如果没有找到结果,我如何让它显示一条消息来代替查询我更新了代码,但它只显示“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";
}
?>
最佳答案
执行如下操作: 在 $query 之后插入:
$rows_get = mysqli_num_rows($query);
if ($rows_get >0)
{
//do all database operation
}else {
echo " No data found";
}
希望这有帮助。
例如修改您的代码..
if ($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.
}
}
关于php - 如果 mysql 不返回结果或为空,我将如何显示消息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53404665/