MySQL PHP:检查行是否存在 [英] MySql php: check if Row exists

问题描述

这可能是一件容易的事，但我是一个业余爱好者，事情对我不起作用.

This is probably an easy thing to do but I'm an amateur and things just aren't working for me.

我只想检查$ lectureName显示的行是否存在.如果某行确实存在$ lectureName，则我希望该函数返回"assigned"(如果没有)，则它应返回"available".这就是我所拥有的.我相当确定这是一团糟.请帮忙.

I just want to check and see if a row exists where the $lectureName shows. If a row does exist with the $lectureName somewhere in it, I want the function to return "assigned" if not then it should return "available". Here's what I have. I'm fairly sure its a mess. Please help.

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";
    }
}

执行此操作后，所有内容都将返回可用状态，即使应返回赋值状态.

When I do this everything return available, even when it should return assigned.

推荐答案

这应该可以解决问题:将结果限制为1行；如果返回一行，则$lectureName为 Assigned ，否则为 Available .

This ought to do the trick: just limit the result to 1 row; if a row comes back the $lectureName is Assigned, otherwise it's Available.

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';
}

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