尝试从php计算mysql db中的行不会返回结果

Question

I'm trying to get the total number of rows in a table using this code:

$count = mysqli_query($con, "SELECT COUNT(*) AS total FROM random_codes");
$count = mysqli_fetch_object($count);
$count = $count->total;

echo "count is $count<br />";

but the echo is always "count is " with no value following. Runing the same SQL code in phpMyAdmin returns the number of rows (above 2000) as total like requested, so the problem is probably in the php code. Dear responders, I'm new to php & sql so please elaborate and/or link me to documentation. Thanks..

EDIT: attempted error checking, no error shows

$result = mysqli_query($con, "SELECT COUNT(*) AS total FROM random_codes");
echo mysqli_error($con);
$obj = mysqli_fetch_object($result);
$count = $obj->total;
echo "count is $count<br />";

Answer 1

I was able to duplicate your problem by turning off PHP's error reporting, and making a bad database connection:

error_reporting(0);

$con = new mysqli("localhost", "baduser", "badpw", "SomeDB");

$count = mysqli_query($con, "SELECT COUNT(*) AS total FROM SomeTable");
echo mysqli_error($con);
$count = mysqli_fetch_object($count);
$count = $count->total;

echo "count is $count<br />";

Output: count is

Try turning on error reporting and double checking your connection information:

error_reporting(E_ALL);

Perhaps that may give you an error such as:

Warning: mysqli::mysqli(): (28000/1045): Access denied for user 'baduser'@'localhost' (using password: YES)

If you are on a shared host, often times they default error reporting to a value that may hide certain errors for you, and you'll need to override that while in development mode.

Answer 2

I think the problem is in the echo statement.What u need to do is use a . operator between "count is" and the variable $count.

echo "count is". $count."<br />";

If this doesnot work try this.

$query=mysql_query("SELECT COUNT(*) AS total FROM random_codes");
$result=mysql_fetch_array($query);
$count=$result['total'];
echo "count is". $count."<br />";