通知,消息和好友请求表出错

时间:2014-03-16 19:55:11

标签: php mysqli

所以我一直试图让我编写的这三个表在我的数据库中创建,但它一直说'#34;消息表没有创建:("。我想知道问题是什么。

$tbl_messages = "CREATE TABLE IF NOT EXISTS usermessages (
                             messageid INT(11) NOT NULL AUTO_INCREMENT,
                             senderid INT(11) NOT NULL,
                             senderusername VARCHAR(16) NOT NULL,
                             receiverid INT(11) NOT NULL,
                             receiverusername VARCHAR(16) NOT NULL,
                             message TEXT NOT NULL,
                             datereceived DATETIME NOT NULL,
                             read ENUM('0', '1') NOT NULL DEFAULT '0',
                             PRIMARY KEY (messageid),
                             UNIQUE KEY(senderusername, receiverusername, senderid, receiverid)
)";

$query = mysqli_query($connect_to_database, $tbl_messages);

if ($query === TRUE) {

    echo "<h2>messages table created successfully :)</h2>";
}else{

    echo "<h2>messages table NOT created :(</h2>";
}

编辑,新代码(2):

$tbl_notifications = "CREATE TABLE IF NOT EXISTS usernotifications (
                  notificationid INT(11) NOT NULL AUTO_INCREMENT,
                  senderid INT(11) NOT NULL,
                  senderusername VARCHAR(16) NOT NULL,
                  receiverid INT(11) NOT NULL,
                  receiverusername VARCHAR(16) NOT NULL,
                  reasonfornotification VARCHAR(255) NOT NULL,
                  read ENUM('0', '1') NOT NULL DEFAULT '0',
                  datereceived DATETIME NOT NULL,
                  PRIMARY KEY (notificationid)
                )";

$query = mysqli_query($connect_to_database, $tbl_notifications);

if ($query === TRUE) {

    echo "<h2>notifications table created successfully :)</h2>";
}else{

    echo "<h2>notifications table NOT created :( </h2>";
}

编辑,新代码:

$tbl_friendrequests = "CREATE TABLE IF NOT EXISTS userfriendrequests (
                      requestid INT(11) NOT NULL AUTO_INCREMENT,
                      senderid INT(11) NOT NULL,
                      senderfirstname VARCHAR(255) NOT NULL,
                      senderlastname VARCHAR(255) NOT NULL,
                      senderusername VARCHAR(16) NOT NULL,
                      receiverid INT(11) NOT NULL,
                      receiverusername VARCHAR(16) NOT NULL,
                      additionalmessage VARCHAR(255) NOT NULL,
                      read ENUM('0', '1') NOT NULL DEFAULT '0',
                      accepted ENUM('0', '1') NOT NULL DEFAULT '0',
                      datereceived DATETIME NOT NULL,
                      PRIMARY KEY (requestid),
                      UNIQUE KEY(senderusername, receiverusername)
                    )";
$query = mysqli_query($connect_to_database, $tbl_friendrequests);

if ($query === TRUE) {

    echo "<h2>friend requests table created successfully :)</h2>";
}else{

    echo "<h2>friend requests table NOT created :( </h2>";
}

0 个答案:

没有答案
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