对此可能有一个简单的答案,但是我不能让我的生活得到这个工作。具有相同名称,不同ID的多个MySQL记录
我使用PHP和MySQL,并有这样的事情设置:
$studentname = mysql_real_escape_string($_POST['sname']);
$studentnumber = mysql_real_escape_string($_POST['snumber']);
$course = mysql_real_escape_string($_POST['courseselect']);
$bike3 = mysql_query("SELECT stoodnumber FROM bikes505 WHERE stoodname='" . $studentname . "'");
$bikestoods = mysql_fetch_array($bike3);
while($row = mysql_fetch_array($stoodlistq))
{
$stoodentname = $row['stoodname'];
$stoodentnumber = $row['stoodnumber'];
//$coursename = $row2['coursename'];
//$coursecode = $row2['coursecode'];
if($studentname == $stoodentname && $studentnumber == $stoodentnumber){
//$success = 1;
//$success = "yupp";
//echo "SUCCESS, WE CAN REGISTER YOOOU!";
echo "<br>";
//echo "INSERT INTO " . $course . "";
switch($course){
case "Biking Safely":
if($studentnumber = $bikestoods[0] or $bikestoods[1]){
echo "Sorry, this student has already registered";
} else if($bikecurrent < $bikemax){
mysql_query("INSERT INTO bikes505 VALUES ('" . $stoodentname . "','" . $stoodentnumber . "')");
echo "Yay, successfully registered " . $stoodentname . " - " . $stoodentnumber . " for " . $course;
echo "<br>";
} else{
echo "Sorry, class is full!";
}
break;
...等等。我遇到的唯一问题是,如果我有两名同名的学生,则列表中的第二位将回显信息不正确。例如,MySQL表有'standname'和'standnumber',如果'Jimmy St.James','1010'和'Jimmy St.James','1090'都是表中的记录,它将会只允许我在课程中注册Jimmy 1010而不注册Jimmy 1090.
我刚刚离开我的验证方式吗?或者我错过了一些非常明显的东西?起初,我认为这只是因为我只使用数组$bikestoods[0]
中的第一项,所以我将其更改为$bikestoods[0] or $bikestoods[1]
,但它仍然无效。
检查它是否在阵列中,在它的任何位置。 PS:按名称检查通常是一个真正的痛苦。真正独特的数据如SSN或其他更好 – Alfabravo 2012-08-15 14:38:30
您正在使用'while($ row = mysql_fetch_array($ standslistq))',但我没有在任何地方看到'$ standslistq'作为MySQL结果。它在哪里定义? – newfurniturey 2012-08-15 14:38:38
对不起,@ newfurniturey它被定义为'$ standinglistq = mysql_query(“SELECT * FROM standsinfo”)'我有几百行代码,所以我不想复制和粘贴所有信息 – WizardOfLoss 2012-08-15 15:07:16