php MySQLi count(*) 总是返回 1
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MySQLi count(*) always returns 1
提问by Lode
I'm trying to count the number of rows in a table and thought that this was the correct way to do that:
我正在尝试计算表中的行数，并认为这是正确的方法：
$result = $db->query("SELECT COUNT(*) FROM `table`;");
$count = $result->num_rows;
But counts always returns (int)1. If I use the same query in phpMyAdmin I get the right result. It sits in a table so I tried testing $count[0]as well, but that returns NULL.
但计数总是返回(int)1。如果我在 phpMyAdmin 中使用相同的查询，我会得到正确的结果。它位于一张桌子上，所以我也尝试进行测试$count[0]，但返回NULL.
What is the right way to do this?
这样做的正确方法是什么？
回答by VolkerK
You have to fetch that one record, it will contain the result of Count()
您必须获取该记录，它将包含 Count() 的结果
$result = $db->query("SELECT COUNT(*) FROM `table`");
$row = $result->fetch_row();
echo '#: ', $row[0];
回答by ErVeY
Always try to do an associative fetch, that way you can easy get what you want in multiple case result
始终尝试进行关联提取，这样您就可以在多种情况下轻松获得您想要的结果
Here's an example
这是一个例子
$result = $mysqli->query("SELECT COUNT(*) AS cityCount FROM myCity")
$row = $result->fetch_assoc();
echo $row['cityCount']." rows in table myCity.";
$result->close();
回答by Stephane Paquet
$result->num_rows; only returns the number of row(s) affected by a query. When you are performing a count(*) on a table it only returns one row so you can not have an other result than 1.
$result->num_rows; 只返回受查询影响的行数。当您在表上执行 count(*) 时，它只返回一行，因此您不能有除 1 之外的其他结果。
回答by Mike Causer
I find this way more readable:
我发现这种方式更具可读性：
$result = $mysqli->query('select count(*) as `c` from `table`');
$count = $result->fetch_object()->c;
echo "there are {$count} rows in the table";
Not that I have anything against arrays...
并不是说我反对数组...
回答by David Sargent
This worked well for me.
这对我来说效果很好。
// Veh Pro Count
$query_tvp = "SELECT count(*) as total from submit";
if ($result_tvp = $mysqli->query("$query_tvp")) {
/* determine number of rows result set */
$total_tvp = $result_tvp->fetch_row();
$total_tvp = $total_tvp['0'];
/* close result set */
$result_tvp->close();
}
echo "Total: $total_tvp";
回答by alex
It's very simple:
这很简单：
$query = "select count(*)as count from users";
$result = $connection->query($query);
$count = $result->fetch_assoc()["count"];
echo $count;
回答by N1ckfm
For me this works, this is when you want to count a repeated elements inside a register.
对我来说这是有效的，这是当你想计算寄存器内重复元素的时候。
For example: number of people who mark 'yes' in a form.
例如：在表格中标记“是”的人数。
table: users
表：用户
register: vote
登记：投票
element of register: yes / no
寄存器元素：是/否
$sql_user = "SELECT COUNT(*) FROM users WHERE vote = 'yes'";
$result1 = $mysqli->query($sql_user);
$row1 = $result1->fetch_row();
$n_yes = $row1['0'];
