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이 문서에서는 PHP 중국어 웹사이트 매뉴얼 풀어 주다
(PHP 5)
mysqli_stmt::$param_count -- mysqli_stmt_param_count — Returns the number of parameter for the given statement
面向对象风格
过程化风格
$stmt
)Returns the number of parameter markers present in the prepared statement.
stmt
仅以过程化样式:由 mysqli_stmt_init() 返回的 statement 标识。
Returns an integer representing the number of parameters.
Example #1 面向对象风格
<?php
$mysqli = new mysqli ( "localhost" , "my_user" , "my_password" , "world" );
if ( mysqli_connect_errno ()) {
printf ( "Connect failed: %s\n" , mysqli_connect_error ());
exit();
}
if ( $stmt = $mysqli -> prepare ( "SELECT Name FROM Country WHERE Name=? OR Code=?" )) {
$marker = $stmt -> param_count ;
printf ( "Statement has %d markers.\n" , $marker );
$stmt -> close ();
}
$mysqli -> close ();
?>
Example #2 过程化风格
<?php
$link = mysqli_connect ( "localhost" , "my_user" , "my_password" , "world" );
if ( mysqli_connect_errno ()) {
printf ( "Connect failed: %s\n" , mysqli_connect_error ());
exit();
}
if ( $stmt = mysqli_prepare ( $link , "SELECT Name FROM Country WHERE Name=? OR Code=?" )) {
$marker = mysqli_stmt_param_count ( $stmt );
printf ( "Statement has %d markers.\n" , $marker );
mysqli_stmt_close ( $stmt );
}
mysqli_close ( $link );
?>
以上例程会输出:
Statement has 2 markers.
[#1] Senthryl [2009-03-10 11:47:26]
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.