PHP准备了日期格式的语句,如
此命令可用于:PHP准备了日期格式的语句,如,php,mysqli,prepared-statement,date-formatting,sql-date-functions,Php,Mysqli,Prepared Statement,Date Formatting,Sql Date Functions,此命令可用于: SELECT `username`, DATE_FORMAT( date_register , '%d %M %Y' ) AS 'date' FROM `client` WHERE email = 'ali@abu.com' LIMIT 1 但我需要做准备好的陈述: SELECT `username`, DATE_FORMAT( date_register , '%d %M %Y' ) AS 'date' FROM `client` WHERE email = ? LIMI
SELECT `username`, DATE_FORMAT( date_register , '%d %M %Y' ) AS 'date' FROM `client` WHERE email = 'ali@abu.com' LIMIT 1
但我需要做准备好的陈述:
SELECT `username`, DATE_FORMAT( date_register , '%d %M %Y' ) AS 'date' FROM `client` WHERE email = ? LIMIT 1
上面这条线不行。正确的语法是什么?提前感谢。以下是正确的语法: 首先为文字日期格式字符串创建变量:
$date_register = '%d %M %Y';
然后准备:
$stmt = mysqli_prepare($con, 'SELECT `username`, DATE_FORMAT( date_register , ? ) AS 'date' FROM `client` WHERE email = ? LIMIT 1') or die(mysqli_error($con));
像往常一样绑定参数:
mysqli_stmt_bind_param($stmt, "ss", $date_register, $email);
执行并存储结果后;像往常一样访问输出:
mysqli_stmt_bind_result($stmt, $username, $date);
继续读,或者。我已经读过了,但没有找到使用with
DATE\u格式和AS
的正确语法示例。