Question

如果没有找到结果，我将如何显示此消息以代替查询，我更新了代码，但仅显示“ N”

<?php

$hostname = "...";
$username = "";
$password = "";
$db = "";

$dbconnect=mysqli_connect($hostname,$username,$password,$db);

if ($dbconnect->connect_error) {
  die("Database connection failed: " . $dbconnect->connect_error);
}
$query=mysqli_query($dbconnect,"SELECT DISTINCT companyname,client_id,feedback,status from review WHERE status=1 ORDER BY RAND() LIMIT 4");
$rows_get = mysqli_num_rows($query);

if ($rows_get >0)
{
$query2=mysqli_query($dbconnect,"SELECT DISTINCT companyname,client_id,feedback,status from review WHERE status=1 ORDER BY RAND() LIMIT 4");
   $row1 = mysqli_fetch_assoc($query2);
   $row2 = mysqli_fetch_assoc($query2);
   $row3 = mysqli_fetch_assoc($query2);
   $row4 = mysqli_fetch_assoc($query2);
   $row5 = mysqli_fetch_assoc($query2);

}else {
   $row1 = "N0 Data";
   $row2 = "N0 Data";
   $row3 = "N0 Data";
   $row4 = "N0 Data";
   $row5 = "N0 Data";
}
?>

Answer 1

执行以下操作：在$ query之后插入以下内容：

$rows_get = mysqli_num_rows($query);

if ($rows_get >0)
{
//do all database operation
}else {
 echo " No data found";
}

希望这会有所帮助。

例如，修改您的代码。

if ($row_get>0){
//i assume you are getting multiple rows
 while ($data =mysqli_fetch_assoc ($query))
{
//run this loop and you will get all you rows.
}
}

如果mysql不返回结果或为空，我将如何显示消息

1 个答案: