[英]PHP MySQLi numrows outputs 1 even if the value is higher or lower than 1
I have this PHP script, which outputs the variable inside the lr
column: 我有这个PHP脚本,该脚本在
lr
列中输出变量:
$conx = mysqli_connect("value here", "value here", "value here", "value here");
$sql = "SELECT lr FROM locations";
if (mysqli_connect_errno($conx)){
die("Failed to connect to MySQL: " . mysqli_connect_error());
}
$query = mysqli_query($conx, $sql);
$numrows = mysqli_num_rows($query);
if ($numrows > 1) {
echo "No locations found";
} elseif ($numrows == '1') {
echo "1 location found";
} elseif ($numrows > 0) {
echo $numrows." locations found";
}
// Free result set
mysqli_free_result($query);
mysqli_close($conx);
Everything seems to work fine, but when it outputs the result it always outputs 1
even if the value is higher or lower than 1
一切似乎都做工精细,但是当它输出的结果总是输出
1
,即使该值高于或低于1
Anyone with a solution? 有人解决吗?
Try this, see if it suits your goal: (this is a very basic example) 试试看,看它是否符合您的目标:(这是一个非常基本的示例)
$conx = mysqli_connect("value here", "value here", "value here", "value here");
$sql = "SELECT lr FROM locations";
if (mysqli_connect_errno($conx)){
die("Failed to connect to MySQL: " . mysqli_connect_error());
}
$query = mysqli_query($conx, $sql);
while($row=mysqli_fetch_array($query);)
{
echo $row['lr']." ";
}
mysqli_close($conx);
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