如何使用php检查mysql查询是否未返回结果(未找到记录） - how to check if mysql query return no result(record not found) using php?

how to check if mysql query return no result(record not found) using php?

我通过textarea将图像文件名传递给php脚本，以在mysqldb中查找每个图像的信息。问题是我试图输出那些在mysql db中找不到的图像文件名，并通知用户哪些图像文件名在mysql中没有找到。我当前的代码无法在db中输出那些丢失的记录，但它正确地输出了在db中找到的那些图像的信息。有人能告诉我我做错了什么吗？

foreach ($lines as $line) {

$line = rtrim($line);

$result = mysqli_query($con,"SELECT ID,name,imgUrl,imgPURL FROM testdb WHERE imgUrl like '%$line'");            

 if (!$result) {
             die('Invalid query: ' . mysql_error());
            }
//echo $result;
  if($result == 0) 
    {
       // image not found, do stuff..
      echo "Not Found Image:".$line; 
    }

while($row = mysqli_fetch_array($result))
  {
  $totalRows++;
  echo "<tr>";
  echo "<td>" . $row['ID'] ."(".$totalRows. ")</td>";
  echo "<td>" . $row['name'] . "</td>";
  echo "<td>" . $row['imgPURL'] . "</td>";
  echo "<td>" . $row['imgUrl'] . "</td>";  echo "</tr>";

}
};
echo "</table>";
echo "<br>totalRows:".$totalRows;

您可以在mysqli 中使用mysqli_num_rows()

if(mysqli_num_rows($result) > 0){
    while($row = mysqli_fetch_array($result))
    {
        $totalRows++;
        echo "<tr>";
        echo "<td>" . $row['ID'] ."(".$totalRows. ")</td>";
        echo "<td>" . $row['name'] . "</td>";
        echo "<td>" . $row['imgPURL'] . "</td>";
        echo "<td>" . $row['imgUrl'] . "</td>";  
        echo "</tr>";         
    }
} else {
    echo "<tr><td colspan='4'>Not Found Image:".$line.'</td></tr>';
}

您想要使用mysqli_num_rows

if(mysqli_num_rows($result)) {
   // Do your while loop here
}

使用mysqli_num_rows比较结果集中的行数。