Problem: I have database where is website column which contains 20results. When I try to count the amount of websites it returns me 50.
问题:我的数据库在哪里是网站列,其中包含20个结果。当我尝试计算网站数量时,它会返回50。
Known Issue:I have 50rows on my database and it returns them all, so it also counts empty spaces, how to prevent counting empty spaces ?
已知问题:我的数据库中有50个并且它们全部返回,因此它还会计算空格,如何防止计算空格?
$query = "SELECT COUNT(website) FROM data";
$result = mysqli_query($connection, $query);
if (!$result) {
die("Database query failed.");
}
while($row = mysqli_fetch_row($result)){
var_dump($row);
echo "<hr />";
}
2 个解决方案
#1
1
Just check if "website" column is not blank in database
只需检查数据库中“网站”列是否为空白
Try this
$query ="SELECT COUNT(website) FROM data WHERE (website != '' OR website IS NOT NULL)";
#2
0
Use WHERE
condition for check for IS NOT NULL
and blank values.
使用WHERE条件检查IS NOT NULL和空值。
$query ="SELECT COUNT(website) FROM data WHERE website != '' or website IS NOT NULL";
#1
1
Just check if "website" column is not blank in database
只需检查数据库中“网站”列是否为空白
Try this
$query ="SELECT COUNT(website) FROM data WHERE (website != '' OR website IS NOT NULL)";
#2
0
Use WHERE
condition for check for IS NOT NULL
and blank values.
使用WHERE条件检查IS NOT NULL和空值。
$query ="SELECT COUNT(website) FROM data WHERE website != '' or website IS NOT NULL";