这肯定是个新手的错误,但我没看出来。下面是我的代码中的一段代码:
$mysqli = mysqli_connect($dbCredentials['hostname'],
$dbCredentials['username'], $dbCredentials['password'],
$dbCredentials['database']);
if ($mysqli->connect_error) {
throw new exception( 'Connect Error (' . $mysqli->connect_errno . ') '
. $mysqli->connect_error);
}
$stmt = $mysqli->prepare("SELECT DISTINCT model FROM vehicle_types
WHERE year = ? AND make = '?' ORDER by model");
$stmt->bind_param('is', $year, $make);
$stmt->execute();
当我回显$year和$make的值时,我看到的是值,但当我运行此脚本时,我得到一个空值,并且以下警告出现在我的日志文件中:
PHP Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement
在本例中,year在数据库中的类型为int(10),我尝试传递一个被强制转换为int的副本,而make是一个使用utf8_unicode_ci编码的varchar(20)。我是不是遗漏了什么?
发布于 2013-02-09 04:58:28
您的预准备语句是错误的,应该是:
$stmt = $mysqli->prepare("SELECT DISTINCT model FROM vehicle_types WHERE year = ? AND make = ? ORDER by model");
单引号就是这么回事?作为值,而不是标记。它将已经是一个字符串,因为您使用bind_param('is'
进行了强制转换
https://stackoverflow.com/questions/14781051
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