Question

我希望脚本在没有找到数据库条目时给出错误。此脚本适用于$ _GET函数。例如，当时。用户输入：index.php？id = 1054355 它找不到任何东西。它会输出一条未找到条目的消息。

我现在的代码：

$userid = mysql_real_escape_string($_GET['id']);

$query = mysql_query("SELECT * FROM nieuws WHERE UserID = '$userid'") or die(mysql_error()); 

//fetch the results / convert results into an array

while($rows = mysql_fetch_array($query)) {

$userid = $rows['UserID'];
$titel = $rows['Titel'];
$body = $rows['Body'];
$door = $rows['Door'];
$bron = $rows['Bron']; 

echo 

 "
 <h1><b>Nieuws - $titel</b></h1>
 <p>&nbsp;</P>
 <h4>$body</h4>
 <p>&nbsp;</p>
 <p>&nbsp;</p>
 <b>Geschreven door: $door</b><p>
 <i>Bronvermelding: $bron</i>
 ";


}

?>

我希望somone可以提出解决方案，

欢呼声。

Answer 1

使用numrows http://tr1.php.net/mysql_num_rows

你必须注意到mysql将很快不再可用。

改用mysqli。

$numRows = mysql_num_rows($query);

if($numRows == 0){
    echo "No record found!";
}

Answer 2

你可以这样做：

$userid = mysql_real_escape_string($_GET['id']);

$query = mysql_query("SELECT * FROM nieuws WHERE UserID = '$userid'") or die(mysql_error()); 

if (mysql_num_rows($result) == 0) {
    echo 'No rows found';
} else {   

    //fetch the results / convert results into an array

    while($rows = mysql_fetch_array($query)) {

      $userid = $rows['UserID'];
      // etc...
    }
}

注意： mysql_*()函数已弃用，很快就会删除。使用mysqli_*()或PDO获取新代码。

Answer 3

$userid = mysql_real_escape_string($_GET['id']);

$query = mysql_query("SELECT * FROM nieuws WHERE UserID = '$userid'") or die(mysql_error()); 

    if(mysql_fetch_array($query ) == false)
        return 'no data found';

//fetch the results / convert results into an array

while($rows = mysql_fetch_array($query)) {

$userid = $rows['UserID'];
$titel = $rows['Titel'];
$body = $rows['Body'];
$door = $rows['Door'];
$bron = $rows['Bron']; 

echo 

 "
 <h1><b>Nieuws - $titel</b></h1>
 <p>&nbsp;</P>
 <h4>$body</h4>
 <p>&nbsp;</p>
 <p>&nbsp;</p>
 <b>Geschreven door: $door</b><p>
 <i>Bronvermelding: $bron</i>
 ";


}

?>

另一种方式：

if(mysql_fetch_row($query ))
 return "no data found!";
else{
// do your stuff
}

在mysqli中

：

if($query->num_rows)
return "no data found!";
    else{
    // do your stuff
    }

Answer 4

你可以这样做：

$result = mysql_query($sql,$con);
$row_count = mysql_num_rows($result);
if($row_count === 0)
{
  echo "nothing found" ;
}
else
{
  // your code for the retrieved result
}

Answer 5

 if(!$query){ //will return true if query was not successful

    echo 'id could not be found';

   }else{

    while($rows = mysql_fetch_array($query)) {

    $userid = $rows['UserID'];
    $titel = $rows['Titel'];
    $body = $rows['Body'];
    $door = $rows['Door'];
    $bron = $rows['Bron']; 

    echo 

     "
     <h1><b>Nieuws - $titel</b></h1>
     <p>&nbsp;</P>
     <h4>$body</h4>
     <p>&nbsp;</p>
     <p>&nbsp;</p>
     <b>Geschreven door: $door</b><p>
     <i>Bronvermelding: $bron</i>
     ";


    }
  }

Answer 6

使用mySQLi（应该使用），

$write = new mysqli($db_hostname,$db_username,$db_password,$db_database); //use your values here
$stmt = $write->prepare("SELECT `UserID`,`Titel`,`Body`,`Door`,`Bron` WHERE `userID` = ?");
$stmt->bind_param('i',$userId); // i = int, s = string.
$stmt->bind_result($userid,$titel,$body,$door,$bron);
$stmt->fetch();
$stmt->execute();
$stmt->close();

这只是一个基本的例子，你可以做很多事情来检查结果，一个这样的方法就是，

if(isset($titel)){//code here to do stuff}

可能值得做的事情是如果没有找到结果，则将用户重定向到通用页面，

if(!isset($titel){header("location:index.php");} //This checks if titel is set, if not then...

正如Awlad Linton所说，你也可以mysqli_num_rows()，这可能更有效率。

没有找到结果时给出错误

6 个答案: