estilo orientado a objetos


<?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) {
    /* this was a select/show or describe query */
    $result = $mysqli->store_result();

    /* process resultset */
    $row = $result->fetch_row();

    /* free resultset */
    $result->close();
}

/* close connection */
$mysqli->close();
?>

Estilo por procedimientos


<?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)) {
    /* this was a select/show or describe query */
    $result = mysqli_store_result($link);

    /* process resultset */
    $row = mysqli_fetch_row($result);

    /* free resultset */
    mysqli_free_result($result);
}

/* close connection */
mysqli_close($link);
?>

mysqli::get_charset

mysqli::$error

[edit] Last updated: Fri, 17 May 2013

add a note User Contributed Notes mysqli::$field_count - [5 notes]

down

Jonathan ¶

6 years ago


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";

down

Typer85 at gmail dot com ¶

6 years ago


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,



?>

down

dedlfix ¶

6 years ago


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

down

jakerosoft at hotmail dot com ¶

7 years ago


<?


$fieldinfo = $result->fetch_field();


if ($fieldinfo & MYSQLI_NOT_NULL_FLAG)  {


  print "not null flag is set";


} else {


  print "not null flag is NOT set";


}


?>

down

Marc-Andr� ¶

7 years ago


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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