MySql php:检查Row是否存在

Question

这可能是一件容易的事情,但我是一个业余爱好者,事情本身并不适合我.

我只想查看$ lectureName显示的行是否存在.如果某个行确实存在且其中包含$ lectureName,我希望该函数返回"已分配",否则它应返回"可用".这就是我所拥有的.我很确定它一团糟.请帮忙.

function checkLectureStatus($lectureName)
{
 $con = connectvar();
 mysql_select_db("mydatabase", $con);
 $result = mysql_query("SELECT * FROM preditors_assigned WHERE lecture_name='$lectureName'");
  while($row = mysql_fetch_array($result));
  {
     if (!$row[$lectureName] == $lectureName)
     {
         mysql_close($con);
         return "Available";
     }
      else
     {
        mysql_close($con);
        return "Assigned";
    }
}

当我这样做时,一切都返回可用,即使它应该返回分配.

Answer 1

检查行是否存在的最简单方法:

$lectureName = mysql_real_escape_string($lectureName);  // SECURITY!
$result = mysql_query("SELECT 1 FROM preditors_assigned WHERE lecture_name='$lectureName' LIMIT 1");
if (mysql_fetch_row($result)) {
    return 'Assigned';
} else {
    return 'Available';
}

无需乱七八糟的数组和字段名称.

Answer 2

这应该是诀窍:将结果限制为1行; 如果一行返回,$lectureName则为"已分配",否则为" 可用".

function checkLectureStatus($lectureName)
{
    $con = connectvar();
    mysql_select_db("mydatabase", $con);
    $result = mysql_query(
        "SELECT * FROM preditors_assigned WHERE lecture_name='$lectureName' LIMIT 1");

    if(mysql_fetch_array($result) !== false)
        return 'Assigned';
    return 'Available';
}

Answer 3

使用mysql_num_rows()来检查行是否可用

$result = mysql_query("SELECT * FROM preditors_assigned WHERE lecture_name='$lectureName' LIMIT 1");
$num_rows = mysql_num_rows($result);

if ($num_rows > 0) {
  // do something
}
else {
  // do something else
}