PHP: Hypertext Preprocessor (original) (raw)

mysqli_stmt::$param_count

mysqli_stmt_param_count

(PHP 5, PHP 7, PHP 8)

mysqli_stmt::$param_count -- mysqli_stmt_param_count — Returns the number of parameters for the given statement

Description

Object-oriented style

Procedural style

Return Values

Returns an integer representing the number of parameters.

Examples

Example #1 Object-oriented style

`<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

if ( stmt=stmt = stmt=mysqli->prepare("SELECT Name FROM Country WHERE Name=? OR Code=?")) {$marker = $stmt->param_count;
printf("Statement has %d markers.\n", $marker);/* close statement /
$stmt->close();
}/
close connection */
$mysqli->close();
?>`

Example #2 Procedural style

`<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

if ( stmt=mysqliprepare(stmt = mysqli_prepare(stmt=mysqliprepare(link, "SELECT Name FROM Country WHERE Name=? OR Code=?")) {$marker = mysqli_stmt_param_count($stmt);
printf("Statement has %d markers.\n", $marker);/* close statement /
mysqli_stmt_close($stmt);
}/
close connection */
mysqli_close($link);
?>`

The above examples will output:

Found A Problem?

Senthryl

16 years ago

`This parameter (and presumably any other parameter in mysqli_stmt) will raise an error with the message "Property access is not allowed yet" if the statement was not prepared properly, or not prepared at all.

To prevent this, always ensure that the return value of the "prepare" statement is true before accessing these properties.

`